# 487     Max Consecutive Ones II

## 487. [Max Consecutive Ones II](https://leetcode.com/problems/max-consecutive-ones-ii/description/)

## 1. Question

**Example 1:**

```
Input: [1,0,1,1,0]

Output: 4

Explanation:
Flip the first zero will get the the maximum number of consecutive 1s. After flipping, the maximum number of consecutive 1s is 4.
```

**Note:**

* The input array will only contain`0`and`1`.
* The length of input array is a positive integer and will not exceed 10,000

**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?

## 2. Implementation

**(1) Two Pointers**

```java
class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int k = 1;
        int zeroes = 0;
        int maxLen = 0, start = 0, end = 0;

        while (end < nums.length) {
            if (nums[end] == 0) {
                ++zeroes;
            }
            ++end;

            while (zeroes > k) {
                if (nums[start] == 0) {
                    --zeroes;
                }
                ++start;
            }
            maxLen = Math.max(maxLen, end - start);
        }
        return maxLen;
    }
}
```

**Follow Up**

```java
class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int k = 1;
        int maxLen = 0;
        Queue<Integer> zeroIndex = new LinkedList<>();

        for (int start = 0, end = 0; end < nums.length; end++) {
            if (nums[end] == 0) {
                zeroIndex.add(end);
            }

            if (zeroIndex.size() > k) {
                start = zeroIndex.remove() + 1;
            }
            maxLen = Math.max(maxLen, end - start + 1);
        }
        return maxLen;
    }
}
```

## 3. Time & Space Complexity

**Two Pointers:** 时间复杂度O(n), 空间复杂度O(1)

Follow up: 时间复杂度O(n), 空间复杂度O(k)


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