**Longest Sublist of 1s After K Sets - Amazon Top Interview Questions**

### Problem Statement :

You are given a list of integers nums containing 1s and 0s and an integer k. Given that you can set at most k 0s to 1s, return the length of the longest sublist containing all 1s. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [1, 1, 1, 0, 0, 1, 0] k = 2 Output 6 Explanation We can set the two middle 0s to 1s and then the list becomes [1, 1, 1, 1, 1, 1, 0].

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums, int k) {
// algorithm
// Use two pointers to keep the track of k zeros which we can flip
int n = nums.size();
int l = 0, r = 0, z = (nums.front() == 0) ? 1 : 0, ans = 0;
while (r < n) {
if (z <= k) {
ans = max(ans, (r - l + 1));
r++;
if (r < n) z += (nums[r] == 0) ? 1 : 0;
} else {
z -= (nums[l] == 0) ? 1 : 0;
l++;
}
}
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] A, int k) {
int res = 0;
int cnt = 0;
for (int i = 0, j = 0; i < A.length; i++) {
if (A[i] == 0)
cnt++;
while (cnt > k) {
if (A[j++] == 0)
cnt--;
}
res = Math.max(res, i - j + 1);
}
return res;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums, k):
zeros = 0
ans = 0
left = 0
for right, n in enumerate(nums):
zeros += n == 0
while zeros > k:
zeros -= nums[left] == 0
left += 1
if right - left + 1 > ans:
ans = right - left + 1
return ans
```

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