LeetCode #567 — MEDIUM

Permutation in String

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.

In other words, return true if one of s1's permutations is the substring of s2.

Example 1:

Input: s1 = "ab", s2 = "eidbaooo"
Output: true
Explanation: s2 contains one permutation of s1 ("ba").

Example 2:

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

Constraints:

1 <= s1.length, s2.length <= 10⁴
s1 and s2 consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise. In other words, return true if one of s1's permutations is the substring of s2.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Two Pointers · Sliding Window

Example 1

"ab"
"eidbaooo"

Example 2

"ab"
"eidboaoo"

Core Insight

What unlocks the optimal approach

Obviously, brute force will result in TLE. Think of something else.
How will you check whether one string is a permutation of another string?
One way is to sort the string and then compare. But, Is there a better way?
If one string is a permutation of another string then they must have one common metric. What is that?
Both strings must have same character frequencies, if one is permutation of another. Which data structure should be used to store frequencies?
What about hash table? An array of size 26?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #567: Permutation in String
class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int need = 0;
        int[] cnt = new int[26];
        for (char c : s1.toCharArray()) {
            if (++cnt[c - 'a'] == 1) {
                ++need;
            }
        }
        int m = s1.length(), n = s2.length();
        for (int i = 0; i < n; ++i) {
            int c = s2.charAt(i) - 'a';
            if (--cnt[c] == 0) {
                --need;
            }
            if (i >= m) {
                c = s2.charAt(i - m) - 'a';
                if (++cnt[c] == 1) {
                    ++need;
                }
            }
            if (need == 0) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #567: Permutation in String
func checkInclusion(s1 string, s2 string) bool {
	need := 0
	cnt := [26]int{}

	for _, c := range s1 {
		if cnt[c-'a']++; cnt[c-'a'] == 1 {
			need++
		}
	}

	m, n := len(s1), len(s2)
	for i := 0; i < n; i++ {
		c := s2[i] - 'a'
		if cnt[c]--; cnt[c] == 0 {
			need--
		}
		if i >= m {
			c = s2[i-m] - 'a'
			if cnt[c]++; cnt[c] == 1 {
				need++
			}
		}
		if need == 0 {
			return true
		}
	}
	return false
}

# Accepted solution for LeetCode #567: Permutation in String
class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        cnt = Counter(s1)
        need = len(cnt)
        m = len(s1)
        for i, c in enumerate(s2):
            cnt[c] -= 1
            if cnt[c] == 0:
                need -= 1
            if i >= m:
                cnt[s2[i - m]] += 1
                if cnt[s2[i - m]] == 1:
                    need += 1
            if need == 0:
                return True
        return False

// Accepted solution for LeetCode #567: Permutation in String
impl Solution {
    pub fn check_inclusion(s1: String, s2: String) -> bool {
        let mut need = 0;
        let mut cnt = vec![0; 26];

        for c in s1.chars() {
            let index = (c as u8 - b'a') as usize;
            if cnt[index] == 0 {
                need += 1;
            }
            cnt[index] += 1;
        }

        let m = s1.len();
        let n = s2.len();
        let s2_bytes = s2.as_bytes();

        for i in 0..n {
            let c = (s2_bytes[i] - b'a') as usize;
            cnt[c] -= 1;
            if cnt[c] == 0 {
                need -= 1;
            }

            if i >= m {
                let c = (s2_bytes[i - m] - b'a') as usize;
                cnt[c] += 1;
                if cnt[c] == 1 {
                    need += 1;
                }
            }

            if need == 0 {
                return true;
            }
        }

        false
    }
}

// Accepted solution for LeetCode #567: Permutation in String
function checkInclusion(s1: string, s2: string): boolean {
    let need = 0;
    const cnt: number[] = Array(26).fill(0);
    const a = 'a'.charCodeAt(0);
    for (const c of s1) {
        if (++cnt[c.charCodeAt(0) - a] === 1) {
            need++;
        }
    }

    const [m, n] = [s1.length, s2.length];
    for (let i = 0; i < n; i++) {
        let c = s2.charCodeAt(i) - a;
        if (--cnt[c] === 0) {
            need--;
        }
        if (i >= m) {
            c = s2.charCodeAt(i - m) - a;
            if (++cnt[c] === 1) {
                need++;
            }
        }
        if (need === 0) {
            return true;
        }
    }
    return false;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.