Leetcode POTD (24 August 2025) – Medium – 1493

1493. Longest Subarray of 1’s After Deleting One Element

Given a binary array nums, you should delete one element from it.

Return the size of the longest non-empty subarray containing only 1‘s in the resulting array. Return 0 if there is no such subarray.

Example 1:

Input: nums = [1,1,0,1]
Output: 3
Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's.

Example 2:

Input: nums = [0,1,1,1,0,1,1,0,1]
Output: 5
Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1].

Example 3:

Input: nums = [1,1,1]
Output: 2
Explanation: You must delete one element.

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

class Solution {
public:
    int longestSubarray(vector<int>& nums) {
        int i = 0;
        int j = 0;
        int last_time_zero_index = -1;
        int longest_subarray = 0;
        while(j<nums.size()){
            if(nums[j]==0){
                i = last_time_zero_index+1;
                last_time_zero_index = j;
            }
            longest_subarray = max(longest_subarray, j-i);
            j++;
        }
        return longest_subarray;
    }
};

Complexity:

Space → O(n)

Time → O(1)

Verdict :- Cheers! We solved POTD.