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1004 - Max Consecutive Ones III (Medium)

https://leetcode.com/problems/max-consecutive-ones-iii/

Problem Statement​

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

Example 1:

Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
Output: 6
Explanation: [1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Example 2:

Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
Output: 10
Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Constraints:

  • 1 <= nums.length <= 10^5
  • nums[i] is either 0 or 1.
  • 0 <= k <= nums.length

Approach 1: Sliding Window​

We are looking for the longest subarray with kk zeros. We can use standard sliding windows to solve it.

Written by @wingkwong
class Solution {
public:
    int longestOnes(vector<int>& nums, int k) {
        // pointer i = window starting point
        // pointer j = window ending point
        int n = nums.size(), ans = 0, i = 0, j = 0;
        while (j < n) {
            // if it is 0, then decrease k by 1
            if (nums[j] == 0) k -= 1;
            // k < 0 means the window includes k zeros
            if (k < 0) {
                // if the starting point of the window is 0,
                // then add 1 to k to reduce the window size by 1 
                if (nums[i] == 0) k++;
                // move pointer i
                i++;
            }
            // move pointer j
            j++;
        }
        return j - i;
    }
};c