Algorithm to find added/removed elements in an array - arrays

I am looking for the most efficent way of solving the following
Problem:
given an array Before = { 8, 7, 2, 1} and an array After ={1, 3, 8, 8}
find the added and the removed elements
the solution is:
added = 3, 8
removed = 7, 2
My idea so far is:
for i = 0 .. B.Lenghtt-1
{
for j= 0 .. A.Lenght-1
{
if A[j] == B[i]
A[j] = 0;
B[i] = 0;
break;
}
}
// B elemnts different from 0 are the Removed elements
// A elemnts different from 0 are the Added elemnts
Does anyone know a better solution perhaps more efficent and that doesn't overwrite the original arrays

Sorting is your friend.
Sort the two arrays (a and b), and then walk them (using x and y as counters). Move down both 1 at a time. You can derive all your tests from there:
if a[x] < b[y], then a[x] was removed (and only increment x)
if a[x] > b[y], then b[y] was added (and only increment y)
(I may have missed an edge case, but you get the general idea.)
(edit: the primary edge case that isn't covered here is handling when you reach the end of one of the arrays before the other, but it's not hard to figure out. :)

You could also use a Dictionary<int, int> and a algorithm similar to this:
foreach i in source_list: dictionary[i]++;
foreach i in dest_list: dictionary[i]--;
The final dictionary tells you which elements were inserted/removed (and how often). This solution should be quite fast even for bigger lists - faster than sorting.

if your language as multiset available (set with count of elements) your question is a standard operation.
B = multiset(Before)
A = multiset(After)
result is A.symdiff(B) (symdiff is union minus intersection and that is exactly what you are looking for to have removed and added).
Obviously you can also get removed only or added only using classical difference between sets.
It can trivially be implemented using hashes and it's O(n) (using sort is slightly less efficient as it is O(n.log(n)) because of the sort itself).

In some sort of C++ pseudo code:
Before.sort();
After.sort();
int i = 0;
int j = 0;
for (; i < Before.size() && j < After.size(); ) {
if (Before[i] < After[j]) {
Removed.add(Before[i]);
++i;
continue;
}
if (Before[i] > After[j]) {
Added.add(After[j]);
++j;
continue;
}
++i;
++j;
}
for (; i < Before.size(); ++i) {
Removed.add(Before[i]);
}
for (; j < After.size(); ++j) {
Added.add(After[j]);
}

This can be solved in linear time.
Create a map for calculating the number of repetitions of each element.
Go through the before array and fill the map.
Go through the after array and decrease the value in the map for each element.
At the end, go through the map and if you find a negative value, that element was added - if positive, that element was removed.
Here is some Java code for this (not tested, just written right now):
Map<Integer, Integer> repetitionMap = new HashMap<Integer, Integer>();
for (int i = 0; i < before.length; i++) {
Integer number = repetitionMap.get(before[i]);
if (number == null) {
repetitionMap.put(before[i], 1);
} else {
repetitionMap.put(before[i], number + 1);
}
}
for (int i = 0; i < after.length; i++) {
Integer number = repetitionMap.get(after[i]);
if (number == null) {
repetitionMap.put(after[i], -1);
} else {
repetitionMap.put(after[i], number - 1);
}
}
Set<Integer> keySet = repetitionMap.keySet();
for (Integer i : keySet) {
Integer number = repetitionMap.get(i);
if (number > 0) {
System.out.println("removed " + number + "times value " + i);
}
if (number < 0) {
System.out.println("added " + number + "times value " + i);
}
}

perl:
#a = ( 8, 7, 2, 2, 1 );
#b = ( 1, 3, 8, 8, 8 );
$d{$_}++ for(#a);
$d{$_}-- for(#b);
print"added = ";
for(keys %d){print "$_ " x (-$d{$_}) if($d{$_}<0)}
print"\n";
print"removed = ";
for(keys %d){print "$_ " x ($d{$_}) if($d{$_}>0)}
print"\n";
result:
$ ./inout.pl
added = 8 8 3
removed = 7 2 2

In Groovy:
def before = [8, 7, 2, 1, 1, 1], after = [1, 3, 8, 8, 8]
def added = before.countBy{it}
def result = after.inject(added){map, a -> map[a] ? map << [(a):map[a] - 1]: map << [(a):-1]}
.inject([:]){m, k, v -> v == 0 ? (m << [:]) : (v < 0 ? m << [(k):"added ${v.abs()} times"] : m << [(k):"removed ${v.abs()} times"])}
println "before $before\nafter $after"
println "result: $result"
Result:
before [8, 7, 2, 1, 1, 1]
after [1, 3, 8, 8, 8]
result: [8:added 2 times, 7:removed 1 times, 2:removed 1 times, 1:removed 2 times, 3:added 1 times]
For countBy I got insipred from Some groovy magic post
In groovy inject is like reduce in other functional languages.
I also refer Groovy collection api slides from Trygve Amundsen with really good table with functional methods
Second solution:
def before = [8, 7, 2, 1, 1, 1], after = [1, 3, 8, 8, 8]
def sb = before.countBy{it}
def sa = after.countBy{it}
def result = sa.inject(sb){m, k, v -> m[k] ? m << [(k): m[k] - v] : m << [(k): -v]}
.inject([:]){m, k, v -> v == 0 ? (m << [:]) : (v < 0 ? m << [(k):"added ${v.abs()} times"] : m << [(k):"removed ${v.abs()} times"])}

Related

Finding the sum of nearest smallest and greatest number in an array

The question is straightforward that we want to find the sum of the nearest smallest and greatest number for every index of an array. But the part which I am not able to get is how to return the resulting array with the sum values on the same index values as the numbers.
Example: [1,5,2,3,8]
The sum for each indexes is as follows:
for index 0 the element is 1 hence sum = (0 + 2) = 2
since it has no smallest number therefore taking the nearest smallest number to be 0.
similarly for index 1 sum = (3 + 8 ) = 11 {3 - nearest smallest number to 5 and 8 nearest largest number}
and so on.
What I did was sort the given array then iterate through it and in every iteration take the sum of arr[i-1] + arr[i+1] elements and storing them in the result/answer array.{ with 0th and last element being dealt with separately }
So basically
If the input array is -> [1,5,2,3,8]
the resultant array will be -> [2,4,7,11,5]
but it is required to be as -> [2,11,4,7,5]
that is the sum of each index element to be at the same index as was the initial number
(I am using C++)
You will get the desired output by running this. It is written in Go.
Copy the original input array
Sort the copied array
Create a Map Sorted Value -> Sorted Index
make an output array with same length as input
Iterate through the Input Array and check the value is largest or smallest and Do the Sum as described in the question
store the Sum in the output array in the same index as input array
package main
import (
"fmt"
"sort"
)
func main() {
indexes := []int{1, 5, 2, 3, 8}
output := findSum(indexes)
fmt.Println(output)
}
func findSum(input []int) []int {
if len(input) < 2 {
return input
}
sorted := make([]int, len(input))
copy(sorted, input)
sort.Slice(sorted, func(i, j int) bool {
return sorted[i]< sorted[j]
})
sortedMap := make(map[int]int)
for i, i2 := range sorted {
sortedMap[i2] = i
}
output := make([]int, len(input))
for j, index := range input {
i := sortedMap[index]
if i == 0 {
output[j] = sorted[i+1]
continue
}
if i == len(input) - 1 {
output[j] = sorted[i-1]
continue
}
output[j] = sorted[i - 1] + sorted[i + 1]
}
return output
}
You can run here
The trick is to create an array of indexes into the original array and sort this, rather than sorting the original array.
You don't mention what language you're using, but here's some Java code to illustrate:
Integer[] arr = {1,5,2,3,8};
System.out.println(Arrays.toString(arr));
Integer[] idx = new Integer[arr.length];
for(int i=0; i<idx.length; i++) idx[i] = i;
System.out.println(Arrays.toString(idx));
Arrays.sort(idx, (a, b) -> arr[a] - arr[b]);
System.out.println(Arrays.toString(idx));
Integer[] res = new Integer[arr.length];
for(int i=0; i<arr.length; i++)
{
res[idx[i]] = (i > 0 ? arr[idx[i-1]] : 0) + (i < arr.length - 1 ? arr[idx[i+1]] : 0);
}
System.out.println(Arrays.toString(res));
Or translated into (possibly non-idiomatic) C++ (Ideone):
int len = 5;
array<int, 5> arr{1,5,2,3,8};
for(auto i : arr) cout << i << " ";
cout << endl;
array<int, 5> idx;
for(int i=0; i<len; i++) idx[i] = i;
for(auto i : idx) cout << i << " ";
cout << endl;
sort(idx.begin(), idx.end(), [&arr](int a, int b) {return arr[a] < arr[b];});
for(auto i : idx) cout << i << " ";
cout << endl;
array<int, 5> res;
for(int i=0; i<len; i++)
res[idx[i]] = (i > 0 ? arr[idx[i-1]] : 0) + (i < len - 1 ? arr[idx[i+1]] : 0);
for(auto i : res) cout << i << " ";
cout << endl;
Output:
[1, 5, 2, 3, 8]
[0, 1, 2, 3, 4]
[0, 2, 3, 1, 4]
[2, 11, 4, 7, 5]

Smallest subarray with sum equal to k

I want to find the length smallest subarray whose sum is equal to k.
Input: arr[] = {2, 4, 6, 10, 2, 1}, K = 12
Output: 2
Explanation:
All possible subarrays with sum 12 are {2, 4, 6} and {10, 2}.
Input: arr[] = { 1, 2, 4, 3, 2, 4, 1 }, K = 7
Output: 2
Here's a solution using JavaScript.
It could be made more efficient, for sure, but I've coded it to work.
function lengthOfShortestSubArrayOfSumK(array, k) {
var combos=[];
for(var i=0; i<Math.pow(2, array.length); i++) {
var bin=("0".repeat(array.length)+i.toString(2)).slice(-array.length).split("");
var ones=bin.reduce((count, digit)=>{count+=digit=="1";return count;},0);
var sum=bin.reduce((sum, digit, index)=>{sum+=digit=="1"?array[index]:0;return sum;},0);
combos.push([bin, ones, sum]);
};
return combos.filter(combo=>combo[2]==k).sort((a, b)=>a[1]-b[1])[0][1];
}
var arraysAndKs=[
{array:[2, 4, 6, 10, 2, 1], k:12},
{array:[1, 2, 4, 3, 2, 4, 1], k:7}
];
for(arrayAndK of arraysAndKs)
console.log("Length of shortest sub array of ["+arrayAndK.array.join(", ")+"] with sum "+arrayAndK.k+" is : "+lengthOfShortestSubArrayOfSumK(arrayAndK.array, arrayAndK.k));
The Binary number between 0 and array.length squared will give us a representation of included array items in the sum.
We count how many "ones" are in that Binary number.
We sum array items masked by those "one"s.
We save into combos array an array of the Binary number, "one"s count, and sum.
We filter combos for sum k, sort by count of "one"s, and retrun the first's "one"s count.
I'm sure this can be translated to any programming language.
You can use an algorithm that finds a subset in size K, and save another variable that stores the number of members that make up such a subarray.
The algorithm for finding a K subarray is:
initialize an array of size K, Each place (idx) indicates whether there is a subarray that amounts to idx (I used a dictionary)
Go over any number (i) in the array, and any sum (j) we can reach in the previous iteration now we can reach j + i.
If in the K place it is marked TRUE, then there is a subarray that amounts to K.
Here's the solution in Python
def foo(arr,k):
dynamic = {0:0}
for i in arr:
temp = {}
for j, l in dynamic.items():
if i + j <= k: # if not it's not interesting us
# choose the smallest subarray
temp[i+j] = min(l+1,dynamic.get(i+j,len(arr)))
dynamic.update(temp)
return dynamic.get(k,-1)
the complexity is O(N*K).
I assumed that the subarray refers to any possible combinations of original array.
Here is a Python code that solves the problem under the condition that the subset must be contiguous:
in O(N) complexity
def shortest_contiguous_subarray(arr,k):
if k in arr:
return 1
n = len(arr)
sub_length = float('inf')
sub = arr[(i:=0)]
j = 1
while j < n:
while sub < k and j < n:
sub += arr[j]
j += 1
while sub > k:
sub -= arr[i]
i += 1
if sub == k:
# print(arr[i:j],j-i)
sub_length = min(sub_length,j-i)
sub -= arr[i]
i += 1
return sub_length if sub_length <= n else -1
This answer works for any array of positive numbers, and can be modified to work with arrays that have zero or negative elements if an O(n) pre-processing pass is performed (1. find the minimum element m, m <= 0, 2. make the whole array positive by adding -m+1 to all elements, 3. solve for sum + n*(1-m))
function search(input, goal) {
let queue = [ { avail: input.slice(), used: [], sum: 0 } ]; // initial state
for (let qi = 0; qi < queue.length; qi ++) {
let s = queue[qi]; // like a pop, but without using O(n) shift
for (let i = 0; i < s.avail.length; i++) {
let e = s.avail[i];
if (s.sum + e > goal) continue; // dead end
if (s.sum + e == goal) return [...s.used, e]; // eureka!!
queue.push({ // keep digging
avail: [...s.avail.slice(0, i), ...s.avail.slice(i+1)],
used: [...s.used, e],
sum: s.sum + e
});
}
}
return undefined; // no subset of input adds up to goal
}
console.log(search([2, 4, 6, 10, 2, 1], 12))
This is a classic breadth-first-search that does a little bit of pruning when it detects that we are already over the target sum. It can be further optimized to avoid exploring the same branch several times (for example, [4,2] is equivalent to [2,4]) - but this would require extra memory to keep a set of "visited" states. Additionally, you could add heuristics to explore more promising branches first.
I have done this by using unordered_map in c++. Hope this helps .
`
/* smallest subarray of sum k*/
#include<bits/stdc++.h>
using namespace std;
int main()
{
vector <int> v = {2,4,6,10,2,12};
int k=12;
unordered_map<int,int>m;
int start=0,end=-1;
int len=0,mini=INT_MAX;
int currsum=0;
for(int i=0;i<v.size();i++){
currsum+=v[i];
if(currsum==k){
start=0,end=i;
len=end-start+1;
mini=min(mini,len);
}
if(v[i]==k){
mini=min(mini,1);
}
if(m.find(currsum-k)!=m.end()){
end=i;
start=m[(currsum-k)]+1;
len=end-start+1;
mini=min(mini,len);
}
m[currsum]=i;
}
cout<<mini;
return 0;
}`
class Solution
{
static int findSubArraySum(int arr[], int N, int k)
{
// code here
// i use prefix sum and hashmap approach
HashMap<Integer, Integer> map = new HashMap<>();
map.put(0,1);
// this is bcoz when 1st element is valid one
int count=0;
int sum=0;
for(int i=0;i<N;i++){
sum += arr[i];
// prefix sum
if(map.containsKey(sum-k)){
count += map.get(sum-k);
}
map.put(sum, map.getOrDefault(sum,0)+1);
}
return count;
}
}
// this approach even for -ve numbers
// i came to dis solution by prefix sum approach
This version finds the entire optimal sub-array, not only its length. It's based on a recursion. It will test each number of the array against the optimal sub-array of the rest.
const bestSum = (targetSum, numbers) => {
var shortestCombination = null
for (var i = 0; i < numbers.length; i++) {
var current = numbers[i];
if (current == 0) {
continue
}
if (current == targetSum) {
return [current]
}
if (current > targetSum) {
continue;
}
// "remove" current from array
numbers[i] = 0;
// now the recursion:
var rest = bestSum(targetSum - current, numbers)
if (rest && (!shortestCombination || rest.length + 1 < shortestCombination.length)) {
shortestCombination = [current].concat(rest);
}
// restore current to array
numbers[i] = current
}
return shortestCombination
}
console.log(bestSum(7, [5, 3, 4, 7])) // Should be 7, not [3, 4]
This is my code in Python 3. I used the same idea of find the longest subarray with a sum equal to K. But in the below code for every prefix sum I am storing the recent index.
def smallestSubArraySumLength(a, n, k):
d=defaultdict(lambda:-1)
d[0]=-1
psum=0
maxl=float('inf')
for i in range(n):
psum+=a[I]
if psum-k in d:
maxl=min(maxl, i-d[psum-k])
d[psum]=i
return maxl

Given an array A of size N, find all combination of four elements in the array whose sum is equal to a given value K

Given an array A of size N, find all combinations of four elements in the array whose sum is equal to a given value K. For example, if the given array is {10, 2, 3, 4, 5, 9, 7, 8} and K = 23, one of the quadruple is “3 5 7 8” (3 + 5 + 7 + 8 = 23).
The output should contain only unique quadruple For example, if the input array is {1, 1, 1, 1, 1, 1} and K = 4, then the output should be only one quadruple {1, 1, 1, 1}
My approach: I tried to solve this problem by storing all the distinct pairs formed from the given array into a hash table (std::unordered_multimap), with their sum as key. Then for each pair sum, I looked for (K - sum) key in the hash table. The problem with this approach is I am getting too many duplicated like (i, j, l, m) and (i, l, j, m) are the same, plus there are duplicates due to the same items in the array. I am not sure what is the optimal way to address that.
The code for the above-mentioned approach is:
#include <iostream>
#include <unordered_map>
#include <tuple>
#include <vector>
int main() {
size_t tc = 0;
std::cin >> tc; //number of test cases
while(tc--) {
size_t n = 0, k = 0;
std::cin >> n >> k;
std::vector<size_t> vec(n);
for (size_t i = 0; i < n; ++i)
std::cin >> vec[i];
std::unordered_multimap<size_t, std::tuple<size_t, size_t>> m;
for (size_t i = 0; i < n - 1; ++i)
for (size_t j = i + 1; j < n; ++j) {
const auto sum = vec[i] + vec[j];
m.emplace(sum, std::make_tuple(i, j));
}
for (size_t i = 0; i < n - 1; ++i)
for (size_t j = i + 1; j < n; ++j) {
const auto sum = vec[i] + vec[j];
auto r = m.equal_range(k - sum);
for (auto it = r.first; it != r.second; ++it) {
if ((i == std::get<0>(it->second))
||(i == std::get<1>(it->second))
||(j == std::get<0>(it->second))
|| (j == std::get<1>(it->second)))
continue;
std::cout << vec[i] << ' ' << vec[j] << ' '
<< vec[std::get<0>(it->second)] << ' '
<< vec[std::get<1>(it->second)] << '$';
}
r = m.equal_range(sum);
for (auto it = r.first; it != r.second; ++it) {
if ((i == std::get<0>(it->second))
&& (j == std::get<1>(it->second))) {
m.erase(it);
break;
}
}
}
std::cout << '\n';
}
return 0;
}
The above code will run as-is in the link mentioned below in the Note.
Note: This problem is taken from https://practice.geeksforgeeks.org/problems/find-all-four-sum-numbers/0
To handle duplicate values in array
Consider [2, 2, 2, 3, 3] with goal 10.
The only solution is the 4-tuple <2,2,3,3>. The main point is to avoid choosing two 2 among three 2.
Let's consider the k-class, the set of tuples in which every tuple contain only k.
e.g: in our array we have the 2-class and 3-class.
The 2-class contains:
<2>
<2,2>
<2,2,2>
while the 3-class contains:
<3>
<3,3>
An idea is to reduce the array of elem (elem being an integer value) to an array of k-class.
idem
[[<2>, <2,2>, <2,2,2>], [<3>, <3,3>]]
We can think of taking the cartesian product between the 2-class set and the 3-class set, and check which result lead to solution.
More concretely, let's take some tuple T, whose last (==rightmost) value is k. (in <2,3,4> rightmost value would be 4).
We can pick any l-class from our array (* l > k) and join the tuples from that l-class to T.
e.g
consider array [2, 9, 9, 3, 3, 4, 6] and tuple <2, 3, 3>
The rightmost value is 3.
The candidate k-class are 4-class, 6-class, 9-class
we can join:
<4>
<6>
<9>
<9,9>
so the next candidates will be:
<2, 3, 3, 4>
<2, 3, 3, 6>
<2, 3, 3, 9>
<2, 3, 3, 9, 9> //that one has too many elem, left for illustration
(* The purpose of l > k is to prevent the permutation. (if <1,2> is solution you don't want <2,1> since addition is commutative))
Algorithm
Foreach tuple, try to create new ones by rightjoining tuples from a "greater" k-class.
discard the resulting ones which have too many elements or whose sum is already too big...
At some point we won't have new candidates, so algorithm will stop
example of cuts:
given array [2,3,7,8,10,11], and tuple <2,3> and S == 13
<2,3,7> is candidate (2+3+7 = 12 < 13)
<2,3,10> is not candidate (2+3+10 = 15 > 13)
<2,3,11> even more so. not
<2,3,8> is not candidate either since the next rightjoin (to reach a 4-tuple) will overflow S
given array [2,3,4,4,4] given tuple <2,3> and candidate <4,4,4>
resulting tuple would be <2,3,4,4,4> which has too many elems, discard it!
Obviously the initialization is
some empty tuple
whose sum is 0
and whose rightmost element is less than any k from the array (you can rightjoin it anybody)
I believe it should not be too hard to translate to C++
class TupleClass {
sum = 0
rightIdx = -1
values = [] // an array of integers (hopefully summing to solution)
//idx is the position of the k-class found in array
add (val, idx) {
const t = new TupleClass()
t.values = this.values.concat(val)
t.sum = this.sum + val
t.rightIdx = idx
return t;
}
toString () {
return `<${this.values.join(',')}>`
}
addTuple (tuple, idx) {
const t = new TupleClass
t.values = this.values.concat(tuple.values)
t.sum = this.sum + tuple.sum
t.rightIdx = idx
return t;
}
get size () {
return this.values.length
}
}
function nodupes (v, S) {
v = v.reduce((acc, klass) => {
acc[klass] = (acc[klass] || {duplicity: 0, klass})
acc[klass].duplicity++
return acc
}, {})
v = Object.values(v).sort((a,b) => a.klass - b.klass).map(({ klass, duplicity }, i) => {
return Array(duplicity).fill(0).reduce((acc, _) => {
const t = acc[acc.length-1].add(klass, i)
acc.push(t)
return acc
}, [new TupleClass()]).slice(1)
})
//v is sorted by k-class asc
//each k-class is an array of tuples with increasing length
//[[<2>, <2,2>, <2,2,2>], [<3>,<3,3>]]
let tuples = [new TupleClass()]
const N = v.length
let nextTuples = []
const solutions = []
while (tuples.length) {
tuples.forEach(tuple => {
//foreach kclass after our rightmost value
for (let j = tuple.rightIdx + 1; j <= N - 1; ++j) {
//foreach tuple of that kclass
for (let tclass of v[j]) {
const nextTuple = tuple.addTuple(tclass, j)
if (nextTuple.sum > S || nextTuple.size > 4) {
break
}
//candidate to solution
if (nextTuple.size == 4) {
if (nextTuple.sum === S) {
solutions.push(nextTuple)
}
//invalid sum so adding more elem won't help, do not push
} else {
nextTuples.push(nextTuple)
}
}
}
})
tuples = nextTuples
nextTuples = []
}
return solutions;
}
const v = [1,1,1,1,1,2,2,2,3,3,3,4,0,0]
const S = 7
console.log('v:', v.join(','), 'and S:',S)
console.log(nodupes(v, 7).map(t=>t.toString()).join('\n'))

Combination (mathematical) of structs [duplicate]

I want to write a function that takes an array of letters as an argument and a number of those letters to select.
Say you provide an array of 8 letters and want to select 3 letters from that. Then you should get:
8! / ((8 - 3)! * 3!) = 56
Arrays (or words) in return consisting of 3 letters each.
Art of Computer Programming Volume 4: Fascicle 3 has a ton of these that might fit your particular situation better than how I describe.
Gray Codes
An issue that you will come across is of course memory and pretty quickly, you'll have problems by 20 elements in your set -- 20C3 = 1140. And if you want to iterate over the set it's best to use a modified gray code algorithm so you aren't holding all of them in memory. These generate the next combination from the previous and avoid repetitions. There are many of these for different uses. Do we want to maximize the differences between successive combinations? minimize? et cetera.
Some of the original papers describing gray codes:
Some Hamilton Paths and a Minimal Change Algorithm
Adjacent Interchange Combination Generation Algorithm
Here are some other papers covering the topic:
An Efficient Implementation of the Eades, Hickey, Read Adjacent Interchange Combination Generation Algorithm (PDF, with code in Pascal)
Combination Generators
Survey of Combinatorial Gray Codes (PostScript)
An Algorithm for Gray Codes
Chase's Twiddle (algorithm)
Phillip J Chase, `Algorithm 382: Combinations of M out of N Objects' (1970)
The algorithm in C...
Index of Combinations in Lexicographical Order (Buckles Algorithm 515)
You can also reference a combination by its index (in lexicographical order). Realizing that the index should be some amount of change from right to left based on the index we can construct something that should recover a combination.
So, we have a set {1,2,3,4,5,6}... and we want three elements. Let's say {1,2,3} we can say that the difference between the elements is one and in order and minimal. {1,2,4} has one change and is lexicographically number 2. So the number of 'changes' in the last place accounts for one change in the lexicographical ordering. The second place, with one change {1,3,4} has one change but accounts for more change since it's in the second place (proportional to the number of elements in the original set).
The method I've described is a deconstruction, as it seems, from set to the index, we need to do the reverse – which is much trickier. This is how Buckles solves the problem. I wrote some C to compute them, with minor changes – I used the index of the sets rather than a number range to represent the set, so we are always working from 0...n.
Note:
Since combinations are unordered, {1,3,2} = {1,2,3} --we order them to be lexicographical.
This method has an implicit 0 to start the set for the first difference.
Index of Combinations in Lexicographical Order (McCaffrey)
There is another way:, its concept is easier to grasp and program but it's without the optimizations of Buckles. Fortunately, it also does not produce duplicate combinations:
The set that maximizes , where .
For an example: 27 = C(6,4) + C(5,3) + C(2,2) + C(1,1). So, the 27th lexicographical combination of four things is: {1,2,5,6}, those are the indexes of whatever set you want to look at. Example below (OCaml), requires choose function, left to reader:
(* this will find the [x] combination of a [set] list when taking [k] elements *)
let combination_maccaffery set k x =
(* maximize function -- maximize a that is aCb *)
(* return largest c where c < i and choose(c,i) <= z *)
let rec maximize a b x =
if (choose a b ) <= x then a else maximize (a-1) b x
in
let rec iterate n x i = match i with
| 0 -> []
| i ->
let max = maximize n i x in
max :: iterate n (x - (choose max i)) (i-1)
in
if x < 0 then failwith "errors" else
let idxs = iterate (List.length set) x k in
List.map (List.nth set) (List.sort (-) idxs)
A small and simple combinations iterator
The following two algorithms are provided for didactic purposes. They implement an iterator and (a more general) folder overall combinations.
They are as fast as possible, having the complexity O(nCk). The memory consumption is bound by k.
We will start with the iterator, which will call a user provided function for each combination
let iter_combs n k f =
let rec iter v s j =
if j = k then f v
else for i = s to n - 1 do iter (i::v) (i+1) (j+1) done in
iter [] 0 0
A more general version will call the user provided function along with the state variable, starting from the initial state. Since we need to pass the state between different states we won't use the for-loop, but instead, use recursion,
let fold_combs n k f x =
let rec loop i s c x =
if i < n then
loop (i+1) s c ##
let c = i::c and s = s + 1 and i = i + 1 in
if s < k then loop i s c x else f c x
else x in
loop 0 0 [] x
In C#:
public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int k)
{
return k == 0 ? new[] { new T[0] } :
elements.SelectMany((e, i) =>
elements.Skip(i + 1).Combinations(k - 1).Select(c => (new[] {e}).Concat(c)));
}
Usage:
var result = Combinations(new[] { 1, 2, 3, 4, 5 }, 3);
Result:
123
124
125
134
135
145
234
235
245
345
Short java solution:
import java.util.Arrays;
public class Combination {
public static void main(String[] args){
String[] arr = {"A","B","C","D","E","F"};
combinations2(arr, 3, 0, new String[3]);
}
static void combinations2(String[] arr, int len, int startPosition, String[] result){
if (len == 0){
System.out.println(Arrays.toString(result));
return;
}
for (int i = startPosition; i <= arr.length-len; i++){
result[result.length - len] = arr[i];
combinations2(arr, len-1, i+1, result);
}
}
}
Result will be
[A, B, C]
[A, B, D]
[A, B, E]
[A, B, F]
[A, C, D]
[A, C, E]
[A, C, F]
[A, D, E]
[A, D, F]
[A, E, F]
[B, C, D]
[B, C, E]
[B, C, F]
[B, D, E]
[B, D, F]
[B, E, F]
[C, D, E]
[C, D, F]
[C, E, F]
[D, E, F]
May I present my recursive Python solution to this problem?
def choose_iter(elements, length):
for i in xrange(len(elements)):
if length == 1:
yield (elements[i],)
else:
for next in choose_iter(elements[i+1:], length-1):
yield (elements[i],) + next
def choose(l, k):
return list(choose_iter(l, k))
Example usage:
>>> len(list(choose_iter("abcdefgh",3)))
56
I like it for its simplicity.
Lets say your array of letters looks like this: "ABCDEFGH". You have three indices (i, j, k) indicating which letters you are going to use for the current word, You start with:
A B C D E F G H
^ ^ ^
i j k
First you vary k, so the next step looks like that:
A B C D E F G H
^ ^ ^
i j k
If you reached the end you go on and vary j and then k again.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
Once you j reached G you start also to vary i.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
...
Written in code this look something like that
void print_combinations(const char *string)
{
int i, j, k;
int len = strlen(string);
for (i = 0; i < len - 2; i++)
{
for (j = i + 1; j < len - 1; j++)
{
for (k = j + 1; k < len; k++)
printf("%c%c%c\n", string[i], string[j], string[k]);
}
}
}
The following recursive algorithm picks all of the k-element combinations from an ordered set:
choose the first element i of your combination
combine i with each of the combinations of k-1 elements chosen recursively from the set of elements larger than i.
Iterate the above for each i in the set.
It is essential that you pick the rest of the elements as larger than i, to avoid repetition. This way [3,5] will be picked only once, as [3] combined with [5], instead of twice (the condition eliminates [5] + [3]). Without this condition you get variations instead of combinations.
Short example in Python:
def comb(sofar, rest, n):
if n == 0:
print sofar
else:
for i in range(len(rest)):
comb(sofar + rest[i], rest[i+1:], n-1)
>>> comb("", "abcde", 3)
abc
abd
abe
acd
ace
ade
bcd
bce
bde
cde
For explanation, the recursive method is described with the following example:
Example: A B C D E
All combinations of 3 would be:
A with all combinations of 2 from the rest (B C D E)
B with all combinations of 2 from the rest (C D E)
C with all combinations of 2 from the rest (D E)
I found this thread useful and thought I would add a Javascript solution that you can pop into Firebug. Depending on your JS engine, it could take a little time if the starting string is large.
function string_recurse(active, rest) {
if (rest.length == 0) {
console.log(active);
} else {
string_recurse(active + rest.charAt(0), rest.substring(1, rest.length));
string_recurse(active, rest.substring(1, rest.length));
}
}
string_recurse("", "abc");
The output should be as follows:
abc
ab
ac
a
bc
b
c
In C++ the following routine will produce all combinations of length distance(first,k) between the range [first,last):
#include <algorithm>
template <typename Iterator>
bool next_combination(const Iterator first, Iterator k, const Iterator last)
{
/* Credits: Mark Nelson http://marknelson.us */
if ((first == last) || (first == k) || (last == k))
return false;
Iterator i1 = first;
Iterator i2 = last;
++i1;
if (last == i1)
return false;
i1 = last;
--i1;
i1 = k;
--i2;
while (first != i1)
{
if (*--i1 < *i2)
{
Iterator j = k;
while (!(*i1 < *j)) ++j;
std::iter_swap(i1,j);
++i1;
++j;
i2 = k;
std::rotate(i1,j,last);
while (last != j)
{
++j;
++i2;
}
std::rotate(k,i2,last);
return true;
}
}
std::rotate(first,k,last);
return false;
}
It can be used like this:
#include <string>
#include <iostream>
int main()
{
std::string s = "12345";
std::size_t comb_size = 3;
do
{
std::cout << std::string(s.begin(), s.begin() + comb_size) << std::endl;
} while (next_combination(s.begin(), s.begin() + comb_size, s.end()));
return 0;
}
This will print the following:
123
124
125
134
135
145
234
235
245
345
static IEnumerable<string> Combinations(List<string> characters, int length)
{
for (int i = 0; i < characters.Count; i++)
{
// only want 1 character, just return this one
if (length == 1)
yield return characters[i];
// want more than one character, return this one plus all combinations one shorter
// only use characters after the current one for the rest of the combinations
else
foreach (string next in Combinations(characters.GetRange(i + 1, characters.Count - (i + 1)), length - 1))
yield return characters[i] + next;
}
}
Simple recursive algorithm in Haskell
import Data.List
combinations 0 lst = [[]]
combinations n lst = do
(x:xs) <- tails lst
rest <- combinations (n-1) xs
return $ x : rest
We first define the special case, i.e. selecting zero elements. It produces a single result, which is an empty list (i.e. a list that contains an empty list).
For n > 0, x goes through every element of the list and xs is every element after x.
rest picks n - 1 elements from xs using a recursive call to combinations. The final result of the function is a list where each element is x : rest (i.e. a list which has x as head and rest as tail) for every different value of x and rest.
> combinations 3 "abcde"
["abc","abd","abe","acd","ace","ade","bcd","bce","bde","cde"]
And of course, since Haskell is lazy, the list is gradually generated as needed, so you can partially evaluate exponentially large combinations.
> let c = combinations 8 "abcdefghijklmnopqrstuvwxyz"
> take 10 c
["abcdefgh","abcdefgi","abcdefgj","abcdefgk","abcdefgl","abcdefgm","abcdefgn",
"abcdefgo","abcdefgp","abcdefgq"]
And here comes granddaddy COBOL, the much maligned language.
Let's assume an array of 34 elements of 8 bytes each (purely arbitrary selection.) The idea is to enumerate all possible 4-element combinations and load them into an array.
We use 4 indices, one each for each position in the group of 4
The array is processed like this:
idx1 = 1
idx2 = 2
idx3 = 3
idx4 = 4
We vary idx4 from 4 to the end. For each idx4 we get a unique combination
of groups of four. When idx4 comes to the end of the array, we increment idx3 by 1 and set idx4 to idx3+1. Then we run idx4 to the end again. We proceed in this manner, augmenting idx3,idx2, and idx1 respectively until the position of idx1 is less than 4 from the end of the array. That finishes the algorithm.
1 --- pos.1
2 --- pos 2
3 --- pos 3
4 --- pos 4
5
6
7
etc.
First iterations:
1234
1235
1236
1237
1245
1246
1247
1256
1257
1267
etc.
A COBOL example:
01 DATA_ARAY.
05 FILLER PIC X(8) VALUE "VALUE_01".
05 FILLER PIC X(8) VALUE "VALUE_02".
etc.
01 ARAY_DATA OCCURS 34.
05 ARAY_ITEM PIC X(8).
01 OUTPUT_ARAY OCCURS 50000 PIC X(32).
01 MAX_NUM PIC 99 COMP VALUE 34.
01 INDEXXES COMP.
05 IDX1 PIC 99.
05 IDX2 PIC 99.
05 IDX3 PIC 99.
05 IDX4 PIC 99.
05 OUT_IDX PIC 9(9).
01 WHERE_TO_STOP_SEARCH PIC 99 COMP.
* Stop the search when IDX1 is on the third last array element:
COMPUTE WHERE_TO_STOP_SEARCH = MAX_VALUE - 3
MOVE 1 TO IDX1
PERFORM UNTIL IDX1 > WHERE_TO_STOP_SEARCH
COMPUTE IDX2 = IDX1 + 1
PERFORM UNTIL IDX2 > MAX_NUM
COMPUTE IDX3 = IDX2 + 1
PERFORM UNTIL IDX3 > MAX_NUM
COMPUTE IDX4 = IDX3 + 1
PERFORM UNTIL IDX4 > MAX_NUM
ADD 1 TO OUT_IDX
STRING ARAY_ITEM(IDX1)
ARAY_ITEM(IDX2)
ARAY_ITEM(IDX3)
ARAY_ITEM(IDX4)
INTO OUTPUT_ARAY(OUT_IDX)
ADD 1 TO IDX4
END-PERFORM
ADD 1 TO IDX3
END-PERFORM
ADD 1 TO IDX2
END_PERFORM
ADD 1 TO IDX1
END-PERFORM.
Another C# version with lazy generation of the combination indices. This version maintains a single array of indices to define a mapping between the list of all values and the values for the current combination, i.e. constantly uses O(k) additional space during the entire runtime. The code generates individual combinations, including the first one, in O(k) time.
public static IEnumerable<T[]> Combinations<T>(this T[] values, int k)
{
if (k < 0 || values.Length < k)
yield break; // invalid parameters, no combinations possible
// generate the initial combination indices
var combIndices = new int[k];
for (var i = 0; i < k; i++)
{
combIndices[i] = i;
}
while (true)
{
// return next combination
var combination = new T[k];
for (var i = 0; i < k; i++)
{
combination[i] = values[combIndices[i]];
}
yield return combination;
// find first index to update
var indexToUpdate = k - 1;
while (indexToUpdate >= 0 && combIndices[indexToUpdate] >= values.Length - k + indexToUpdate)
{
indexToUpdate--;
}
if (indexToUpdate < 0)
yield break; // done
// update combination indices
for (var combIndex = combIndices[indexToUpdate] + 1; indexToUpdate < k; indexToUpdate++, combIndex++)
{
combIndices[indexToUpdate] = combIndex;
}
}
}
Test code:
foreach (var combination in new[] {'a', 'b', 'c', 'd', 'e'}.Combinations(3))
{
System.Console.WriteLine(String.Join(" ", combination));
}
Output:
a b c
a b d
a b e
a c d
a c e
a d e
b c d
b c e
b d e
c d e
Here is an elegant, generic implementation in Scala, as described on 99 Scala Problems.
object P26 {
def flatMapSublists[A,B](ls: List[A])(f: (List[A]) => List[B]): List[B] =
ls match {
case Nil => Nil
case sublist#(_ :: tail) => f(sublist) ::: flatMapSublists(tail)(f)
}
def combinations[A](n: Int, ls: List[A]): List[List[A]] =
if (n == 0) List(Nil)
else flatMapSublists(ls) { sl =>
combinations(n - 1, sl.tail) map {sl.head :: _}
}
}
If you can use SQL syntax - say, if you're using LINQ to access fields of an structure or array, or directly accessing a database that has a table called "Alphabet" with just one char field "Letter", you can adapt following code:
SELECT A.Letter, B.Letter, C.Letter
FROM Alphabet AS A, Alphabet AS B, Alphabet AS C
WHERE A.Letter<>B.Letter AND A.Letter<>C.Letter AND B.Letter<>C.Letter
AND A.Letter<B.Letter AND B.Letter<C.Letter
This will return all combinations of 3 letters, notwithstanding how many letters you have in table "Alphabet" (it can be 3, 8, 10, 27, etc.).
If what you want is all permutations, rather than combinations (i.e. you want "ACB" and "ABC" to count as different, rather than appear just once) just delete the last line (the AND one) and it's done.
Post-Edit: After re-reading the question, I realise what's needed is the general algorithm, not just a specific one for the case of selecting 3 items. Adam Hughes' answer is the complete one, unfortunately I cannot vote it up (yet). This answer's simple but works only for when you want exactly 3 items.
I had a permutation algorithm I used for project euler, in python:
def missing(miss,src):
"Returns the list of items in src not present in miss"
return [i for i in src if i not in miss]
def permutation_gen(n,l):
"Generates all the permutations of n items of the l list"
for i in l:
if n<=1: yield [i]
r = [i]
for j in permutation_gen(n-1,missing([i],l)): yield r+j
If
n<len(l)
you should have all combination you need without repetition, do you need it?
It is a generator, so you use it in something like this:
for comb in permutation_gen(3,list("ABCDEFGH")):
print comb
https://gist.github.com/3118596
There is an implementation for JavaScript. It has functions to get k-combinations and all combinations of an array of any objects. Examples:
k_combinations([1,2,3], 2)
-> [[1,2], [1,3], [2,3]]
combinations([1,2,3])
-> [[1],[2],[3],[1,2],[1,3],[2,3],[1,2,3]]
Lets say your array of letters looks like this: "ABCDEFGH". You have three indices (i, j, k) indicating which letters you are going to use for the current word, You start with:
A B C D E F G H
^ ^ ^
i j k
First you vary k, so the next step looks like that:
A B C D E F G H
^ ^ ^
i j k
If you reached the end you go on and vary j and then k again.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
Once you j reached G you start also to vary i.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
...
function initializePointers($cnt) {
$pointers = [];
for($i=0; $i<$cnt; $i++) {
$pointers[] = $i;
}
return $pointers;
}
function incrementPointers(&$pointers, &$arrLength) {
for($i=0; $i<count($pointers); $i++) {
$currentPointerIndex = count($pointers) - $i - 1;
$currentPointer = $pointers[$currentPointerIndex];
if($currentPointer < $arrLength - $i - 1) {
++$pointers[$currentPointerIndex];
for($j=1; ($currentPointerIndex+$j)<count($pointers); $j++) {
$pointers[$currentPointerIndex+$j] = $pointers[$currentPointerIndex]+$j;
}
return true;
}
}
return false;
}
function getDataByPointers(&$arr, &$pointers) {
$data = [];
for($i=0; $i<count($pointers); $i++) {
$data[] = $arr[$pointers[$i]];
}
return $data;
}
function getCombinations($arr, $cnt)
{
$len = count($arr);
$result = [];
$pointers = initializePointers($cnt);
do {
$result[] = getDataByPointers($arr, $pointers);
} while(incrementPointers($pointers, count($arr)));
return $result;
}
$result = getCombinations([0, 1, 2, 3, 4, 5], 3);
print_r($result);
Based on https://stackoverflow.com/a/127898/2628125, but more abstract, for any size of pointers.
Here you have a lazy evaluated version of that algorithm coded in C#:
static bool nextCombination(int[] num, int n, int k)
{
bool finished, changed;
changed = finished = false;
if (k > 0)
{
for (int i = k - 1; !finished && !changed; i--)
{
if (num[i] < (n - 1) - (k - 1) + i)
{
num[i]++;
if (i < k - 1)
{
for (int j = i + 1; j < k; j++)
{
num[j] = num[j - 1] + 1;
}
}
changed = true;
}
finished = (i == 0);
}
}
return changed;
}
static IEnumerable Combinations<T>(IEnumerable<T> elements, int k)
{
T[] elem = elements.ToArray();
int size = elem.Length;
if (k <= size)
{
int[] numbers = new int[k];
for (int i = 0; i < k; i++)
{
numbers[i] = i;
}
do
{
yield return numbers.Select(n => elem[n]);
}
while (nextCombination(numbers, size, k));
}
}
And test part:
static void Main(string[] args)
{
int k = 3;
var t = new[] { "dog", "cat", "mouse", "zebra"};
foreach (IEnumerable<string> i in Combinations(t, k))
{
Console.WriteLine(string.Join(",", i));
}
}
Hope this help you!
Another version, that forces all the first k to appear firstly, then all the first k+1 combinations, then all the first k+2 etc.. It means that if you have sorted array, the most important on the top, it would take them and expand gradually to the next ones - only when it is must do so.
private static bool NextCombinationFirstsAlwaysFirst(int[] num, int n, int k)
{
if (k > 1 && NextCombinationFirstsAlwaysFirst(num, num[k - 1], k - 1))
return true;
if (num[k - 1] + 1 == n)
return false;
++num[k - 1];
for (int i = 0; i < k - 1; ++i)
num[i] = i;
return true;
}
For instance, if you run the first method ("nextCombination") on k=3, n=5 you'll get:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
But if you'll run
int[] nums = new int[k];
for (int i = 0; i < k; ++i)
nums[i] = i;
do
{
Console.WriteLine(string.Join(" ", nums));
}
while (NextCombinationFirstsAlwaysFirst(nums, n, k));
You'll get this (I added empty lines for clarity):
0 1 2
0 1 3
0 2 3
1 2 3
0 1 4
0 2 4
1 2 4
0 3 4
1 3 4
2 3 4
It's adding "4" only when must to, and also after "4" was added it adds "3" again only when it must to (after doing 01, 02, 12).
Array.prototype.combs = function(num) {
var str = this,
length = str.length,
of = Math.pow(2, length) - 1,
out, combinations = [];
while(of) {
out = [];
for(var i = 0, y; i < length; i++) {
y = (1 << i);
if(y & of && (y !== of))
out.push(str[i]);
}
if (out.length >= num) {
combinations.push(out);
}
of--;
}
return combinations;
}
Clojure version:
(defn comb [k l]
(if (= 1 k) (map vector l)
(apply concat
(map-indexed
#(map (fn [x] (conj x %2))
(comb (dec k) (drop (inc %1) l)))
l))))
Algorithm:
Count from 1 to 2^n.
Convert each digit to its binary representation.
Translate each 'on' bit to elements of your set, based on position.
In C#:
void Main()
{
var set = new [] {"A", "B", "C", "D" }; //, "E", "F", "G", "H", "I", "J" };
var kElement = 2;
for(var i = 1; i < Math.Pow(2, set.Length); i++) {
var result = Convert.ToString(i, 2).PadLeft(set.Length, '0');
var cnt = Regex.Matches(Regex.Escape(result), "1").Count;
if (cnt == kElement) {
for(int j = 0; j < set.Length; j++)
if ( Char.GetNumericValue(result[j]) == 1)
Console.Write(set[j]);
Console.WriteLine();
}
}
}
Why does it work?
There is a bijection between the subsets of an n-element set and n-bit sequences.
That means we can figure out how many subsets there are by counting sequences.
e.g., the four element set below can be represented by {0,1} X {0, 1} X {0, 1} X {0, 1} (or 2^4) different sequences.
So - all we have to do is count from 1 to 2^n to find all the combinations. (We ignore the empty set.) Next, translate the digits to their binary representation. Then substitute elements of your set for 'on' bits.
If you want only k element results, only print when k bits are 'on'.
(If you want all subsets instead of k length subsets, remove the cnt/kElement part.)
(For proof, see MIT free courseware Mathematics for Computer Science, Lehman et al, section 11.2.2. https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/ )
short python code, yielding index positions
def yield_combos(n,k):
# n is set size, k is combo size
i = 0
a = [0]*k
while i > -1:
for j in range(i+1, k):
a[j] = a[j-1]+1
i=j
yield a
while a[i] == i + n - k:
i -= 1
a[i] += 1
All said and and done here comes the O'caml code for that.
Algorithm is evident from the code..
let combi n lst =
let rec comb l c =
if( List.length c = n) then [c] else
match l with
[] -> []
| (h::t) -> (combi t (h::c))#(combi t c)
in
combi lst []
;;
Here is a method which gives you all combinations of specified size from a random length string. Similar to quinmars' solution, but works for varied input and k.
The code can be changed to wrap around, ie 'dab' from input 'abcd' w k=3.
public void run(String data, int howMany){
choose(data, howMany, new StringBuffer(), 0);
}
//n choose k
private void choose(String data, int k, StringBuffer result, int startIndex){
if (result.length()==k){
System.out.println(result.toString());
return;
}
for (int i=startIndex; i<data.length(); i++){
result.append(data.charAt(i));
choose(data,k,result, i+1);
result.setLength(result.length()-1);
}
}
Output for "abcde":
abc abd abe acd ace ade bcd bce bde cde
Short javascript version (ES 5)
let combine = (list, n) =>
n == 0 ?
[[]] :
list.flatMap((e, i) =>
combine(
list.slice(i + 1),
n - 1
).map(c => [e].concat(c))
);
let res = combine([1,2,3,4], 3);
res.forEach(e => console.log(e.join()));
Another python recusive solution.
def combination_indicies(n, k, j = 0, stack = []):
if len(stack) == k:
yield list(stack)
return
for i in range(j, n):
stack.append(i)
for x in combination_indicies(n, k, i + 1, stack):
yield x
stack.pop()
list(combination_indicies(5, 3))
Output:
[[0, 1, 2],
[0, 1, 3],
[0, 1, 4],
[0, 2, 3],
[0, 2, 4],
[0, 3, 4],
[1, 2, 3],
[1, 2, 4],
[1, 3, 4],
[2, 3, 4]]
I created a solution in SQL Server 2005 for this, and posted it on my website: http://www.jessemclain.com/downloads/code/sql/fn_GetMChooseNCombos.sql.htm
Here is an example to show usage:
SELECT * FROM dbo.fn_GetMChooseNCombos('ABCD', 2, '')
results:
Word
----
AB
AC
AD
BC
BD
CD
(6 row(s) affected)
Here is my proposition in C++
I tried to impose as little restriction on the iterator type as i could so this solution assumes just forward iterator, and it can be a const_iterator. This should work with any standard container. In cases where arguments don't make sense it throws std::invalid_argumnent
#include <vector>
#include <stdexcept>
template <typename Fci> // Fci - forward const iterator
std::vector<std::vector<Fci> >
enumerate_combinations(Fci begin, Fci end, unsigned int combination_size)
{
if(begin == end && combination_size > 0u)
throw std::invalid_argument("empty set and positive combination size!");
std::vector<std::vector<Fci> > result; // empty set of combinations
if(combination_size == 0u) return result; // there is exactly one combination of
// size 0 - emty set
std::vector<Fci> current_combination;
current_combination.reserve(combination_size + 1u); // I reserve one aditional slot
// in my vector to store
// the end sentinel there.
// The code is cleaner thanks to that
for(unsigned int i = 0u; i < combination_size && begin != end; ++i, ++begin)
{
current_combination.push_back(begin); // Construction of the first combination
}
// Since I assume the itarators support only incrementing, I have to iterate over
// the set to get its size, which is expensive. Here I had to itrate anyway to
// produce the first cobination, so I use the loop to also check the size.
if(current_combination.size() < combination_size)
throw std::invalid_argument("combination size > set size!");
result.push_back(current_combination); // Store the first combination in the results set
current_combination.push_back(end); // Here I add mentioned earlier sentinel to
// simplyfy rest of the code. If I did it
// earlier, previous statement would get ugly.
while(true)
{
unsigned int i = combination_size;
Fci tmp; // Thanks to the sentinel I can find first
do // iterator to change, simply by scaning
{ // from right to left and looking for the
tmp = current_combination[--i]; // first "bubble". The fact, that it's
++tmp; // a forward iterator makes it ugly but I
} // can't help it.
while(i > 0u && tmp == current_combination[i + 1u]);
// Here is probably my most obfuscated expression.
// Loop above looks for a "bubble". If there is no "bubble", that means, that
// current_combination is the last combination, Expression in the if statement
// below evaluates to true and the function exits returning result.
// If the "bubble" is found however, the ststement below has a sideeffect of
// incrementing the first iterator to the left of the "bubble".
if(++current_combination[i] == current_combination[i + 1u])
return result;
// Rest of the code sets posiotons of the rest of the iterstors
// (if there are any), that are to the right of the incremented one,
// to form next combination
while(++i < combination_size)
{
current_combination[i] = current_combination[i - 1u];
++current_combination[i];
}
// Below is the ugly side of using the sentinel. Well it had to haave some
// disadvantage. Try without it.
result.push_back(std::vector<Fci>(current_combination.begin(),
current_combination.end() - 1));
}
}
Here is a code I recently wrote in Java, which calculates and returns all the combination of "num" elements from "outOf" elements.
// author: Sourabh Bhat (heySourabh#gmail.com)
public class Testing
{
public static void main(String[] args)
{
// Test case num = 5, outOf = 8.
int num = 5;
int outOf = 8;
int[][] combinations = getCombinations(num, outOf);
for (int i = 0; i < combinations.length; i++)
{
for (int j = 0; j < combinations[i].length; j++)
{
System.out.print(combinations[i][j] + " ");
}
System.out.println();
}
}
private static int[][] getCombinations(int num, int outOf)
{
int possibilities = get_nCr(outOf, num);
int[][] combinations = new int[possibilities][num];
int arrayPointer = 0;
int[] counter = new int[num];
for (int i = 0; i < num; i++)
{
counter[i] = i;
}
breakLoop: while (true)
{
// Initializing part
for (int i = 1; i < num; i++)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i] = counter[i - 1] + 1;
}
// Testing part
for (int i = 0; i < num; i++)
{
if (counter[i] < outOf)
{
continue;
} else
{
break breakLoop;
}
}
// Innermost part
combinations[arrayPointer] = counter.clone();
arrayPointer++;
// Incrementing part
counter[num - 1]++;
for (int i = num - 1; i >= 1; i--)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i - 1]++;
}
}
return combinations;
}
private static int get_nCr(int n, int r)
{
if(r > n)
{
throw new ArithmeticException("r is greater then n");
}
long numerator = 1;
long denominator = 1;
for (int i = n; i >= r + 1; i--)
{
numerator *= i;
}
for (int i = 2; i <= n - r; i++)
{
denominator *= i;
}
return (int) (numerator / denominator);
}
}

Pseudocode - Arrays - Maxrun

I'm trying to solve a problem but I have difficulties with algorithms.
I have to write pseudocode for an iterative algorithm maxRun(A) that takes an array A of integer as input and return the maximal length of a run in A.
The subarray A[k...l] is a run if A[j] <= A[j + 1] for all j where k <= j < l. So it is a non decreasing segment of A.
Ex. A = [1,5,2,3,4,1], the max length would be 3 [2,3,4].
Thanks.
Simple Java implementation:
public class FindRun {
public static int maxRun(int[] a) {
int max = 0;
int index = 0;
int previous = a[0] + 1;
int run = 0;
while (index < a.length) {
if (a[index] >= previous) {
run++;
} else {
max = Math.max(max, run);
run = 1;
}
previous = a[index];
index++;
}
return Math.max(max, run);
}
public static void main(String[] args) {
System.out.println(maxRun(new int[] { 1, 5, 2, 3, 4, 1 }));
}
}
Here's a solution similar to #Michael's, in Ruby:
a = [1,5,2,3,4,1]
r = [1]
(1...a.size).each { |i| r << ((a[i] == a[i-1] + 1) ? r[i-1] + 1 : 1) }
r.max #=> 3
r.index(r.max) #=> 4
indicating the the maximum run is of length 3 and ends at a offset 4; that is, the run 2,3,4. I will now explain the algorithm I used. For those who don't know Ruby, this will also give you a taste of the language:
r = [1] creates an array with one element, whose value is 1. This is read, "The longest run ending at a offset 0 is of length 1.
(1...a.size) is the sequence 1, 2, 3, 4, 5. Three dots between 1 and a.size means the sequence ends with a.size - 1, which is 5.
each causes the following block, enclosed by {} to be executed once for each element of the sequence. The block variable i (in |i|) represents the sequence element.
r << x means add x to the end of the array r.
the expression to the right of r << says, "if the element of a at index i is one greater than element of a at index i-1, then the length of the run ending at index i is one greater than the length of the run ending at index i-1; else, a new run begins, whose length at offset i is 1.
After each is finished:
# r => [1, 1, 1, 2, 3, 1]
All that is required now is to find the element of r whose value is greatest:
r.max #=> 3
and the associated index:
r.index(r.max) #=> 4
Actually, the code above would more typically be written like this:
(1...a.size).each_with_object([1]) {|i,r| r << a[i] == a[i-1]+1 ? r[i-1]+1 : 1}
start, length = r.index(r.max) + 1 - r.max, r.max #=> 2, 3
Alternatively, we could have made r a hash (call it h) rather than an array, and written:
(1...a.size).each_with_object({0 => 1}) {|i,h|
h[i] = a[i] == a[i-1]+1 ? h[i-1]+1 : 1}.max_by {|_,v| v} #=> [4, 3]

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