Longest Subarray of 1's After Deleting One Element
Find the longest subarray of 1s left after deleting exactly one element, using a sliding window that allows at most one zero inside it.
Problem Understanding
Longest Subarray of 1's After Deleting One Element (LeetCode 1493): given a binary array nums, delete exactly one element and return the length of the longest non-empty subarray containing only 1s in the result. Return 0 if there is none.
Examples: [1, 1, 0, 1] → 3; [0, 1, 1, 1, 0, 1, 1, 0, 1] → 5; [1, 1, 1] → 2.
The Intuition: A Window That Tolerates One Zero
After deleting one element, a run of 1s in the result came from a stretch of the original with at most one zero, the one you deleted.
- Grow the window to the right.
- Shrink from the left whenever it holds a second zero.
- Measure length - 1, the 1s left after the deletion.
It is Max Consecutive Ones III with k = 1, counting the deletion.
Interactive Walkthrough
AT MOST ONE ZERO IN THE WINDOW
GROW RIGHT
nums (window outlined)
lo hi
0
1
2
3
4
5
6
7
8
zeros
1
length − 1
0
best
0
nums[0] = 0 joins the window. Zeros in it: 1
Next
Window [0, 0] has 1 elements; deleting one leaves 0
nums
The outlined cells are the window, and a zero inside it is tinted: that is the element to delete. Try All ones to see why the answer is n - 1.
Dry Run Table
Input: [0, 1, 1, 1, 0, 1, 1, 0, 1].
| hi | Window | zeros | length − 1 | best |
|---|---|---|---|---|
| 0 to 3 | [0, 3] | 1 | 3 | 3 |
| 4 | [1, 4] after shrinking | 1 | 3 | 3 |
| 5 | [1, 5] | 1 | 4 | 4 |
| 6 | [1, 6] | 1 | 5 | 5 |
| 7 | [5, 7] after shrinking | 1 | 2 | 5 |
| 8 | [5, 8] | 1 | 3 | 5 |
Answer: 5, from the window [1, 6] with the zero at index 4 deleted.
The Solution Template
A variable-size window with a zero counter; the answer is the largest hi - lo.
Longest Subarray of 1's After Deleting One Element Code
1var longestSubarray = function(nums) {2 let lo = 0, zeros = 0, best = 0;3 for (let hi = 0; hi < nums.length; hi++) {4 if (nums[hi] === 0) zeros++;5 while (zeros > 1) {6 if (nums[lo] === 0) zeros--;7 lo++;8 }9 best = Math.max(best, hi - lo);10 }11 return best;12};
Edge Cases & Common Mistakes
- All ones: answer n - 1, because the deletion is mandatory.
- All zeros: answer 0.
- Returning the window length: forgets the deleted element; use hi - lo, not hi - lo + 1.
- Shrinking with if instead of while: works here because zeros rises by at most 1 per step, but while is the safe general form.
Summary
Slide a window that holds at most one zero and record its length minus one. O(n) time, O(1) space.
More practice: DSA interview questions with answers · patterns cheat sheet