This problem was asked to me in Amazon interview -
Given a array of positive integers, you have to find the smallest positive integer that can not be formed from the sum of numbers from array.
Example:
Array:[4 13 2 3 1]
result= 11 { Since 11 was smallest positive number which can not be formed from the given array elements }
What i did was :
sorted the array
calculated the prefix sum
Treverse the sum array and check if next element is less than 1
greater than sum i.e. A[j]<=(sum+1). If not so then answer would
be sum+1
But this was nlog(n) solution.
Interviewer was not satisfied with this and asked a solution in less than O(n log n) time.
There's a beautiful algorithm for solving this problem in time O(n + Sort), where Sort is the amount of time required to sort the input array.
The idea behind the algorithm is to sort the array and then ask the following question: what is the smallest positive integer you cannot make using the first k elements of the array? You then scan forward through the array from left to right, updating your answer to this question, until you find the smallest number you can't make.
Here's how it works. Initially, the smallest number you can't make is 1. Then, going from left to right, do the following:
If the current number is bigger than the smallest number you can't make so far, then you know the smallest number you can't make - it's the one you've got recorded, and you're done.
Otherwise, the current number is less than or equal to the smallest number you can't make. The claim is that you can indeed make this number. Right now, you know the smallest number you can't make with the first k elements of the array (call it candidate) and are now looking at value A[k]. The number candidate - A[k] therefore must be some number that you can indeed make with the first k elements of the array, since otherwise candidate - A[k] would be a smaller number than the smallest number you allegedly can't make with the first k numbers in the array. Moreover, you can make any number in the range candidate to candidate + A[k], inclusive, because you can start with any number in the range from 1 to A[k], inclusive, and then add candidate - 1 to it. Therefore, set candidate to candidate + A[k] and increment k.
In pseudocode:
Sort(A)
candidate = 1
for i from 1 to length(A):
if A[i] > candidate: return candidate
else: candidate = candidate + A[i]
return candidate
Here's a test run on [4, 13, 2, 1, 3]. Sort the array to get [1, 2, 3, 4, 13]. Then, set candidate to 1. We then do the following:
A[1] = 1, candidate = 1:
A[1] ≤ candidate, so set candidate = candidate + A[1] = 2
A[2] = 2, candidate = 2:
A[2] ≤ candidate, so set candidate = candidate + A[2] = 4
A[3] = 3, candidate = 4:
A[3] ≤ candidate, so set candidate = candidate + A[3] = 7
A[4] = 4, candidate = 7:
A[4] ≤ candidate, so set candidate = candidate + A[4] = 11
A[5] = 13, candidate = 11:
A[5] > candidate, so return candidate (11).
So the answer is 11.
The runtime here is O(n + Sort) because outside of sorting, the runtime is O(n). You can clearly sort in O(n log n) time using heapsort, and if you know some upper bound on the numbers you can sort in time O(n log U) (where U is the maximum possible number) by using radix sort. If U is a fixed constant, (say, 109), then radix sort runs in time O(n) and this entire algorithm then runs in time O(n) as well.
Hope this helps!
Use bitvectors to accomplish this in linear time.
Start with an empty bitvector b. Then for each element k in your array, do this:
b = b | b << k | 2^(k-1)
To be clear, the i'th element is set to 1 to represent the number i, and | k is setting the k-th element to 1.
After you finish processing the array, the index of the first zero in b is your answer (counting from the right, starting at 1).
b=0
process 4: b = b | b<<4 | 1000 = 1000
process 13: b = b | b<<13 | 1000000000000 = 10001000000001000
process 2: b = b | b<<2 | 10 = 1010101000000101010
process 3: b = b | b<<3 | 100 = 1011111101000101111110
process 1: b = b | b<<1 | 1 = 11111111111001111111111
First zero: position 11.
Consider all integers in interval [2i .. 2i+1 - 1]. And suppose all integers below 2i can be formed from sum of numbers from given array. Also suppose that we already know C, which is sum of all numbers below 2i. If C >= 2i+1 - 1, every number in this interval may be represented as sum of given numbers. Otherwise we could check if interval [2i .. C + 1] contains any number from given array. And if there is no such number, C + 1 is what we searched for.
Here is a sketch of an algorithm:
For each input number, determine to which interval it belongs, and update corresponding sum: S[int_log(x)] += x.
Compute prefix sum for array S: foreach i: C[i] = C[i-1] + S[i].
Filter array C to keep only entries with values lower than next power of 2.
Scan input array once more and notice which of the intervals [2i .. C + 1] contain at least one input number: i = int_log(x) - 1; B[i] |= (x <= C[i] + 1).
Find first interval that is not filtered out on step #3 and corresponding element of B[] not set on step #4.
If it is not obvious why we can apply step 3, here is the proof. Choose any number between 2i and C, then sequentially subtract from it all the numbers below 2i in decreasing order. Eventually we get either some number less than the last subtracted number or zero. If the result is zero, just add together all the subtracted numbers and we have the representation of chosen number. If the result is non-zero and less than the last subtracted number, this result is also less than 2i, so it is "representable" and none of the subtracted numbers are used for its representation. When we add these subtracted numbers back, we have the representation of chosen number. This also suggests that instead of filtering intervals one by one we could skip several intervals at once by jumping directly to int_log of C.
Time complexity is determined by function int_log(), which is integer logarithm or index of the highest set bit in the number. If our instruction set contains integer logarithm or any its equivalent (count leading zeros, or tricks with floating point numbers), then complexity is O(n). Otherwise we could use some bit hacking to implement int_log() in O(log log U) and obtain O(n * log log U) time complexity. (Here U is largest number in the array).
If step 1 (in addition to updating the sum) will also update minimum value in given range, step 4 is not needed anymore. We could just compare C[i] to Min[i+1]. This means we need only single pass over input array. Or we could apply this algorithm not to array but to a stream of numbers.
Several examples:
Input: [ 4 13 2 3 1] [ 1 2 3 9] [ 1 1 2 9]
int_log: 2 3 1 1 0 0 1 1 3 0 0 1 3
int_log: 0 1 2 3 0 1 2 3 0 1 2 3
S: 1 5 4 13 1 5 0 9 2 2 0 9
C: 1 6 10 23 1 6 6 15 2 4 4 13
filtered(C): n n n n n n n n n n n n
number in
[2^i..C+1]: 2 4 - 2 - - 2 - -
C+1: 11 7 5
For multi-precision input numbers this approach needs O(n * log M) time and O(log M) space. Where M is largest number in the array. The same time is needed just to read all the numbers (and in the worst case we need every bit of them).
Still this result may be improved to O(n * log R) where R is the value found by this algorithm (actually, the output-sensitive variant of it). The only modification needed for this optimization is instead of processing whole numbers at once, process them digit-by-digit: first pass processes the low order bits of each number (like bits 0..63), second pass - next bits (like 64..127), etc. We could ignore all higher-order bits after result is found. Also this decreases space requirements to O(K) numbers, where K is number of bits in machine word.
If you sort the array, it will work for you. Counting sort could've done it in O(n), but if you think in a practically large scenario, range can be pretty high.
Quicksort O(n*logn) will do the work for you:
def smallestPositiveInteger(self, array):
candidate = 1
n = len(array)
array = sorted(array)
for i in range(0, n):
if array[i] <= candidate:
candidate += array[i]
else:
break
return candidate
Related
There are n positive numbers (A1 , ... An) on a circle, how do we divide this circle into subsegments with sum greater or equal to m so that number of subsegments are maximum in O(n) , or O(nlogn)
eg :
n = 6 m = 6
3 1 2 3 6 3
ANS = 3
since we can divide the array into three subsegments{[2,4],[5,5],[6,1]}
If there is at least one number greater or equal to m in the array, just cut the array into smallest possible pieces starting from one of these numbers. Otherwise (and if sum of numbers is at least 2*m) use a pointer-chasing algorithm.
This algorithm uses 2 additional arrays: L for chain lengths (initially zero) and S for starting indices (initially equal to own indices: 0, 1, 2, ...). And 2 array indices: F and B (initially zero).
Increment F while sum between F and B is less than m. Then increment B while sum between F and B is greater than m (but stop when is is still greater or equal to m).
Update arrays: L[F] = 1 + L[B], S[F] = S[B].
Repeat steps 1,2 while F<n. While incrementing F on step 1, copy most recently updated values to L[F] and S[F].
Reset F to zero.
Increment F while sum of elements before F and after B is less than m. Then increment B while sum before F and after B is greater than m (but stop when is is still greater or equal to m).
If F <= S[B] use L[B] + 1 to update maximum number of subsegments.
Repeat steps 5,6 while B<n.
I was trying to do my friends problem set from a few years ago to sharpen up my knowledge about data structures etc. I came across this problem, and I'm not really sure where to start. Hopefully someone could help me out!
We are given n unsorted arrays, each array has n elements. Ex.
3 1 2
7 6 9
4 9 12
Now, say we take one element from each array, and add them up. Lets just call the sum of these elements an "n-sum".
I need to devise an algorithm that gives us the n smallest "n-sums" (duplicates are allowed).
In our above ex, the answer would be:
11, 12, 12
# 11 comes from: 1 (first array) + 6 (second array) + 4 (third array)
# 12 comes from: 2 (first array) + 6 (second array) + 4 (third array)
# 12 comes from: 1 (first array) + 7 (second array) + 4 (third array)
One of the suggestions given were to use a priority queue.
Thanks!
The time is at least O (n^2): You must visit all array elements, because if all elements were equal to 1000 except on in each row being 0, you would have to look at the n elements equal to 0, or you couldn't find the smallest sum.
Sort each row, taking O (n^2 log n) steps. In each row, subtract the first element from all elements in the row, so the first element in each row is 0; after you found the smallest sums you can compensate for that. Your example turns into
3 1 2 -> 1 2 3 -> 0 1 2
7 6 9 -> 6 7 9 -> 0 1 3
4 9 12 -> 4 9 12-> 0 5 7
Now finding all sums ≤ K can be done in m steps if there are m sums: In the first row, pick all values in turn as long as they are ≤ K. In the second row, pick all values in turn as long as the sum from two rows is ≤ K and so on. Since each row starts with 0, no time is wasted.
For example, sums ≤ 5 are: 0+0+0, 0+0+5, 0+1+0, 0+3+0, 1+0+0, 1+1+0, 1+3+0, 2+0+0, 2+1+0, 2+3+0. Many more than the three that we needed. If we stop after finding 3 sums ≤ 5, we know very quickly "there are at least 3 sums ≤ 5". We need to have an early stop, because in the general case there could be n^n possible sums.
If you pick K = "largest element in the second column", then you know there are at least n+1 sums with a value ≤ K, because you can pick all 0's, or all 0's except one value from the second column. In your example, K = 5 (we know that worked). Let X be the value where there are n sums ≤ X but fewer than n sums ≤ X - 1. We find X with binary search between 0 and K, and then we find the sums. Example:
K = 5 is known to be big enough. We try K = 2, and find 4 sums (actually we stop at 3 sums). Too many. We try K = 1, and there are three solutions 0+0+0, 0+1+0 and 1+0+0. We try K = 0, but only one solution.
This part goes very quick, so we'd try to reduce the time used for sorting. We notice that in this case looking at the first two columns was enough. We can in each row find the two smallest items, and in this case that would be enough. If the two smallest items are not enough to determine the n smallest sums, find the third smallest item etc. where needed. For example, since the 2nd largest item of the last row is 5, we wouldn't need the third item of the row, because even the 5 is not element of a sum if K ≤ 4.
Given an array of N positive elements. Lets suppose we list all N × (N+1) / 2 non-empty continuous subarrays of the array A and then replaced all the subarrays with the maximum element present in the respective subarray. So now we have N × (N+1) / 2 elements where each element is maximum among its subarray.
Now we are having Q queries, where each query is one of 3 types :
1 K : We need to count of numbers strictly greater than K among those N × (N+1) / 2 elements.
2 K : We need to count of numbers strictly less than K among those N × (N+1) / 2 elements.
3 K : We need to count of numbers equal to K among those N × (N+1) / 2 elements.
Now main problem am facing is N can be upto 10^6. So i can't generate all those N × (N+1) / 2 elements. Please help to solve this porblem.
Example : Let N=3 and we have Q=2. Let array A be [1,2,3] then all sub arrays are :
[1] -> [1]
[2] -> [2]
[3] -> [3]
[1,2] -> [2]
[2,3] -> [3]
[1,2,3] -> [3]
So now we have [1,2,3,2,3,3]. As Q=2 so :
Query 1 : 3 3
It means we need to tell count of numbers equal to 3. So answer is 3 as there are 3 numbers equal to 3 in the generated array.
Query 2 : 1 4
It means we need to tell count of numbers greater than 4. So answer is 0 as no one is greater than 4 in generated array.
Now both N and Q can be up to 10^6. So how to solve this problem. Which data structure should be suitable to solve it.
I believe I have a solution in O(N + Q*log N) (More about time complexity). The trick is to do a lot of preparation with your array before even the first query arrives.
For each number, figure out where is the first number on left / right of this number that is strictly bigger.
Example: for array: 1, 8, 2, 3, 3, 5, 1 both 3's left block would be position of 8, right block would be the position of 5.
This can be determined in linear time. This is how: Keep a stack of previous maximums in a stack. If a new maximum appears, remove maximums from the stack until you get to a element bigger than or equal to the current one. Illustration:
In this example, in the stack is: [15, 13, 11, 10, 7, 3] (you will of course keep the indexes, not the values, I will just use value for better readability).
Now we read 8, 8 >= 3 so we remove 3 from stack and repeat. 8 >= 7, remove 7. 8 < 10, so we stop removing. We set 10 as 8's left block, and add 8 to the maximums stack.
Also, whenever you remove from the stack (3 and 7 in this example), set the right block of removed number to the current number. One problem though: right block would be set to the next number bigger or equal, not strictly bigger. You can fix this with simply checking and relinking right blocks.
Compute what number is how many times a maximum of some subsequence.
Since for each number you now know where is the next left / right bigger number, I trust you with finding appropriate math formula for this.
Then, store the results in a hashmap, key would be a value of a number, and value would be how many times is that number a maximum of some subsequence. For example, record [4->12] would mean that number 4 is the maximum in 12 subsequences.
Lastly, extract all key-value pairs from the hashmap into an array, and sort that array by the keys. Finally, create a prefix sum for the values of that sorted array.
Handle a request
For request "exactly k", just binary search in your array, for more/less thank``, binary search for key k and then use the prefix array.
This answer is an adaptation of this other answer I wrote earlier. The first part is exactly the same, but the others are specific for this question.
Here's an implemented a O(n log n + q log n) version using a simplified version of a segment tree.
Creating the segment tree: O(n)
In practice, what it does is to take an array, let's say:
A = [5,1,7,2,3,7,3,1]
And construct an array-backed tree that looks like this:
In the tree, the first number is the value and the second is the index where it appears in the array. Each node is the maximum of its two children. This tree is backed by an array (pretty much like a heap tree) where the children of the index i are in the indexes i*2+1 and i*2+2.
Then, for each element, it becomes easy to find the nearest greater elements (before and after each element).
To find the nearest greater element to the left, we go up in the tree searching for the first parent where the left node has value greater and the index lesser than the argument. The answer must be a child of this parent, then we go down in the tree looking for the rightmost node that satisfies the same condition.
Similarly, to find the nearest greater element to the right, we do the same, but looking for a right node with an index greater than the argument. And when going down, we look for the leftmost node that satisfies the condition.
Creating the cumulative frequency array: O(n log n)
From this structure, we can compute the frequency array, that tells how many times each element appears as maximum in the subarray list. We just have to count how many lesser elements are on the left and on the right of each element and multiply those values. For the example array ([1, 2, 3]), this would be:
[(1, 1), (2, 2), (3, 3)]
This means that 1 appears only once as maximum, 2 appears twice, etc.
But we need to answer range queries, so it's better to have a cumulative version of this array, that would look like:
[(1, 1), (2, 3), (3, 6)]
The (3, 6) means, for example, that there are 6 subarrays with maxima less than or equal to 3.
Answering q queries: O(q log n)
Then, to answer each query, you just have to make binary searches to find the value you want. For example. If you need to find the exact number of 3, you may want to do: query(F, 3) - query(F, 2). If you want to find those lesser than 3, you do: query(F, 2). If you want to find those greater than 3: query(F, float('inf')) - query(F, 3).
Implementation
I've implemented it in Python and it seems to work well.
import sys, random, bisect
from collections import defaultdict
from math import log, ceil
def make_tree(A):
n = 2**(int(ceil(log(len(A), 2))))
T = [(None, None)]*(2*n-1)
for i, x in enumerate(A):
T[n-1+i] = (x, i)
for i in reversed(xrange(n-1)):
T[i] = max(T[i*2+1], T[i*2+2])
return T
def print_tree(T):
print 'digraph {'
for i, x in enumerate(T):
print ' ' + str(i) + '[label="' + str(x) + '"]'
if i*2+2 < len(T):
print ' ' + str(i)+ '->'+ str(i*2+1)
print ' ' + str(i)+ '->'+ str(i*2+2)
print '}'
def find_generic(T, i, fallback, check, first, second):
j = len(T)/2+i
original = T[j]
j = (j-1)/2
#go up in the tree searching for a value that satisfies check
while j > 0 and not check(T[second(j)], original):
j = (j-1)/2
#go down in the tree searching for the left/rightmost node that satisfies check
while j*2+1<len(T):
if check(T[first(j)], original):
j = first(j)
elif check(T[second(j)], original):
j = second(j)
else:
return fallback
return j-len(T)/2
def find_left(T, i, fallback):
return find_generic(T, i, fallback,
lambda a, b: a[0]>b[0] and a[1]<b[1], #value greater, index before
lambda j: j*2+2, #rightmost first
lambda j: j*2+1 #leftmost second
)
def find_right(T, i, fallback):
return find_generic(T, i, fallback,
lambda a, b: a[0]>=b[0] and a[1]>b[1], #value greater or equal, index after
lambda j: j*2+1, #leftmost first
lambda j: j*2+2 #rightmost second
)
def make_frequency_array(A):
T = make_tree(A)
D = defaultdict(lambda: 0)
for i, x in enumerate(A):
left = find_left(T, i, -1)
right = find_right(T, i, len(A))
D[x] += (i-left) * (right-i)
F = sorted(D.items())
for i in range(1, len(F)):
F[i] = (F[i][0], F[i-1][1] + F[i][1])
return F
def query(F, n):
idx = bisect.bisect(F, (n,))
if idx>=len(F): return F[-1][1]
if F[idx][0]!=n: return 0
return F[idx][1]
F = make_frequency_array([1,2,3])
print query(F, 3)-query(F, 2) #3 3
print query(F, float('inf'))-query(F, 4) #1 4
print query(F, float('inf'))-query(F, 1) #1 1
print query(F, 2) #2 3
You problem can be divided into several steps:
For each element of initial array calculate the number of "subarrays" where current element is maximum. This will involve a bit of combinatorics. First you need for each element to know index of previous and next element that is bigger than current element. Then calculate the number of subarrays as (i - iprev) * (inext - i). Finding iprev and inext requires two traversals of the initial array: in forward and backward order. For iprev you need to traverse you array left to right. During the traversal maintain the BST that contains the biggest of the previous elements along with their index. For each element of original array, find the minimal element in BST that is bigger than current. It's index, stored as value, will be iprev. Then remove from BST all elements that are smaller that current. This operation should be O(logN), as you are removing whole subtrees. This step is required, as current element you are about to add will "override" all element that are less than it. Then add current element to BST with it's index as value. At each point of time, BST will store the descending subsequence of previous elements where each element is bigger than all it's predecessors in array (for previous elements {1,2,44,5,2,6,26,6} BST will store {44,26,6}). The backward traversal to find inext is similar.
After previous step you'll have pairs K→P where K is the value of some element from the initial array and P is the number of subarrays where this element is maxumum. Now you need to group this pairs by K. This means calculating sum of P values of the equal K elements. Be careful about the corner cases when two elements could have share the same subarrays.
As Ritesh suggested: Put all grouped K→P into an array, sort it by K and calculate cumulative sum of P for each element in one pass. It this case your queries will be binary searches in this sorted array. Each query will be performed in O(log(N)) time.
Create a sorted value-to-index map. For example,
[34,5,67,10,100] => {5:1, 10:3, 34:0, 67:2, 100:4}
Precalculate the queries in two passes over the value-to-index map:
Top to bottom - maintain an augmented tree of intervals. Each time an index is added,
split the appropriate interval and subtract the relevant segments from the total:
indexes intervals total sub-arrays with maximum greater than
4 (0,3) 67 => 15 - (4*5/2) = 5
2,4 (0,1)(3,3) 34 => 5 + (4*5/2) - 2*3/2 - 1 = 11
0,2,4 (1,1)(3,3) 10 => 11 + 2*3/2 - 1 = 13
3,0,2,4 (1,1) 5 => 13 + 1 = 14
Bottom to top - maintain an augmented tree of intervals. Each time an index is added,
adjust the appropriate interval and add the relevant segments to the total:
indexes intervals total sub-arrays with maximum less than
1 (1,1) 10 => 1*2/2 = 1
1,3 (1,1)(3,3) 34 => 1 + 1*2/2 = 2
0,1,3 (0,1)(3,3) 67 => 2 - 1 + 2*3/2 = 4
0,1,3,2 (0,3) 100 => 4 - 4 + 4*5/2 = 10
The third query can be pre-calculated along with the second:
indexes intervals total sub-arrays with maximum exactly
1 (1,1) 5 => 1
1,3 (3,3) 10 => 1
0,1,3 (0,1) 34 => 2
0,1,3,2 (0,3) 67 => 3 + 3 = 6
Insertion and deletion in augmented trees are of O(log n) time-complexity. Total precalculation time-complexity is O(n log n). Each query after that ought to be O(log n) time-complexity.
Problem from :
https://www.hackerrank.com/contests/epiccode/challenges/white-falcon-and-sequence.
Visit link for references.
I have a sequence of integers (-10^6 to 10^6) A. I need to choose two contiguous disjoint subsequences of A, let's say x and y, of the same size, n.
After that you will calculate the sum given by ∑x(i)y(n−i+1) (1-indexed)
And I have to choose x and y such that sum is maximised.
Eg:
Input:
12
1 7 4 0 9 4 0 1 8 8 2 4
Output: 120
Where x = {4,0,9,4}
y = {8,8,2,4}
∑x(i)y(n−i+1)=4×4+0×2+9×8+4×8=120
Now, the approach that I was thinking of for this is something in lines of O(n^2) which is as follows:
Initialise two variables l = 0 and r = N-1. Here, N is the size of the array.
Now, for l=0, I will calculate the sum while (l<r) which basically refers to the subsequences that will start from the 0th position in the array. Then, I will increment l and decrement r in order to come up with subsequences that start from the above position + 1 and on the right hand side, start from right-1.
Is there any better approach that I can use? Anything more efficient? I thought of sorting but we cannot sort numbers since that will change the order of the numbers.
To answer the question we first define S(i, j) to be the max sum of multlying the two sub-sequence items, for sub-array A[i...j] when the sub-sequence x starts at position i, and sub-sequence y ends on position j.
For example, if A=[1 7 4 0 9 4 0 1 8 8 2 4], then S(1, 2)=1*7=7 and S(2, 5)=7*9+4*0=63.
The recursive rule to compute S is: S(i, j)=max(0, S(i+1, j-1)+A[i]*A[j]), and the end condition is S(i, j)=0 iff i>=j.
The requested final answer is simply the maximum value of S(i, j) for all combinations of i=1..N, j=1..N, since one of the S(i ,j) values will correspond to the max x,y sub-sequences, and thus will be equal the maximum value for the whole array. The complexity of computing all such S(i, j) values is O(N^2) using dynamic programming, since in the course of computing S(i, j) we will also compute the values of up to N other S(i', j') values, but ultimately each combination will be computed only once.
def max_sum(l):
def _max_sub_sum(i, j):
if m[i][j]==None:
v=0
if i<j:
v=max(0, _max_sub_sum(i+1, j-1)+l[i]*l[j])
m[i][j]=v
return m[i][j]
n=len(l)
m=[[None for i in range(n)] for j in range(n)]
v=0
for i in range(n):
for j in range(i, n):
v=max(v, _max_sub_sum(i, j))
return v
WARNING:
This method assumes the numbers are non-negative so this solution does not answer the poster's actual problem now it has been clarified that negative input values are allowed.
Trick 1
Assuming the numbers are always non-negative, it is always best to make the sequences as wide as possible given the location where they meet.
Trick 2
We can change the sum into a standard convolution by summing over all values of i. This produces twice the desired result (as we get both the product of x with y, and y with x), but we can divide by 2 at the end to get the original answer.
Trick 3
You are now attempting to find the maximum of a convolution of a signal with itself. There is a standard method for doing this which is to use the fast fourier transform. Some libraries will have this built in, e.g. in Scipy there is fftconvolve.
Python code
Note that you don't allow the central value to be reused (e.g. for a sequance 1,3,2 we can't make x 1,3 and y 3,1) so we need to examine alternate values of the convolved output.
We can now compute the answer in Python via:
import scipy.signal
A = [1, 7, 4, 0, 9, 4, 0, 1, 8, 8, 2, 4]
print max(scipy.signal.fftconvolve(A,A)[1::2]) / 2
Given an array of positive integers a I want to output array of integers b so that b[i] is the closest number to a[i] that is smaller then a[i], and is in {a[0], ... a[i-1]}. If such number doesn't exist, then b[i] = -1.
Example:
a = 2 1 7 5 7 9
b = -1 -1 2 2 5 7
b[0] = -1 since there is no number that is smaller than 2
b[1] = -1 since there is no number that is smaller than 1 from {2}
b[2] = 2, closest number to 7 that is smaller than 7 from {2,1} is 2
b[3] = 2, closest number to 5 that is smaller than 5 from {2,1,7} is 2
b[4] = 5, closest number to 7 that is smaller than 7 from {2,1,7,5} is 5
I was thinking about implementing balanced binary tree, however it will require a lot of work. Is there an easier way of doing this?
Here is one approach:
for i ← 1 to i ← (length(A)-1) {
// A[i] is added in the sorted sequence A[0, .. i-1] save A[i] to make a hole at index j
item = A[i]
j = i
// keep moving the hole to next smaller index until A[j - 1] is <= item
while j > 0 and A[j - 1] > item {
A[j] = A[j - 1] // move hole to next smaller index
j = j - 1
}
A[j] = item // put item in the hole
// if there are elements to the left of A[j] in sorted sequence A[0, .. i-1], then store it in b
// TODO : run loop so that duplicate entries wont hamper results
if j > 1
b[i] = A[j-1]
else
b[1] = -1;
}
Dry run:
a = 2 1 7 5 7 9
a[1] = 2
its straight forward, set b[1] to -1
a[2] = 1
insert into subarray : [1 ,2]
any elements before 1 in sorted array ? no.
So set b[2] to -1 . b: [-1, -1]
a[3] = 7
insert into subarray : [1 ,2, 7]
any elements before 7 in sorted array ? yes. its 2
So set b[3] to 2. b: [-1, -1, 2]
a[4] = 5
insert into subarray : [1 ,2, 5, 7]
any elements before 5 in sorted array ? yes. its 2
So set b[4] to 2. b: [-1, -1, 2, 2]
and so on..
Here's a sketch of a (nearly) O(n log n) algorithm that's somewhere in between the difficulty of implementing an insertion sort and balanced binary tree: Do the problem backwards, use merge/quick sort, and use binary search.
Pseudocode:
let c be a copy of a
let b be an array sized the same as a
sort c using an O(n log n) algorithm
for i from a.length-1 to 1
binary search over c for key a[i] // O(log n) time
remove the item found // Could take O(n) time
if there exists an item to the left of that position, b[i] = that item
otherwise, b[i] = -1
b[0] = -1
return b
There's a few implementation details that can make this have poor runtime.
For instance, since you have to remove items, doing this on a regular array and shifting things around will make this algorithm still take O(n^2) time. So, you could store key-value pairs instead. One would be the key, and the other would be the number of those keys (kind of like a multiset implemented on an array). "Removing" one would just be subtracting the second item from the pair and so on.
Eventually you will be left with a bunch of 0-value keys. This would eventually make the if there exists an item to the left take roughly O(n) time, and therefore, the entire algorithm would degrade to a O(n^2) for that reason. So another optimization might be to batch remove all of them periodically. For instance, when 1/2 of them are 0-values, perform a pruning.
The ideal option might be to implement another data structure that has a much more favorable remove time. Something along the lines of a modified unrolled linked list with indices could work, but it would certainly increase the implementation complexity of this approach.
I've actually implemented this. I used the first two optimizations above (storing key-value pairs for compression, and pruning when 1/2 of them are 0s). Here's some benchmarks to compare using an insertion sort derivative to this one:
a.length This method Insert sort Method
100 0.0262ms 0.0204ms
1000 0.2300ms 0.8793ms
10000 2.7303ms 75.7155ms
100000 32.6601ms 7740.36 ms
300000 98.9956ms 69523.6 ms
1000000 333.501 ms ????? Not patient enough
So, as you can see, this algorithm grows much, much slower than the insertion sort method I posted before. However, it took 73 lines of code vs 26 lines of code for the insertion sort method. So in terms of simplicity, the insertion sort method might still be the way to go if you don't have time requirements/the input is small.
You could treat it like an insertion sort.
Pseudocode:
let arr be one array with enough space for every item in a
let b be another array with, again, enough space for all elements in a
For each item in a:
perform insertion sort on item into arr
After performing the insertion, if there exists a number to the left, append that to b.
Otherwise, append -1 to b
return b
The main thing you have to worry about is making sure that you don't make the mistake of reallocating arrays (because it would reallocate n times, which would be extremely costly). This will be an implementation detail of whatever language you use (std::vector's reserve for C++ ... arr.reserve(n) for D ... ArrayList's ensureCapacity in Java...)
A potential downfall with this approach compared to using a binary tree is that it's O(n^2) time. However, the constant factors using this method vs binary tree would make this faster for smaller sizes. If your n is smaller than 1000, this would be an appropriate solution. However, O(n log n) grows much slower than O(n^2), so if you expect a's size to be significantly higher and if there's a time limit that you are likely to breach, you might consider a more complicated O(n log n) algorithm.
There are ways to slightly improve the performance (such as using a binary insertion sort: using binary search to find the position to insert into), but generally they won't improve performance enough to matter in most cases since it's still O(n^2) time to shift elements to fit.
Consider this:
a = 2 1 7 5 7 9
b = -1 -1 2 2 5 7
c 0 1 2 3 4 5 6 7 8 9
0 - - - - - - - - - -
Where the index of C is value of a[i] such that 0,3,4,6,8 would have null values.
and the 1st dimension of C contains the highest to date closest value to a[i]
So in step by a[3] we have the following
c 0 1 2 3 4 5 6 7 8 9
0 - -1 -1 - - 2 - 2 - -
and by step a[5] we have the following
c 0 1 2 3 4 5 6 7 8 9
0 - -1 -1 - - 2 - 5 - 7
This way when we get to the 2nd 7 at a[4] we know that 2 is the largest value to date and all we need to do is loop back through a[i-1] until we encounter a 7 again comparing the a[i] value to that in c[7] if bigger, replace c[7]. Once a[i-1] = the 7 we put c[7] into b[i] and move on to next a[i].
The main downfalls to this approach that I can see are:
footprint size depending on how big the c[] needs to be dimensioned..
the fact that you have to revisit elements of a[] that you've already touched. If the distribution of data is such that there are significant spaces between the two 7's then keeping track of the highest value as you go would presumably be faster. Alternatively it might be better to gather statistics on the a[i] up front to know what distributions exist and then use a hybrid method maintaining the max until such time that no more instances of that number are in the statistics.