1000 - 1099
1004 - Max Consecutive Ones III (Medium)
Problem Link
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 <= 10^5nums[i]is either0or1.0 <= k <= nums.length
Approach 1: Sliding Window
We are looking for the longest subarray with zeros. We can use standard sliding windows to solve it.
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