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487. Max Consecutive Ones II

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Description

Given a binary array nums, return the maximum number of consecutive 1's in the array if you can flip at most one 0.

 

Example 1:

Input: nums = [1,0,1,1,0]
Output: 4
Explanation: Flip the first zero will get the maximum number of consecutive 1s. After flipping, the maximum number of consecutive 1s is 4.

Example 2:

Input: nums = [1,0,1,1,0,1]
Output: 4

 

Constraints:

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

 

Follow up: What if the input numbers come in one by one as an infinite stream? In other words, you can't store all numbers coming from the stream as it's too large to hold in memory. Could you solve it efficiently?

Solutions

Python3

class Solution:
    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
        n = len(nums)
        prefix = [0] * n
        suffix = [0] * n
        res = 0
        for i in range(n):
            if i == 0:
                prefix[i] = nums[i]
            else:
                prefix[i] = 0 if nums[i] == 0 else prefix[i - 1] + 1
            res = max(res, prefix[i])

        for i in range(n - 1, -1, -1):
            if i == n - 1:
                suffix[i] = nums[i]
            else:
                suffix[i] = 0 if nums[i] == 0 else suffix[i + 1] + 1

        for i in range(n):
            if nums[i] == 0:
                t = 1
                if i > 0:
                    t += prefix[i - 1]
                if i < n - 1:
                    t += suffix[i + 1]
                res = max(res, t)
        return res

Java

• Two Pointers, time complexity: O(n²), memory complexity: O(1)

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int n = nums.length;
        int res = 0;
        for (int i = 0; i < n; ++i) {
            int cnt = 1;
            int j = i;
            while (j < n && (cnt > 0 || nums[j] == 1)) {
                if (nums[j] == 0) --cnt;
                ++j;
            }
            res = Math.max(res, j - i);
        }
        return res;
    }
}

• Prefix Array & Suffix Array, time complexity: O(n), memory complexity: O(n)

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int n = nums.length;

        int[] prefix = new int[n];
        int[] suffix = new int[n];

        int res = 0;
        for (int i = 0; i < n; ++i) {
            if (i == 0) prefix[0] = nums[0];
            else prefix[i] = nums[i] == 0 ? 0 : prefix[i - 1] + 1;
            res = Math.max(res, prefix[i]);
        }

        for (int i = n - 1; i >= 0; --i) {
            if (i == n - 1) suffix[n - 1] = nums[n - 1];
            else suffix[i] = nums[i] == 0 ? 0 : suffix[i + 1] + 1;
        }

        for (int i = 0; i < n; ++i) {
            if (nums[i] == 0) {
                int t = 1;
                if (i > 0) t += prefix[i - 1];
                if (i < n - 1) t += suffix[i + 1];
                res = Math.max(res, t);
            }
        }
        return res;
    }
}

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