Approach Summary
Use a fixed-size window equal to s1.length. Compare character frequency maps of the window and s1.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(n^2) | O(n) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(n) | O(1) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(n^2)
Space: O(n)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
public boolean checkInclusion(String s1, String s2) {
int n1 = s1.length();
int n2 = s2.length();
if(n1 > n2){
return false;
}
int s1Freq[] = new int[26];
int s2Freq[] = new int[26];
int k=n1;
for(int i=0; i<k; i++){
s1Freq[s1.charAt(i) - 'a']++;
s2Freq[s2.charAt(i) - 'a']++;
}
if(Arrays.equals(s1Freq, s2Freq)){
return true;
}
int left = 0;
for(int right=k; right<n2; right++){
s2Freq[s2.charAt(left) - 'a']--;
left++;
s2Freq[s2.charAt(right) - 'a']++;
if(Arrays.equals(s1Freq, s2Freq)){
return true;
}
}
return false;
}
}class Solution {
public:
bool checkInclusion(string s1, string s2) {
int n1 = s1.size();
int n2 = s2.size();
if(n1 > n2){
return false;
}
int k = n1;
vector<int> s1Freq(26, 0);
vector<int> s2Freq(26, 0);
for(int i=0; i<k; i++){
s1Freq[s1[i] - 'a']++;
s2Freq[s2[i] - 'a']++;
}
if(s1Freq == s2Freq){
return true;
}
int left = 0;
for(int right = k; right<n2; right++){
s2Freq[s2[left] - 'a']--;
left++;
s2Freq[s2[right] - 'a']++;
if(s1Freq == s2Freq){
return true;
}
}
return false;
}
};class Solution:
def permutation_in_string(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Permutation in String
*/
function permutationInString(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(n)
Space: O(1)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public boolean checkInclusion(String s1, String s2) {
int n1 = s1.length();
int n2 = s2.length();
if(n1 > n2){
return false;
}
int s1Freq[] = new int[26];
int s2Freq[] = new int[26];
int k=n1;
for(int i=0; i<k; i++){
s1Freq[s1.charAt(i) - 'a']++;
s2Freq[s2.charAt(i) - 'a']++;
}
if(Arrays.equals(s1Freq, s2Freq)){
return true;
}
int left = 0;
for(int right=k; right<n2; right++){
s2Freq[s2.charAt(left) - 'a']--;
left++;
s2Freq[s2.charAt(right) - 'a']++;
if(Arrays.equals(s1Freq, s2Freq)){
return true;
}
}
return false;
}
}class Solution {
public:
bool checkInclusion(string s1, string s2) {
int n1 = s1.size();
int n2 = s2.size();
if(n1 > n2){
return false;
}
int k = n1;
vector<int> s1Freq(26, 0);
vector<int> s2Freq(26, 0);
for(int i=0; i<k; i++){
s1Freq[s1[i] - 'a']++;
s2Freq[s2[i] - 'a']++;
}
if(s1Freq == s2Freq){
return true;
}
int left = 0;
for(int right = k; right<n2; right++){
s2Freq[s2[left] - 'a']--;
left++;
s2Freq[s2[right] - 'a']++;
if(s1Freq == s2Freq){
return true;
}
}
return false;
}
};class Solution:
def permutation_in_string(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Permutation in String
*/
function permutationInString(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Sliding Window Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Sliding Window pattern.
Framework: Sliding Window • Interview Prep: Tested at FAANG •
Explore all Sliding Window problems & invariants →
How to Recognize This Pattern
- Fixed window = length of pattern
- Check if window is anagram
Complexity Analysis
Time Complexity
O(n)
Space Complexity
O(1)
Tags
String Hash Map Sliding Window
Frequently Asked Questions
What is the optimal time and space complexity for Permutation in String?
The optimal solution for Permutation in String achieves O(n) runtime complexity and requires O(1) auxiliary space. Time Complexity is O(n) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(1) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Permutation in String requires the Sliding Window pattern?
In technical interviews, look for key problem signals: Fixed window = length of pattern; Check if window is anagram. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Sliding Window strategy.
What critical edge cases must be handled in Permutation in String?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Permutation in String?
Permutation in String is a Medium-difficulty question tagged with medium recurrence. It is actively asked in technical screening and onsite interview loops at Microsoft, Amazon.