Max Consecutive Ones III

Max Consecutive Ones III - Problem

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 zeros to ones.

You need to find the longest contiguous subarray that contains only 1's after flipping at most k zeros.

Example: If nums = [1,1,1,0,0,0,1,1,1,1,0] and k = 2, you can flip 2 zeros to get a maximum of 6 consecutive ones.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2

› Output: 6

💡 Note: We can flip the two zeros at indices 3 and 4 to get [1,1,1,1,1,0,1,1,1,1,0]. The longest sequence of 1's is from index 0 to 5, which has length 6.

Example 2 — All Ones

$ 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

💡 Note: We can flip 3 zeros to create the longest consecutive sequence. The optimal flips give us a sequence of length 10.

Example 3 — Edge Case

$ Input: nums = [1,1,1,1], k = 0

› Output: 4

💡 Note: No zeros to flip needed. The entire array is already all 1's, so the answer is 4.

Constraints

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

Visualization

Tap to expand

Asked in

G Google 45 a Amazon 38 M Microsoft 32 f Facebook 28

The sliding window technique is optimal for this problem. Use two pointers to maintain a window with at most k zeros, expanding right and contracting left as needed. Best approach is sliding window with Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Check All Subarrays

⏱️ Time: O(n²) Space: O(1)

For each starting position, expand the window and count zeros. If zeros exceed k, stop. Track the maximum valid window size.

Sliding Window (Two Pointers)

⏱️ Time: O(n) Space: O(1)

Maintain a sliding window using left and right pointers. Expand right pointer and count zeros. When zeros exceed k, shrink from left until valid again.

Brute Force - Check All Subarrays — Algorithm Steps

For each starting index
Expand window while counting zeros
Stop when zeros > k
Track maximum length

Visualization

Tap to expand

Step-by-Step Walkthrough

Start Position

Begin at each index

Expand Window

Count zeros while expanding

Stop & Track

Stop when zeros > k, track max length

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize, int k) {
    int maxLength = 0;
    
    for (int i = 0; i < numsSize; i++) {
        int zeros = 0;
        for (int j = i; j < numsSize; j++) {
            if (nums[j] == 0) zeros++;
            if (zeros <= k) {
                int length = j - i + 1;
                if (length > maxLength) maxLength = length;
            } else {
                break;
            }
        }
    }
    
    return maxLength;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[10000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    
    printf("%d\n", solution(nums, numsSize, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops - outer loop n times, inner loop up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for counters and max length

✓ Linear Space

87.6K Views

High Frequency

~15 min Avg. Time

2.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler