A sliding window is a way to maintain information about a contiguous range as its boundaries move—not a template that fits every subarray problem. Before coding, define the range, state what your data structure represents, and explain why each pointer movement preserves correctness. That discipline separates a valid linear-time window from a two-pointer loop that can silently miss answers.
Start by defining the window and its invariant
For an inclusive window [left, right], the elements currently under consideration are exactly those from index left through right. The maintained state—perhaps a sum, character frequencies, or candidate extrema—must describe exactly those elements. State the validity condition too: for example, “the window contains no repeated characters” or “the window has at most K distinct values.”
An invariant is the fact that remains true at a particular point in the algorithm. A useful formulation is: “After each update, the stored state describes exactly the current range; after shrinking, the selected validity condition is restored.” Be precise about when this is true. Some algorithms allow a temporarily invalid window while the right boundary advances, then restore validity before measuring or returning a candidate.
- Range: Are endpoints inclusive? Is the window required to have a particular size?
- State: What exactly do the sum, counts, deque, or map record?
- Validity: Which condition makes the current range eligible?
- Movement: What justifies moving the right or left boundary?
This framing follows the pointer-and-state approach described in the LeetCode community tutorial on sliding-window patterns.
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Choose fixed-size or variable-size windows
The first recognition question is whether the range length is prescribed or must change to meet a condition. Fixed-size windows slide by a known amount; variable-size windows expand and shrink according to validity.
| Pattern | Invariant or state | Recognition cue | Correctness check |
|---|---|---|---|
| Fixed-size | The current range has exactly k elements; its summary reflects those elements. |
Every subarray or substring of length k, or one result per such range. |
Emit the first result only after collecting k elements; on each slide, remove the departing contribution and add the entering one. |
| Variable-size, longest valid range | The range satisfies the constraint after shrinking. | Longest or maximum length under an at-most condition. | Show why adding at the right can violate validity, why removing at the left can restore it, and record length only when valid. |
| Variable-size, shortest covering range | The range contains the required values or frequencies. | Minimum range that covers a target. | Define coverage, including required multiplicities; record a valid candidate before shrinking makes it invalid. |
The official LeetCode Sliding Window Maximum problem defines a window of size k that moves from left to right. Its example, nums = [1,3,-1,-3,5,3,6,7] and k = 3, produces [3,3,5,5,6,7]. Each output is the maximum of one contiguous three-element range. That is the fixed-size model: the length never changes, and the next range starts one index later.
Maintain state so updates stay cheap and visible
Fixed-size sums
For a sum over a fixed-size window, calculate the initial sum once. When the window moves right, add the entering value and subtract the departing value. The invariant is that the running sum equals the sum of exactly the current k elements. Recomputing each sum from scratch is unnecessary.
Frequency maps and distinct counts
For an anagram check, duplicate-free substring, or at-most-K-distinct constraint, keep a frequency map for characters or values currently inside the window. On insertion, increment the entering item’s count; on removal, decrement the departing item’s count and delete it when that count reaches zero. If you maintain a distinct-item total, increment it when a count changes from zero to one and decrement it when a count changes from one to zero.
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Keep the meaning of counters explicit. “Number of distinct keys” is not the same as “number of matches,” and a coverage test with repeated required characters must account for their multiplicities. The community tutorial describes frequency-map and at-most/exactly-K patterns; the LeetCode study guide also categorizes fixed- and variable-size approaches.
Know why each boundary moves
In the usual variable-window pattern, the right boundary advances to include new data. If the expanded range is invalid, move the left boundary forward, removing each departed item from the maintained state, until validity is restored. For a longest-valid-range problem, update the best length only after restoration. For a shortest-covering-range problem, record valid candidates while coverage still holds, then shrink to look for a shorter one.
The key correctness requirement is monotonic behavior that matches those movements. For example, with nonnegative values, extending a range cannot reduce its sum; if a sum limit is exceeded, removing values from the left can reduce it. Likewise, adding characters can introduce a duplicate, and removing from the left can eventually eliminate it. This gives a reason to advance the left edge rather than trying every possible start.
A proof sketch should explain why the movement does not skip an optimum. In a longest-valid-range problem, if a left boundary has been discarded because the range ending at the current right boundary is invalid, later extensions do not make that discarded start valid when the constraint’s violations persist under extension. Therefore, it need not be reconsidered. That reasoning is specific to the condition and its monotonic behavior; it is not automatic for every problem described as “subarray” or “substring.”
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Example: longest substring without repeated characters
- Extend
rightby one character and increment its frequency. - If that character makes the range contain a duplicate, advance
leftand decrement frequencies until no duplicate remains. - With validity restored, compare
right - left + 1with the best length so far.
The current frequency map must always describe the inclusive range from left through right. Shrinking continues until the duplicate is removed—not for an arbitrary number of steps. This expand-and-repair pattern is described in the LeetCode community tutorial.
Use a deque when the window needs extrema
A sum or distinct count does not tell you the maximum or minimum in a range. If a problem repeatedly asks for a window’s maximum, maintain candidate indices in a decreasing-value deque: the index at the front points to the largest current candidate. Remove indices from the front once they fall left of the window. When a new value arrives, remove smaller-or-equal candidates from the back because the new value is at least as useful and will remain in the range longer.
The deque contains indices, not just values, so the algorithm can identify expired entries and compare their positions with the current boundaries. After expiration and domination removals, its front is the current maximum. For a minimum, reverse the value ordering.
Each index is appended once and removed at most once, either when it expires from the front or is dominated from the back. Thus the deque operations take amortized O(n) time over an input of length n; the cited Doocs LeetCode Wiki solution gives O(n) time and O(k) space for Sliding Window Maximum.
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For a variable-size constraint involving both maximum and minimum—such as requiring their difference to stay under a limit—maintain both a decreasing deque for maxima and an increasing deque for minima. The fronts provide the two extrema needed to test validity, while expired indices must be removed from each deque. The community tutorial describes this two-queue approach.
Independent reader supportYour contribution helps us test, update, and keep practical guides available for everyone.Recognize when an ordinary window is not justified
A standard window works when the problem’s validity condition supports the chosen boundary movements. It is not enough that the input asks about a contiguous range. If extending the right edge can move a sum either up or down, “shrink while the sum is too large” may discard a start that would later lead to an answer.
Counterexample: Subarray Sum Equals K with negative values
With negative numbers, extending a range can increase or decrease its sum. The sum therefore does not provide a predictable monotone boundary for a conventional grow-and-shrink rule. For this exact-target counting problem, use prefix sums and a hash map: as you scan, a current prefix sum p forms a subarray summing to k with every earlier prefix equal to p - k. Store counts of earlier prefix sums so each matching start contributes to the answer. The community tutorial recommends this approach for Subarray Sum Equals K in this setting.
When more state is required
Even when a window is valid, the state must be sufficient to test the condition. A running sum cannot answer a maximum query; a distinct count cannot tell whether a required frequency is covered. Add the appropriate structure—frequency counts, monotonic queues, or another state representation—or use a different algorithm if the movement rule still cannot be proven.
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Explain correctness and complexity in an interview
For a two-pointer algorithm in which each pointer only moves forward, each element enters once and leaves at most once. If each state update is O(1), or is amortized O(1), the total pointer and update work is O(n). State this condition rather than claiming that every sliding-window solution is linear: expensive updates, unsuitable state, or invalid pointer movements change the analysis. Hash-map operations also have guarantees that depend on the language and implementation.
For a frequency-map window, explain what each key and count represents, when keys are added or removed, and exactly how the validity test changes. For a deque solution, explain why each candidate is inserted once and can be removed only once. In either case, name the auxiliary state: a frequency map can grow with the number of distinct values in the window; the maximum deque uses O(k) space for a fixed window of size k in the cited solution.
A concise interview explanation can follow this order:
- Identify the contiguous range and whether its size is fixed or variable.
- State the invariant and the data structure that maintains it.
- Explain what event advances each pointer and how state is updated.
- Justify that the validity rule makes those movements safe and cannot skip an optimum.
- Give time and auxiliary-space complexity for the specific implementation.
Sliding window is useful when you can maintain the needed state incrementally and justify a one-way movement rule. Without both, a familiar-looking loop is only a guess.
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