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Max Consecutive Ones III in C++
Suppose we have an array A of 0s and 1s, we can update up to K values from 0 to 1. We have to find the length of the longest (contiguous) subarray that contains only 1s. So if A = [1,1,1,0,0,0,1,1,1,1,0] and k = 2, then the output will be 6, So if we flip 2 0s, the array can be of like [1,1,1,0,0,1,1,1,1,1,1], the length of longest sequence of 1s is 6.
To solve this, we will follow these steps −
- set ans := 0, j := 0 and n := size of array
- for i in range 0 to n – 1
- if A[i] is 0, then decrease k by 1
- while j <= i and k < 0
- if A[j] = 0, then increase k by 1
- increase j by 1
- ans := max of i – j + 1, ans
- return ans
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int longestOnes(vector<int>& A, int k) {
int ans = 0;
int j = 0;
int n = A.size();
for(int i = 0; i < n; i++){
if(A[i] == 0) k--;
while(j <= i && k <0){
if(A[j] == 0){
k++;
}
j++;
}
ans = max(i - j + 1, ans);
}
return ans;
}
};
main(){
vector<int> v = {1,1,1,0,0,0,1,1,1,1,0};
Solution ob;
cout <<(ob.longestOnes(v, 3));
}Input
[1,1,1,0,0,0,1,1,1,1,0] 3
Output
10
Updated on: 30-Apr-2020
405 Views
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