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

Solution 1: Enumeration

We can enumerate each position $i$ to be deleted, then calculate the number of consecutive 1s on the left and right, and finally take the maximum value.

Specifically, we use two arrays $left$ and $right$ of length $n+1$, where $left[i]$ represents the number of consecutive 1s ending with $nums[i-1]$, and $right[i]$ represents the number of consecutive 1s starting with $nums[i]$.

The final answer is $\max_{0 \leq i < n} {left[i] + right[i+1]}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

Python3JavaC++GoTypeScript

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

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

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

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

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;
}

Solution 2: Two Pointers

The problem is actually asking us to find the longest subarray that contains at most one $0$. The remaining length after deleting one element from this subarray is the answer.

Therefore, we can use two pointers $j$ and $i$ to point to the left and right boundaries of the subarray, initially $j = 0$, $i = 0$. In addition, we use a variable $cnt$ to record the number of $0$s in the subarray.

Next, we move the right pointer $i$. If $nums[i] = 0$, then $cnt$ is incremented by $1$. When $cnt > 1$, we need to move the left pointer $j$ until $cnt \leq 1$. Then, we update the answer, i.e., $ans = \max(ans, i - j)$. Continue to move the right pointer $i$ until $i$ reaches the end of the array.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.