487 Max Consecutive Ones II

Problem:

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

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:

1. The input array will only contain 0 and 1.
2. 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?

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int res = 0, zero = 0, left = 0, k = 1;
        for (int right = 0; right < nums.size(); ++right) {
            if (nums[right] == 0) ++zero;
            while (zero > k) {
                if (nums[left++] == 0) --zero;
            }
            res = max(res, right - left + 1);
        }
        return res;
    }
};

========= (can only be used for a at most 1 flip. Generic solution is above. Above can be used

for multiple flips)

 class Solution {

public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int res = 0, cur = 0, cnt = 0;
        for (int num : nums) {
            ++cnt;
            if (num == 0) {
                cur = cnt;
                cnt = 0;
            } 
            res = max(res, cnt + cur);
        }
        return res;
    }
};

=========

The above method cannot be used in the case of follow up, because nums[left] needs to access the previous number. We can save all the positions of 0 we encountered, so that we know where to move when we need to move left:
 
Solution three:
class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int res = 0, left = 0, k = 1;
        queue<int> q;
        for (int right = 0; right < nums.size(); ++right) {
            if (nums[right] == 0) q.push(right);
            if (q.size() > k) {
                left = q.front() + 1; q.pop();
            }
            res = max(res, right - left + 1);
        }
        return res;
    }
};