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

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

Difficulty: Medium

Topics: Array, Dynamic Programming, Sliding Window, Biweekly Contest 29

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.

Hint:

1. Maintain a sliding window where there is at most one zero in it.

Solution:

We need to find the longest contiguous subarray of 1's after deleting exactly one element from the given binary array. The solution involves using a sliding window technique to maintain a window that contains at most one zero. The key insight is that by allowing at most one zero in the window, we can simulate the deletion of that zero to obtain a contiguous segment of 1's. The length of this segment will be the window length minus one, accounting for the deleted element.

Approach

1. Sliding Window Technique: We maintain a window defined by two pointers, left and right. The right pointer expands the window by moving forward, while the left pointer contracts the window if the number of zeros exceeds one.
2. Zero Counter: We keep track of the number of zeros in the current window. If the count exceeds one, we move the left pointer forward until the zero count is reduced to one.
3. Max Length Calculation: For each valid window (containing at most one zero), the potential length of the contiguous 1's after deleting one element is right - left. We update the maximum length found during the iteration.

Let's implement this solution in PHP: 1493. Longest Subarray of 1's After Deleting One Element

<?php
/**
 * @param Integer[] $nums
 * @return Integer
 */
function longestSubarray($nums) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo longestSubarray([1,1,0,1]) . "\n"; // 3
echo longestSubarray([0,1,1,1,0,1,1,0,1]) . "\n"; // 5
echo longestSubarray([1,1,1]) . "\n"; // 2
?>

Explanation:

1. Initialization: We initialize left pointer and zeros counter to zero. The max_len variable will store the result.
2. Expanding Window: The right pointer traverses each element in the array. For each element, if it is zero, we increment the zeros counter.
3. Contracting Window: If the number of zeros exceeds one, we move the left pointer forward. If the element at left is zero, we decrement the zeros counter. This ensures the window always contains at most one zero.
4. Updating Result: For each valid window, the length of contiguous 1's after deleting one element is right - left. We update max_len if this value is greater than the current maximum.
5. Result: After processing all elements, max_len holds the length of the longest contiguous subarray of 1's after deleting one element.

This approach efficiently processes the array in linear time, making it optimal for large inputs. The sliding window ensures that we only traverse each element once, maintaining an O(n) time complexity. The space complexity is O(1) as we use only a few variables.

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