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Imagine your study streak :
If bad days exceed K:
You don’t restart from beginning
You just remove earlier days from your streak (move left)
This way:
You always maintain the best valid streak without repeating work

The brute force method wastes time by restarting again and again.
The sliding window approach fixes this by saying:
“Don’t restart, just adjust the current window.”
Instead of checking every subarray:
Think like this:
You have a window [left → right]
You are allowed at most K zeros inside this window
| left = 0 zero_count = 0 |
| If nums[right] == 0 → increase zero count |
| max_length = max(max_length, right - left + 1) |
| nums = [1,0,1,1,0,0,1] k = 2 |
| Step | Window | Zeros | Action |
| Start | [1] | 0 | Expand |
| → | [1,0] | 1 | Expand |
| → | [1,0,1,1] | 1 | Expand |
| → | [1,0,1,1,0] | 2 | Expand |
| → | [1,0,1,1,0,0] | 3 | ❌ Shrink |
| Adjust | Valid again | 2 | Continue |
Final Answer = 5
Code
| def longestOnes(nums, k): left = 0 zero_count = 0 max_length = 0 for right in range(len(nums)): if nums[right] == 0: zero_count += 1 while zero_count > k: if nums[left] == 0: zero_count -= 1 left += 1 max_length = max(max_length, right - left + 1) return max_length |
Each element is visited at most twice
✔ No repeated work
✔ Real-time adjustment
✔ Works efficiently for large inputs
✔ Most important pattern for exams/interviews
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