What is the brute force approach and complexity for Permutation in String?

For each length-m window, build a fresh count array and compare with time complexity O(N · M) and space complexity O(1).

What is the optimal approach and complexity for Permutation in String?

Track how many of the 26 letters already have the right count with optimal time complexity O(N + M) and space complexity O(1).

How do you identify the pattern for Permutation in String?

Look for key signals in the prompt: "permutation of s1" indicates Order doesn't matter, only letter counts → compare count arrays.; "substring of s2" indicates Contiguous range of s2 → window.; "(implied) same length as s1" indicates A permutation keeps the length, so the window size is fixed at s1.length..

Permutation in String

Medium·LeetCode #567 Pro OnlyMedium

LeetCode

Problem Statement

You are given two strings s1 and s2.

Return true if some rearrangement of s1 (a permutation) appears inside s2 as a contiguous substring. Otherwise return false.

In other words: is there a substring of s2 that uses exactly the same letters as s1, the same number of times each, in any order?

Input

Two strings s1 and s2 of lowercase English letters.

Output

true if a permutation of s1 is a substring of s2, otherwise false.

Example 1

Input: s1 = "ab", s2 = "eidbaooo"

Output: true

s2 contains "ba" (positions 3–4), which is a rearrangement of "ab".

Example 2

Input: s1 = "ab", s2 = "eidboaoo"

Output: false

There is a 'b' and an 'a' in s2, but they are never next to each other, so no length-2 substring has exactly one 'a' and one 'b'.

Constraints

• 1 ≤ s1.length, s2.length ≤ 10^4
• s1 and s2 contain only lowercase English letters

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“permutation of s1”

→ Order doesn't matter, only letter counts → compare count arrays.

“substring of s2”

→ Contiguous range of s2 → window.

“(implied) same length as s1”

→ A permutation keeps the length, so the window size is fixed at s1.length.

Approach Complexity Ladder

Brute Force — count every window from scratchO(N · M)O(1)

For each length-m window, build a fresh count array and compare

Better — slide the counts, compare 26 each stepO(26 · N)O(1)

Update two counts per slide, then compare the two 26-length arrays

Optimal — fixed window + matches counterO(N + M)O(1)

Track how many of the 26 letters already have the right count

Brute Force Intuition

A permutation of s1 must have length m = s1.length. So check every substring of s2 of length m: count its letters from scratch and compare with the counts of s1.

Recognition rule for this sub-pattern:

Window size is fixed at K. Each slide adds one element on the right and removes one on the left — update the aggregate in O(1).

Remember this · Sliding Window

Trigger

Contiguous subarray / substring + longest, shortest, maximum, count, or "of size K".

Think

Can I keep a valid range and move it, instead of checking every subarray from scratch?

Key question

When is my current window valid?

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