The idea in one pictureCore idea
Instead of recomputing each window from scratch, you update it: add the element that enters on the right, remove the one that leaves on the left. Each element enters and leaves once, so the whole scan is O(n).
Window of size 3 sliding right
Fixed-size windowPattern
The window size k is given. Slide a window of size k across the array and compute a result at each position.
function maxSumSubarray(nums: number[], k: number): number {
let windowSum = 0;
// First window
for (let i = 0; i < k; i++) windowSum += nums[i];
let maxSum = windowSum;
// Slide window
for (let i = k; i < nums.length; i++) {
windowSum += nums[i] - nums[i - k];
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}Variable-size windowPattern
Here the size changes. Expand the right edge until the window breaks the rule, then shrink the left edge until the rule holds again.
Longest substring without repeats: "abcabcbb"
function lengthOfLongestSubstring(s: string): number {
const seen = new Set<string>();
let left = 0, maxLen = 0;
for (let right = 0; right < s.length; right++) {
while (seen.has(s[right])) {
seen.delete(s[left]);
left++;
}
seen.add(s[right]);
maxLen = Math.max(maxLen, right - left + 1);
}
return maxLen;
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graph LR
A["expand: right++<br/>add s[right]"] --> B{"window<br/>still valid?"}
B -->|yes| C["record answer"]
B -->|no| D["shrink: remove s[left]<br/>left++"]
D --> B
C --> A
class A blue
class B yellow
class C green
class D red
classDef orange fill:#ffd8a8,stroke:#1e1e1e,color:#1e1e1e
classDef blue fill:#a5d8ff,stroke:#1e1e1e,color:#1e1e1e
classDef green fill:#b2f2bb,stroke:#1e1e1e,color:#1e1e1e
classDef red fill:#ffc9c9,stroke:#1e1e1e,color:#1e1e1e
classDef violet fill:#d0bfff,stroke:#1e1e1e,color:#1e1e1e
classDef yellow fill:#ffec99,stroke:#1e1e1e,color:#1e1e1e
classDef grey fill:#e9ecef,stroke:#1e1e1e,color:#1e1e1eWhen to use a sliding windowWhen to use
Common patterns
| Problem | Window | What you track |
|---|---|---|
| Max / min sum of a subarray of size k | Fixed | A running sum |
| Longest substring with at most k distinct characters | Variable | Character counts; shrink when distinct > k |
| Minimum window substring | Variable | Counts of needed characters; shrink to minimise |
| Permutation in string | Fixed (size of s1) | Character frequencies to compare |
Common mistakesPitfalls
In the interviewIn the interview
Interviewers often push back on the nested loop. That sentence is the answer they are waiting for.
Frequently asked questionsFAQ
When should you use a sliding window?
Use it on contiguous subarray or substring problems where the constraint involves a sum, length, or distinct-count bound. The moving-pointer pattern removes repeated left-side work, converting O(n^2) brute force into a single O(n) pass with O(1) or O(k) extra space.
What is the difference between a fixed and a variable window?
A fixed window keeps a constant length and shifts one element at a time, while a variable window grows and shrinks both ends to satisfy a constraint like sum below a target. Both share the same right-pointer expansion and left-pointer shrink mechanics.
How do you track character counts in a sliding window string problem?
Use a frequency map or an integer array of size 26 or 128 for characters. Add the incoming right character and remove the outgoing left character on each move, then compare the counts to the constraint. This keeps the whole scan O(n) with O(1) space for bounded alphabets.
Now explain sliding window to me out loud
I'll give you interview problems and ask follow-ups, like a real interviewer.