Permutation in String - Practice Coding Problems

Permutation in String - Problem

Given two strings s1 and s2, determine if s2 contains any permutation of s1 as a substring. A permutation is any rearrangement of the characters in a string.

What is a permutation? A permutation of a string is formed by rearranging its characters. For example, "abc" has 6 permutations: "abc", "acb", "bac", "bca", "cab", "cba".

Your task: Return true if any substring of s2 is a permutation of s1, otherwise return false.

Key insight: Instead of generating all permutations, we can check if any substring of s2 has the same character frequencies as s1.

Input & Output

example_1.py — Basic permutation found

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

› Output: true

💡 Note: s2 contains "ba" which is a permutation of s1 = "ab"

example_2.py — No permutation exists

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

› Output: false

💡 Note: No substring of s2 is a permutation of s1. "bo", "oa", "ao", "oo" don't match "ab"

example_3.py — Longer pattern

$ Input: s1 = "adc", s2 = "dcda"

› Output: true

💡 Note: s2 contains "dcd" + "a" = "dcda", but looking at substrings of length 3: "dcd" is not a permutation, but "cda" is a permutation of "adc"

Constraints

1 ≤ s1.length, s2.length ≤ 10⁴
s1 and s2 consist of lowercase English letters only
Time limit: 1 second per test case

Visualization

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Understanding the Visualization

Setup Recipe

Count required ingredients from s1: {a: 1, b: 1}

First Measurement

Check first window 'ei': {e: 1, i: 1} - doesn't match

Slide Window

Move to 'id': remove 'e', add 'd' → {i: 1, d: 1}

Continue Sliding

Move to 'db': remove 'i', add 'b' → {d: 1, b: 1}

Found Match!

Move to 'ba': remove 'd', add 'a' → {b: 1, a: 1} ✓

Key Takeaway

🎯 Key Insight: Use sliding window with frequency maps to achieve O(n) time complexity instead of generating all permutations!

Asked in

f Facebook 45 ⊞ Microsoft 38 a Amazon 32 G Google 28

The optimal solution uses a sliding window approach with frequency counting. Instead of generating all permutations, we track character frequencies in a window of size |s1| and slide it across s2. This achieves O(n) time complexity by efficiently updating frequency counts as the window moves.

Common Approaches

Approach	Time	Space	Notes
✓ Sliding Window with Frequency Map (Optimal)	O(n + m)	O(1)	Use sliding window technique with character frequency counting for O(n) solution
Brute Force (Check All Substrings)	O(n * m * log m)	O(m)	Check every substring of s2 with length equal to s1 by sorting and comparing

Sliding Window with Frequency Map (Optimal) — Algorithm Steps

Create frequency map for s1
Initialize sliding window of size len(s1) on s2
Compare frequency maps
Slide window: add new character, remove old character
Check frequencies after each slide

Visualization

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Step-by-Step Walkthrough

Initialize

Create frequency map for s1 and initial window

Compare

Check if current window matches s1 frequencies

Slide

Add new character, remove old character

Update

Update frequency counts efficiently

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool compareMaps(int freq1[], int freq2[]) {
    for (int i = 0; i < 26; i++) {
        if (freq1[i] != freq2[i]) {
            return false;
        }
    }
    return true;
}

bool checkInclusion(char *s1, char *s2) {
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    
    if (len1 > len2) {
        return false;
    }
    
    // Frequency arrays for lowercase letters
    int s1Freq[26] = {0};
    int windowFreq[26] = {0};
    
    // Count frequencies in s1
    for (int i = 0; i < len1; i++) {
        s1Freq[s1[i] - 'a']++;
    }
    
    // Initialize window
    for (int i = 0; i < len1; i++) {
        windowFreq[s2[i] - 'a']++;
    }
    
    // Check initial window
    if (compareMaps(s1Freq, windowFreq)) {
        return true;
    }
    
    // Slide the window
    for (int i = len1; i < len2; i++) {
        // Add new character
        windowFreq[s2[i] - 'a']++;
        
        // Remove old character
        windowFreq[s2[i - len1] - 'a']--;
        
        // Check if frequencies match
        if (compareMaps(s1Freq, windowFreq)) {
            return true;
        }
    }
    
    return false;
}

int main() {
    char s1[1001], s2[10001];
    scanf("%s", s1);
    scanf("%s", s2);
    
    printf("%s\n", checkInclusion(s1, s2) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

One pass through s2 (n) and s1 (m) with constant time operations per character

✓ Linear Growth

Space Complexity

O(1)

Frequency maps store at most 26 characters (constant space)

✓ Linear Space

52.4K Views

High Frequency

~25 min Avg. Time

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Sliding Window with Frequency Map (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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