LeetCode Patterns: Sliding Window Technique

If you have ever caught yourself writing a double for loop to inspect every contiguous subarray, sliding window is probably the optimisation you are missing. It turns an $O(nk)$ or $O(n^2)$ scan into a single linear pass by reusing the work it has already done. This article walks through the technique from first principles, then drills four canonical LeetCode problems plus a monotonic-deque variant.

1. The Idea in One Picture

A sliding window is a contiguous range [left, right] over an array or string. Instead of recomputing everything when the range moves, we add the element entering on the right and remove the element leaving on the left. Each element is touched at most twice, so the total cost is $O(n)$.

There are two flavours:

• Fixed-size window — width is a constant k and both pointers advance together.
• Variable-size window — right expands to grow the window, left contracts to repair it when an invariant breaks.

A picture is worth a thousand words. Below, the green cells are the live window and the right column shows the incremental sum update.

The orange cell is entering the window; the pink one is leaving. Notice the right column: each step is one subtraction and one addition, never a fresh sum() call. That is the entire trick.

2. Fixed-Size Window — Maximum Sum Subarray of Size K

Problem. Given an array arr and an integer k, find the maximum sum of any contiguous subarray of length k.

Naive approach. Sum every length-k slice — $O(nk)$.

Sliding window. Compute the first window’s sum, then slide one position at a time, subtracting the element that leaves and adding the element that enters.

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def max_sum_subarray(arr, k):
    """Return the maximum sum of any length-k contiguous subarray."""
    if len(arr) < k:
        return None

    window_sum = sum(arr[:k])     # first window
    best = window_sum

    for right in range(k, len(arr)):
        # entering: arr[right]    leaving: arr[right - k]
        window_sum += arr[right] - arr[right - k]
        best = max(best, window_sum)

    return best

Java

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public int maxSumSubarray(int[] arr, int k) {
    if (arr.length < k) return Integer.MIN_VALUE;
    int windowSum = 0;
    for (int i = 0; i < k; i++) windowSum += arr[i];
    int best = windowSum;
    for (int right = k; right < arr.length; right++) {
        windowSum += arr[right] - arr[right - k];
        best = Math.max(best, windowSum);
    }
    return best;
}

Complexity. Time $O(n)$ — one addition and one subtraction per slide. Space $O(1)$.

Why it works. The window’s sum is a function of its multiset of elements. When the window shifts by one, the multiset changes by exactly one removal and one insertion, so we only need to apply that delta. This incremental update generalises to running counts, frequency maps, and even monotonic deques (Section 6).

3. Variable-Size Window — Two Postures

When the window size is not given, we use two pointers and let the problem dictate the policy:

• Find the longest valid window: keep expanding right; whenever the window goes invalid, contract left until it is valid again, then record the length.
• Find the shortest valid window: keep expanding right until the window becomes valid; then contract left as far as possible while staying valid, recording the length on each shrink.

The picture below traces the longest-substring-without-repeating-characters algorithm on "abcabcbb". Green = valid (expanding), pink = duplicate detected, purple = contracting to repair.

And here is what the window length looks like over time. Green bands are expansion phases, purple bands are contraction phases.

The crucial invariant: left only ever moves right. So even though there is a while inside a for, the inner loop’s total work across the whole run is bounded by n. That is the amortised-$O(n)$ argument.

4. LeetCode 3 — Longest Substring Without Repeating Characters

Problem. Given a string s, return the length of the longest substring with all distinct characters.

Pattern. Variable window, find longest. The invariant is “all characters in the window are unique”. When s[right] is already in the window, slide left past its previous occurrence.

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def length_of_longest_substring(s):
    """LC 3 — longest substring with distinct characters."""
    last_seen = {}        # char -> most recent index
    left = 0
    best = 0

    for right, ch in enumerate(s):
        if ch in last_seen and last_seen[ch] >= left:
            # jump left past the previous occurrence in O(1)
            left = last_seen[ch] + 1
        last_seen[ch] = right
        best = max(best, right - left + 1)

    return best

Java

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public int lengthOfLongestSubstring(String s) {
    Map<Character, Integer> lastSeen = new HashMap<>();
    int left = 0, best = 0;
    for (int right = 0; right < s.length(); right++) {
        char c = s.charAt(right);
        if (lastSeen.containsKey(c) && lastSeen.get(c) >= left) {
            left = lastSeen.get(c) + 1;
        }
        lastSeen.put(c, right);
        best = Math.max(best, right - left + 1);
    }
    return best;
}

Complexity. Time $O(n)$ — each character is inserted into and removed from the map at most once. Space $O(\min(n, |\Sigma|))$ where $|\Sigma|$ is the alphabet size.

Two implementation styles. The version above uses jump-left — left leaps directly past the previous occurrence. The textbook alternative is step-left: hold a frequency map and while count[s[right]] > 1: drop(s[left]); left += 1. Both are $O(n)$; jump-left is one line shorter, step-left generalises better to “at most k” variants.

5. LeetCode 76 — Minimum Window Substring

Problem. Given s and t, return the shortest substring of s that contains every character of t (with multiplicities). Return "" if no such window exists.

Pattern. Variable window, find shortest. The invariant is “the window contains all required characters at the required counts”. We expand until the invariant holds, then aggressively contract.

The cleanest way to track validity is a valid counter — the number of distinct required characters whose count in the window has reached the required threshold. The window is valid iff valid == len(need). This avoids comparing two hash maps on every step.

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from collections import Counter

def min_window(s, t):
    """LC 76 — minimum window substring containing all chars of t."""
    if not s or not t or len(s) < len(t):
        return ""

    need = Counter(t)
    have = {}
    valid = 0                     # distinct chars whose count meets need
    left = 0
    best_len = float("inf")
    best_start = 0

    for right, ch in enumerate(s):
        if ch in need:
            have[ch] = have.get(ch, 0) + 1
            if have[ch] == need[ch]:
                valid += 1

        # shrink while still valid
        while valid == len(need):
            if right - left + 1 < best_len:
                best_len = right - left + 1
                best_start = left

            drop = s[left]
            left += 1
            if drop in need:
                if have[drop] == need[drop]:
                    valid -= 1            # about to fall below threshold
                have[drop] -= 1

    return "" if best_len == float("inf") else s[best_start:best_start + best_len]

Java

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public String minWindow(String s, String t) {
    if (s.length() < t.length()) return "";
    Map<Character, Integer> need = new HashMap<>();
    for (char c : t.toCharArray()) need.merge(c, 1, Integer::sum);

    Map<Character, Integer> have = new HashMap<>();
    int valid = 0, left = 0;
    int bestLen = Integer.MAX_VALUE, bestStart = 0;

    for (int right = 0; right < s.length(); right++) {
        char c = s.charAt(right);
        if (need.containsKey(c)) {
            have.merge(c, 1, Integer::sum);
            if (have.get(c).intValue() == need.get(c).intValue()) valid++;
        }
        while (valid == need.size()) {
            if (right - left + 1 < bestLen) {
                bestLen = right - left + 1;
                bestStart = left;
            }
            char d = s.charAt(left++);
            if (need.containsKey(d)) {
                if (have.get(d).intValue() == need.get(d).intValue()) valid--;
                have.merge(d, -1, Integer::sum);
            }
        }
    }
    return bestLen == Integer.MAX_VALUE ? "" : s.substring(bestStart, bestStart + bestLen);
}

Complexity. Time $O(|s| + |t|)$ — both pointers advance monotonically. Space $O(|\Sigma|)$.

The subtle line is if have[drop] == need[drop]: valid -= 1 — we decrement valid before the have count actually falls below the threshold, because we’re about to make it fall. Get the order wrong and the counter desyncs.

6. LeetCode 567 — Permutation in String

Problem. Return True iff some substring of s2 is a permutation of s1.

Pattern. Fixed-size window of width len(s1). Two strings are permutations of each other iff their character-frequency vectors are equal. So we maintain a length-26 frequency vector for the current window in s2 and compare to the frequency vector of s1.

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def check_inclusion(s1, s2):
    """LC 567 — does s2 contain a permutation of s1?"""
    n, m = len(s1), len(s2)
    if n > m:
        return False

    need = [0] * 26
    have = [0] * 26
    for c in s1:
        need[ord(c) - ord('a')] += 1
    for c in s2[:n]:
        have[ord(c) - ord('a')] += 1

    # how many of the 26 buckets currently match
    matches = sum(1 for i in range(26) if need[i] == have[i])

    for right in range(n, m):
        if matches == 26:
            return True

        r = ord(s2[right]) - ord('a')
        l = ord(s2[right - n]) - ord('a')

        # element entering on the right
        have[r] += 1
        if have[r] == need[r]:
            matches += 1
        elif have[r] == need[r] + 1:
            matches -= 1

        # element leaving on the left
        have[l] -= 1
        if have[l] == need[l]:
            matches += 1
        elif have[l] == need[l] - 1:
            matches -= 1

    return matches == 26

Java

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public boolean checkInclusion(String s1, String s2) {
    int n = s1.length(), m = s2.length();
    if (n > m) return false;
    int[] need = new int[26], have = new int[26];
    for (int i = 0; i < n; i++) {
        need[s1.charAt(i) - 'a']++;
        have[s2.charAt(i) - 'a']++;
    }
    int matches = 0;
    for (int i = 0; i < 26; i++) if (need[i] == have[i]) matches++;
    for (int right = n; right < m; right++) {
        if (matches == 26) return true;
        int r = s2.charAt(right) - 'a';
        int l = s2.charAt(right - n) - 'a';
        have[r]++;
        if (have[r] == need[r]) matches++;
        else if (have[r] == need[r] + 1) matches--;
        have[l]--;
        if (have[l] == need[l]) matches++;
        else if (have[l] == need[l] - 1) matches--;
    }
    return matches == 26;
}

Complexity. Time $O(m)$ — every slide is a constant amount of work because we maintain matches incrementally instead of rescanning all 26 buckets. Space $O(1)$.

The same skeleton solves LC 438 (Find All Anagrams) verbatim — just collect the start indices instead of returning early.

7. Variant: Sliding Window Maximum (Monotonic Deque)

When the window asks for the min or max of its elements, the simple “add / remove” trick is not enough — removing the current max means we need to know the next-best element instantly. The standard tool is a monotonic deque that stores values (or indices) in strictly decreasing order. The front of the deque is always the current window max.

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from collections import deque

def max_sliding_window(nums, k):
    """LC 239 — max in each window of size k."""
    dq = deque()      # holds indices; values at these indices are decreasing
    out = []

    for i, x in enumerate(nums):
        # 1. evict indices whose value is dominated by x
        while dq and nums[dq[-1]] < x:
            dq.pop()
        dq.append(i)

        # 2. evict the front if it has slid out of the window
        if dq[0] <= i - k:
            dq.popleft()

        # 3. once the window is full, the front is the max
        if i >= k - 1:
            out.append(nums[dq[0]])

    return out

Each index is pushed and popped at most once, so total work is $O(n)$ — the same linear bound as the basic patterns.

8. When to Reach for Sliding Window

The decision tree below summarises which flavour fits which problem shape:

Reach for sliding window when all of the following are true:

1. The answer is over a contiguous range (subarray / substring).
2. The window’s value can be maintained incrementally as the boundaries move (running sum, frequency map, monotonic deque, …).
3. The objective rewards either a specific size (fixed k) or a boundary condition (longest valid / shortest valid).

If any of those fails — non-contiguous picks, requiring a global view, or needing to undo arbitrary insertions — try prefix sums, two pointers on sorted data, dynamic programming, or a heap instead.

9. Templates Worth Memorising

Fixed-size window

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def fixed_window(arr, k):
    state = init_state(arr[:k])           # e.g. sum, frequency map
    best = report(state)
    for right in range(k, len(arr)):
        state = add(state, arr[right])
        state = remove(state, arr[right - k])
        best = better(best, report(state))
    return best

Variable window — longest valid

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def longest_valid(arr):
    left = 0
    state = empty_state()
    best = 0
    for right, x in enumerate(arr):
        state = add(state, x)
        while not is_valid(state):
            state = remove(state, arr[left])
            left += 1
        best = max(best, right - left + 1)
    return best

Variable window — shortest valid

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def shortest_valid(arr):
    left = 0
    state = empty_state()
    best = float("inf")
    for right, x in enumerate(arr):
        state = add(state, x)
        while is_valid(state):
            best = min(best, right - left + 1)
            state = remove(state, arr[left])
            left += 1
    return best if best != float("inf") else 0

10. Common Pitfalls

• Off-by-one in length. Window length is right - left + 1, not right - left.
• Updating valid after the count change. Increment valid exactly when have[c] first equals need[c]; decrement before it falls below. The order matters.
• Using list.pop(0) for fixed windows. That is $O(k)$. Use collections.deque (or simple index arithmetic) for $O(1)$.
• Recomputing the window from scratch. The whole point is incremental update — if your inner loop scans the window, you have lost the asymptotic improvement.
• Negative numbers in “sum ≥ target”. Contracting may not reduce the sum, so the “shortest valid” template doesn’t apply directly. Reach for prefix sums + monotonic deque, or a different framing.

11. Practice Set

12. Summary

Sliding window is fundamentally an amortisation argument: instead of recomputing each window’s value from scratch, you maintain a tiny piece of state and update it as the window moves by one position. Two pointers, both monotonically increasing, give you an $O(n)$ algorithm where the brute force is $O(n^2)$.

The mental checklist when you see a contiguous-range problem:

1. Is the window size fixed or variable?
2. If variable, am I after the longest or the shortest valid window?
3. What state do I maintain — a sum, a frequency map, a monotonic deque?
4. What is the invariant that determines validity, and which pointer’s move can break it?

Internalise those four questions and most sliding window problems collapse to filling in a template.