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| def longestOnes(nums, k): left = 0 zero_count = 0 max_length = 0 for right in range(len(nums)): # Step 1: Expand window if nums[right] == 0: zero_count += 1 # Step 2: Shrink window if invalid while zero_count > k: if nums[left] == 0: zero_count -= 1 left += 1 # Step 3: Update maximum length max_length = max(max_length, right - left + 1) return max_length |
| if nums[right] == 0: zero_count += 1 |
???? We count it because it may need flipping
| while zero_count > k: |
???? Too many zeros → not allowed
| if nums[left] == 0: zero_count -= 1 left += 1 |
???? Remove elements until window becomes valid again
| max_length = max(max_length, right - left + 1) |
???? Store the largest valid window
| nums = [1,0,1,1,0,0,1] k = 2 Expand window → count zeros When zeros > 2 → shrink Keep updating max length |
???? Final Output = 5
✔ Use while loop for shrinking
✔ Always maintain ≤ K zeros
✔ Update answer after fixing window
✔ No need to actually flip bits
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