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.


n ≤ 100,000 where n is the length of nums

Example 1


nums = [1, 1, 1, 0, 0, 1, 0]

k = 2




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));
            if (r < n) z += (nums[r] == 0) ? 1 : 0;
        } else {
            z -= (nums[l] == 0) ? 1 : 0;
    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)
            while (cnt > k) {
                if (A[j++] == 0)
            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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