1004. Max Consecutive Ones III LeetCode Solution

In this guide, you will get 1004. Max Consecutive Ones III LeetCode Solution with the best time and space complexity. The solution to Max Consecutive Ones III problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Max Consecutive Ones III solution in C++
Max Consecutive Ones III solution in Java
Max Consecutive Ones III solution in Python
Additional Resources

1004. Max Consecutive Ones III LeetCode Solution image

Problem Statement of Max Consecutive Ones III

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 <= 105
nums[i] is either 0 or 1.
0 <= k <= nums.length

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(1)

1004. Max Consecutive Ones III LeetCode Solution in C++

class Solution {
 public:
  int longestOnes(vector<int>& nums, int k) {
    int ans = 0;

    for (int l = 0, r = 0; r < nums.size(); ++r) {
      if (nums[r] == 0)
        --k;
      while (k < 0)
        if (nums[l++] == 0)
          ++k;
      ans = max(ans, r - l + 1);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

1004. Max Consecutive Ones III LeetCode Solution in Java

class Solution {
  public int longestOnes(int[] nums, int k) {
    int ans = 0;

    for (int l = 0, r = 0; r < nums.length; ++r) {
      if (nums[r] == 0)
        --k;
      while (k < 0)
        if (nums[l++] == 0)
          ++k;
      ans = Math.max(ans, r - l + 1);
    }

    return ans;
  }
}
// code provided by PROGIEZ

1004. Max Consecutive Ones III LeetCode Solution in Python

class Solution:
  def longestOnes(self, nums: list[int], k: int) -> int:
    ans = 0

    l = 0
    for r, num in enumerate(nums):
      if num == 0:
        k -= 1
      while k < 0:
        if nums[l] == 0:
          k += 1
        l += 1
      ans = max(ans, r - l + 1)

    return ans
 # code by PROGIEZ