You’re in a coding interview. The interviewer says: “Find all start indices of anagrams of p in s.” This is a fixed-size sliding window with character frequency comparison. As the window slides, you add the incoming character and remove the outgoing one, then check if counts match.

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The Problem

Given two strings s and p, return a list of all start indices of p’s anagrams in s.

Example

Input: s = "cbaebabacd", p = "abc" Output: [0, 6]

Already comfortable with the solution? Practice it in a mock interview →

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What Interviewers Are Actually Testing

Skill	What Interviewers Observe
Fixed-size sliding window	Do you slide a window of size len(p) across s?
Efficient comparison	Do you use a matches counter for O(1) per step?
Add-and-remove	Do you increment/decrement counts correctly?

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Step 1: The Key Insight

A fixed-size sliding window of length len(p). Track a matches counter — the number of characters (out of 26) whose counts currently agree between the window and p. When matches == 26, the window is an anagram.

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Python Implementation

def findAnagrams(s: str, p: str) -> list[int]:
    if len(p) > len(s):
        return []

    p_count = [0] * 26
    s_count = [0] * 26

    for c in p:
        p_count[ord(c) - ord('a')] += 1
    for c in s[:len(p)]:
        s_count[ord(c) - ord('a')] += 1

    matches = sum(1 for i in range(26) if p_count[i] == s_count[i])
    result = [0] if matches == 26 else []

    for i in range(len(p), len(s)):
        incoming = ord(s[i]) - ord('a')
        outgoing = ord(s[i - len(p)]) - ord('a')

        s_count[incoming] += 1
        if s_count[incoming] == p_count[incoming]:
            matches += 1
        elif s_count[incoming] == p_count[incoming] + 1:
            matches -= 1

        s_count[outgoing] -= 1
        if s_count[outgoing] == p_count[outgoing]:
            matches += 1
        elif s_count[outgoing] == p_count[outgoing] - 1:
            matches -= 1

        if matches == 26:
            result.append(i - len(p) + 1)

    return result

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Time & Space Complexity

Aspect	Complexity
Time	O(n) — O(1) per step
Space	O(1) — two 26-element arrays

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Common Interview Mistakes

1. Recomputing frequency counts from scratch each window — O(n × L) instead of O(n).
2. Not deleting zero-count entries in Counter comparisons — breaks equality.
3. Off-by-one in start index. Window ending at i starts at i - len(p) + 1.

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What a Strong Candidate Sounds Like

“Fixed-size sliding window of length len(p). I track a matches variable — how many of the 26 characters have equal counts. When I slide, I update two counts and adjust matches. When matches == 26, it’s an anagram. O(1) per step, O(n) total.”

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Related Problems to Practice

• Valid Anagram — Same frequency check, no window.
• Longest Substring Without Repeating Characters — Variable-size sliding window.
• Minimum Window Substring — Harder variable window.

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Practice This in a Mock Interview

Start a mock interview for Find All Anagrams in a String on Intervu.

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Further reading:

• How to Prepare for a Coding Interview in 2026, the complete roadmap
• Why LeetCode alone isn’t enough, and what to practice instead
• Practice any LeetCode problem as a live mock interview
• The Grind 75 Pathway, a structured study plan with AI practice links
• Browse all walkthroughs on GitHub