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

Description

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.

Solutions

  • class Solution {
    public int longestSubarray(int[] nums) {
        int n = nums.length;
        int[] left = new int[n];
        int[] right = new int[n];
        for (int i = 1; i < n; ++i) {
            if (nums[i - 1] == 1) {
                left[i] = left[i - 1] + 1;
            }
        }
        for (int i = n - 2; i >= 0; --i) {
            if (nums[i + 1] == 1) {
                right[i] = right[i + 1] + 1;
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans = Math.max(ans, left[i] + right[i]);
        }
        return ans;
    }
}

  • class Solution {
public:
    int longestSubarray(vector<int>& nums) {
        int n = nums.size();
        vector<int> left(n);
        vector<int> right(n);
        for (int i = 1; i < n; ++i) {
            if (nums[i - 1] == 1) {
                left[i] = left[i - 1] + 1;
            }
        }
        for (int i = n - 2; ~i; --i) {
            if (nums[i + 1] == 1) {
                right[i] = right[i + 1] + 1;
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans = max(ans, left[i] + right[i]);
        }
        return ans;
    }
};

  • class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        n = len(nums)
        left = [0] * n
        right = [0] * n
        for i in range(1, n):
            if nums[i - 1] == 1:
                left[i] = left[i - 1] + 1
        for i in range(n - 2, -1, -1):
            if nums[i + 1] == 1:
                right[i] = right[i + 1] + 1
        return max(a + b for a, b in zip(left, right))


  • func longestSubarray(nums []int) int {
	n := len(nums)
	left := make([]int, n)
	right := make([]int, n)
	for i := 1; i < n; i++ {
		if nums[i-1] == 1 {
			left[i] = left[i-1] + 1
		}
	}
	for i := n - 2; i >= 0; i-- {
		if nums[i+1] == 1 {
			right[i] = right[i+1] + 1
		}
	}
	ans := 0
	for i := 0; i < n; i++ {
		ans = max(ans, left[i]+right[i])
	}
	return ans
}

  • function longestSubarray(nums: number[]): number {
    const n = nums.length;
    const left: number[] = Array(n + 1).fill(0);
    const right: number[] = Array(n + 1).fill(0);
    for (let i = 1; i <= n; ++i) {
        if (nums[i - 1]) {
            left[i] = left[i - 1] + 1;
        }
    }
    for (let i = n - 1; ~i; --i) {
        if (nums[i]) {
            right[i] = right[i + 1] + 1;
        }
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans = Math.max(ans, left[i] + right[i + 1]);
    }
    return ans;
}

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