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Shortest Subarray with Sum at Least K in C++
Suppose we have an array A. We have to find the length of the shortest, non-empty, contiguous subarray of A whose sum is at least K. If there is no such subarray, then return -1.
So, if the input is like [5,3,-2,2,1] and k = 6, then the output will be 2, as we can see (5+3) >= 6
To solve this, we will follow these steps −
n := size of A
ans := n + 1, j := 0, sum := 0
Define one deque dq
for initialize i := 0, when i < n, update (increase i by 1), do −
if i > 0, then −
A[i] := A[i] + A[i - 1]
if A[i] >= K, then −
ans := minimum of ans and i + 1
while (not dq is empty and of A[i] - first element A[dq] >= K), do −
ans := minimum of ans and first element of i - dq
delete front element from dq
while (not dq is empty and A[i] <= last element of A[dq], do −
delete last element from dq
insert i at the end of dq
return (if ans is same as n + 1, then -1, otherwise ans)
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int shortestSubarray(vector<int> &A, int K) {
int n = A.size();
int ans = n + 1;
int j = 0;
int sum = 0;
deque<int> dq;
for (int i = 0; i < n; i++) {
if (i > 0)
A[i] += A[i - 1];
if (A[i] >= K) {
ans = min(ans, i + 1);
}
while (!dq.empty() && A[i] - A[dq.front()] >= K) {
ans = min(ans, i - dq.front());
dq.pop_front();
}
while (!dq.empty() && A[i] <= A[dq.back()])
dq.pop_back();
dq.push_back(i);
}
return ans == n + 1 ? -1 : ans;
}
};
main(){
Solution ob;
vector<int> v = {5,3,-2,2,1};
cout << (ob.shortestSubarray(v, 6));
}Input
{5,3,-2,2,1}, 6Output
2
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