487-max-consecutive-ones-ii

Question

https://leetcode.com/problems/max-consecutive-ones-ii/description/

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

Example:

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?

Thought Process

1. Two Pointers
   1. We use one pointer to move though the elements. and another for the last zero
   2. When we encounter a 1, we just keep increasing the length
   3. When we encounter a 0, we deduct the last position of zero to get the length of sequence 1's ending with current 0
   4. Time complexity O(n)
   5. Space complexity O(1)
2. Two Pointers
   1. We can save the unnecessary use of math.max by doing it when num is 0
   2. Time complexity O(n)
   3. Space complexity O(1)

Solution

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int lastZero = -1, len = 0, cur = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 1) {
                cur++;
            } else {
                cur = i - lastZero;
                lastZero = i;
            }
            len = Math.max(cur, len);
        }
        return len;
    }
}

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int len = 0, pre = 0, cur = 0;
        boolean zero = false;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 1) {
                cur++;
            } else {
                zero = true;
                len = Math.max(len, pre + 1 + cur);
                pre = cur;
                cur = 0;
            }
        }
        return Math.max(len, zero? pre + 1 + cur : cur);
    }
}

Additional