Longest Run of Ones After One Deletion

medium array sliding window

Problem

Exactly one element must be deleted from a binary array. Return the longest run of 1s achievable in the result. Use a sliding window that allows at most one zero inside; the answer is (window length) − 1.

Inputnums = [1,1,0,1,1,1,0,1]

Output5

Window [0..5] has length 6 with one zero — after deletion run length = 5.

nums (binary, comma-separated)

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

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

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

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

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