Maximize Number of 1's Visualization & Animation

Finds the maximum consecutive 1s achievable by flipping at most k zeros using a variable sliding window; O(n).

## What is it? Given a binary array and an integer `k`, find the maximum number of consecutive 1s you can obtain by flipping at most `k` zeros. Sliding window technique. ## How it works - Use `left` and `right` pointers; count zeros in the current window - Expand `right`: if `arr[right] == 0`, increment `zeroCount` - If `zeroCount > k`: shrink by moving `left` forward until `zeroCount <= k` - Track `maxLen = max(maxLen, right - left + 1)` at each step - Return `maxLen` ## When to use - "Allow at most k changes" problems on binary arrays - Maximum window with a limited number of "bad" elements - LeetCode 1004 — Max Consecutive Ones III ## Key Points - O(n) time, O(1) space - Variant of the variable-size sliding window template - When `arr[left] == 0` and we move left past it, decrement `zeroCount`

Category: algorithms

Difficulty: intermediate

Time Complexity: O(n)

Space Complexity: O(1)

Maximize Number of 1's

intermediate

Finds the maximum consecutive 1s achievable by flipping at most k zeros using a variable sliding window; O(n).

GFGLeetCode
zeros 0/2
len 0
max 0
New Max!
1
[0]
0
[1]
1
[2]
1
[3]
0
[4]
1
[5]
0
[6]
1
[7]
1
[8]
1 in window
0 in window (will flip)
Result window
Outside window
Init: left=0, right=-1, zeros=0. Expand right and allow at most 2 zeros in the window.