Longest Run of Ones Allowing k Flips

medium array sliding window

Problem

Given a binary array and an integer k, return the longest contiguous block of 1s that can be created by flipping at most k zeros to 1s. Slide a window with at most k zeros inside, shrinking from the left whenever the zero budget is exceeded.

Inputnums = [1,1,0,1,0,0,1,1,1], k = 2

Output6

Flip the two zeros at indices 2 and 4 (or 4 and 5) to get a run of length 6.

nums (binary, comma-separated)

def longest_ones(nums, k):
    l = zeros = best = 0
    for r, x in enumerate(nums):
        if x == 0: zeros += 1
        while zeros > k:
            if nums[l] == 0: zeros -= 1
            l += 1
        if r - l + 1 > best: best = r - l + 1
    return best

function longestOnes(nums, k) {
  let l = 0, zeros = 0, best = 0;
  for (let r = 0; r < nums.length; r++) {
    if (nums[r] === 0) zeros++;
    while (zeros > k) {
      if (nums[l] === 0) zeros--;
      l++;
    }
    if (r - l + 1 > best) best = r - l + 1;
  }
  return best;
}

class Solution {
    public int longestOnes(int[] nums, int k) {
        int l = 0, zeros = 0, best = 0;
        for (int r = 0; r < nums.length; r++) {
            if (nums[r] == 0) zeros++;
            while (zeros > k) {
                if (nums[l] == 0) zeros--;
                l++;
            }
            if (r - l + 1 > best) best = r - l + 1;
        }
        return best;
    }
}

int longestOnes(vector<int>& nums, int k) {
    int l = 0, zeros = 0, best = 0;
    for (int r = 0; r < (int)nums.size(); r++) {
        if (nums[r] == 0) zeros++;
        while (zeros > k) {
            if (nums[l] == 0) zeros--;
            l++;
        }
        if (r - l + 1 > best) best = r - l + 1;
    }
    return best;
}

Time: O(n) Space: O(1)