Find All Anagrams in a String

Problem

Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.

Examples

The brute-force approach

For each starting index i in s, extract the substring of length len(p) and check whether it's an anagram of p by sorting both and comparing.

result = []
sorted_p = sorted(p)

for i in range(len(s) - len(p) + 1):
    window = s[i : i + len(p)]        # ← extract substring
    if sorted(window) == sorted_p:    # ← sort and compare: O(k log k)
        result.append(i)

return result

Sorting a window of length k costs O(k log k), and you do it for every position in s, so the total is O(n × k log k). For a pattern of length 26, that's already slow, and every new starting index requires sorting the window from scratch. The insight is that sliding the window one step right only changes two characters: one falls off the left, one enters from the right. You don't need to re-sort everything.

Spotting the pattern

This is a Sliding Window problem. The key question to ask yourself:

How do I check every length-k window of s against p without re-examining all k characters at each position?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.