Smallest subarray with sum equal to k - arrays

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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;
}`

Answer 5

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

Answer 6

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]

Answer 7

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