Given two strings txt and pat having lowercase letters, the task is to check if any permutation of pat is a substring of txt.
Examples:
Input: txt = "geeks", pat = "eke"
Output: true
Explanation: "eek" is a permutation of "eke" which exists in "geeks".Input: txt = "programming", pat = "rain"
Output: false
Explanation: No permutation of "rain" exists as a substring in "programming".
Try It Yourself
Table of Content
[Naive Approach] Checking all possible substrings
The idea is to iterate over all possible substrings of txt having the same length as pat. For each substring of txt, check if pat is a permutation of the substring. Two strings are said to be permutations of each other if they contain the same characters with the same frequencies, but possibly in a different order. So, we can match the frequency of each character in both the strings to check if they are permutations of each other or not.
// C++ Program to check if any permutation of pattern
// is substring by checking all possible substrings
#include <iostream>
#include <vector>
using namespace std;
const int MAX_CHAR = 26;
bool arePermutations(string &txt, int startIdx, string &pat) {
vector<int> freq(MAX_CHAR, 0);
for (int i = 0; i < pat.length(); i++) {
// Decrement the count of character in txt
freq[txt[startIdx + i] - 'a'] -= 1;
// Increment the count of character in pat
freq[pat[i] - 'a'] += 1;
}
for (int i = 0; i < MAX_CHAR; i++) {
if (freq[i] != 0)
return false;
}
return true;
}
bool search(string &txt, string &pat) {
int n = txt.length();
int m = pat.length();
for (int i = 0; i < n - m + 1; i++) {
// Check if substring in txt starting from i
// is a permutation of pat
if (arePermutations(txt, i, pat))
return true;
}
return false;
}
int main() {
string txt = "geeks";
string pat = "eke";
if (search(txt, pat))
cout << "true";
else
cout << "false";
return 0;
}
// C Program to check if any permutation of pattern
// is substring by checking all possible substrings
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#define MAX_CHAR 26
bool arePermutations(char *txt, int startIdx, char *pat) {
int freq[MAX_CHAR] = {0};
int i;
for (i = 0; i < strlen(pat); i++) {
// Increment the count of character in txt
freq[txt[startIdx + i] - 'a'] += 1;
// Decrement the count of character in pat
freq[pat[i] - 'a'] -= 1;
}
for (i = 0; i < MAX_CHAR; i++) {
if (freq[i] != 0)
return false;
}
return true;
}
int search(char *txt, char *pat) {
int n = strlen(txt);
int m = strlen(pat);
int i;
for (i = 0; i < n - m + 1; i++) {
// Check if substring in txt starting from i
// is a permutation of pat
if (arePermutations(txt, i, pat))
return true;
}
return false;
}
int main() {
char txt[] = "geeks";
char pat[] = "eke";
if (search(txt, pat))
printf("true\n");
else
printf("false\n");
return 0;
}
// Java Program to check if any permutation of pattern
// is substring by checking all possible substrings
class GfG {
static final int MAX_CHAR = 26;
static boolean arePermutations(String txt, int startIdx, String pat) {
int[] freq = new int[MAX_CHAR];
for (int i = 0; i < pat.length(); i++) {
// Increment the count of character in txt
freq[txt.charAt(startIdx + i) - 'a'] += 1;
// Decrement the count of character in pat
freq[pat.charAt(i) - 'a'] -= 1;
}
for (int i = 0; i < MAX_CHAR; i++) {
if (freq[i] != 0)
return false;
}
return true;
}
static boolean search(String txt, String pat) {
int n = txt.length();
int m = pat.length();
for (int i = 0; i < n - m + 1; i++) {
// Check if substring in txt starting from i
// is a permutation of pat
if (arePermutations(txt, i, pat))
return true;
}
return false;
}
public static void main(String[] args) {
String txt = "geeks";
String pat = "eke";
if (search(txt, pat))
System.out.println("true");
else
System.out.println("false");
}
}
# Python Program to check if any permutation of pattern
# is substring by checking all possible substrings
MAX_CHAR = 26
def arePermutations(txt, startIdx, pat):
freq = [0] * MAX_CHAR
for i in range(len(pat)):
# Increment the count of character in txt
freq[ord(txt[startIdx + i]) - ord('a')] += 1
# Decrement the count of character in pat
freq[ord(pat[i]) - ord('a')] -= 1
for i in range(MAX_CHAR):
if freq[i] != 0:
return False
return True
def search(txt, pat):
n = len(txt)
m = len(pat)
for i in range(n - m + 1):
# Check if substring in txt starting from i
# is a permutation of pat
if arePermutations(txt, i, pat):
return True
return False
if __name__ == "__main__":
txt = "geeks"
pat = "eke"
if search(txt, pat):
print("true")
else:
print("false")
// C# Program to check if any permutation of pattern
// is substring by checking all possible substrings
using System;
class GfG {
const int MAX_CHAR = 26;
static bool arePermutations(string txt, int startIdx, string pat) {
int[] freq = new int[MAX_CHAR];
for (int i = 0; i < pat.Length; i++) {
// Increment the count of character in txt
freq[txt[startIdx + i] - 'a'] += 1;
// Decrement the count of character in pat
freq[pat[i] - 'a'] -= 1;
}
for (int i = 0; i < MAX_CHAR; i++) {
if (freq[i] != 0)
return false;
}
return true;
}
static bool search(string txt, string pat) {
int n = txt.Length;
int m = pat.Length;
for (int i = 0; i < n - m + 1; i++) {
// Check if substring in txt starting from i
// is a permutation of pat
if (arePermutations(txt, i, pat))
return true;
}
return false;
}
static void Main() {
string txt = "geeks";
string pat = "eke";
if (search(txt, pat))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
// JavaScript Program to check if any permutation of pattern
// is substring by checking all possible substrings
const MAX_CHAR = 26;
function arePermutations(txt, startIdx, pat) {
let freq = new Array(MAX_CHAR).fill(0);
for (let i = 0; i < pat.length; i++) {
// Increment the count of character in txt
freq[txt.charCodeAt(startIdx + i) - 'a'.charCodeAt(0)] += 1;
// Decrement the count of character in pat
freq[pat.charCodeAt(i) - 'a'.charCodeAt(0)] -= 1;
}
for (let i = 0; i < MAX_CHAR; i++) {
if (freq[i] !== 0)
return false;
}
return true;
}
function search(txt, pat) {
let n = txt.length;
let m = pat.length;
for (let i = 0; i < n - m + 1; i++) {
// Check if substring in txt starting from i
// is a permutation of pat
if (arePermutations(txt, i, pat))
return true;
}
return false;
}
// Driver Code
let txt = "geeks";
let pat = "eke";
if (search(txt, pat))
console.log("true");
else
console.log("false");
Output
true
Time Complexity: O(n*m), where n is the length of txt and m is the length of pat.
Auxiliary Space: O(MAX_CHAR), where MAX_CHAR = 26 as txt and pat have lowercase letters only.
[Expected Approach] Using Sliding Window
The idea is similar to the previous approach where we check if pat is a permutation of any substring of txt but instead of comparing each substring from the beginning, we can use Sliding Window Technique and slide a window of the same size as pat. When a new character is added to the window, we increase its frequency by 1 and when a character is removed from the window, we decrease its frequency by 1.
// C++ Program to check if any permutation of string
// is substring using Sliding Window Technique
#include <iostream>
#include <vector>
using namespace std;
const int MAX_CHAR = 26;
// check if all characters have 0 frequency
bool check(vector<int> &freq) {
for(int i = 0; i < MAX_CHAR; i++) {
if(freq[i] != 0)
return false;
}
return true;
}
bool search(string &txt, string &pat) {
int n = txt.length();
int m = pat.length();
vector<int> freq(MAX_CHAR, 0);
// construct the first window
for(int i = 0; i < m; i++) {
freq[txt[i] - 'a'] += 1;
freq[pat[i] - 'a'] -= 1;
}
// Check for first window
if(check(freq))
return true;
for (int i = m; i < n; i++) {
// Add the ith character into the window
freq[txt[i] - 'a'] += 1;
// Remove the (i - m)th character from the window
freq[txt[i - m] - 'a'] -= 1;
// Check for the current window
if (check(freq))
return true;
}
return false;
}
int main() {
string txt = "geeks";
string pat = "eke";
if (search(txt, pat))
cout << "true";
else
cout << "false";
return 0;
}
// C Program to check if any permutation of string
// is substring using Sliding Window Technique
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#define MAX_CHAR 26
// check if all characters have 0 frequency
bool check(int freq[]) {
for (int i = 0; i < MAX_CHAR; i++) {
if (freq[i] != 0)
return false;
}
return true;
}
bool search(char *txt, char *pat) {
int n = strlen(txt);
int m = strlen(pat);
// construct the first window
int freq[MAX_CHAR] = {0};
for (int i = 0; i < m; i++) {
freq[txt[i] - 'a'] += 1;
freq[pat[i] - 'a'] -= 1;
}
// Check for first window
if (check(freq))
return true;
for (int i = m; i < n; i++) {
// Add the ith character into the window
freq[txt[i] - 'a'] += 1;
// Remove the (i - m)th character from the window
freq[txt[i - m] - 'a'] -= 1;
// Check for the current window
if (check(freq))
return true;
}
return false;
}
int main() {
char txt[] = "geeks";
char pat[] = "eke";
if (search(txt, pat))
printf("true");
else
printf("false");
return 0;
}
// Java Program to check if any permutation of string
// is substring using Sliding Window Technique
import java.util.*;
class GfG {
static final int MAX_CHAR = 26;
// check if all characters have 0 frequency
static boolean check(int[] freq) {
for (int i = 0; i < MAX_CHAR; i++) {
if (freq[i] != 0)
return false;
}
return true;
}
static boolean search(String txt, String pat) {
int n = txt.length();
int m = pat.length();
// construct the first window
int[] freq = new int[MAX_CHAR];
Arrays.fill(freq, 0);
for (int i = 0; i < m; i++) {
freq[txt.charAt(i) - 'a'] += 1;
freq[pat.charAt(i) - 'a'] -= 1;
}
// Check for first window
if (check(freq))
return true;
for (int i = m; i < n; i++) {
// Add the ith character into the window
freq[txt.charAt(i) - 'a'] += 1;
// Remove the (i - m)th character from the window
freq[txt.charAt(i - m) - 'a'] -= 1;
// Check for the current window
if (check(freq))
return true;
}
return false;
}
public static void main(String[] args) {
String txt = "geeks";
String pat = "eke";
if (search(txt, pat))
System.out.println("true");
else
System.out.println("false");
}
}
# Python Program to check if any permutation of string
# is substring using Sliding Window Technique
MAX_CHAR = 26
# check if all characters have 0 frequency
def check(freq):
for i in range(MAX_CHAR):
if freq[i] != 0:
return False
return True
def search(txt, pat):
n = len(txt)
m = len(pat)
# construct the first window
freq = [0] * MAX_CHAR
for i in range(m):
freq[ord(txt[i]) - ord('a')] += 1
freq[ord(pat[i]) - ord('a')] -= 1
# Check for first window
if check(freq):
return True
for i in range(m, n):
# Add the ith character into the window
freq[ord(txt[i]) - ord('a')] += 1
# Remove the (i - m)th character from the window
freq[ord(txt[i - m]) - ord('a')] -= 1
# Check for the current window
if check(freq):
return True
return False
if __name__ == "__main__":
txt = "geeks"
pat = "eke"
if search(txt, pat):
print("true")
else:
print("false")
// C# Program to check if any permutation of string
// is substring using Sliding Window Technique
using System;
class GfG {
const int MAX_CHAR = 26;
// check if all characters have 0 frequency
static bool check(int[] freq) {
for (int i = 0; i < MAX_CHAR; i++) {
if (freq[i] != 0)
return false;
}
return true;
}
static bool search(string txt, string pat) {
int n = txt.Length;
int m = pat.Length;
// construct the first window
int[] freq = new int[MAX_CHAR];
for (int i = 0; i < m; i++) {
freq[txt[i] - 'a'] += 1;
freq[pat[i] - 'a'] -= 1;
}
// Check for first window
if (check(freq))
return true;
for (int i = m; i < n; i++) {
// Add the ith character into the window
freq[txt[i] - 'a'] += 1;
// Remove the (i - m)th character from the window
freq[txt[i - m] - 'a'] -= 1;
// Check for the current window
if (check(freq))
return true;
}
return false;
}
static void Main() {
string txt = "geeks";
string pat = "eke";
if (search(txt, pat))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
// JavaScript Program to check if any permutation of string
// is substring using Sliding Window Technique
const MAX_CHAR = 26;
// check if all characters have 0 frequency
function check(freq) {
for (let i = 0; i < MAX_CHAR; i++) {
if (freq[i] !== 0)
return false;
}
return true;
}
function search(txt, pat) {
let n = txt.length;
let m = pat.length;
// construct the first window
let freq = new Array(MAX_CHAR).fill(0);
for (let i = 0; i < m; i++) {
freq[txt.charCodeAt(i) - 'a'.charCodeAt(0)] += 1;
freq[pat.charCodeAt(i) - 'a'.charCodeAt(0)] -= 1;
}
// Check for first window
if (check(freq))
return true;
for (let i = m; i < n; i++) {
// Add the ith character into the window
freq[txt.charCodeAt(i) - 'a'.charCodeAt(0)] += 1;
// Remove the (i - m)th character from the window
freq[txt.charCodeAt(i - m) - 'a'.charCodeAt(0)] -= 1;
// Check for the current window
if (check(freq))
return true;
}
return false;
}
// Driver Code
let txt = "geeks";
let pat = "eke";
if (search(txt, pat))
console.log("true");
else
console.log("false");
Output
true
Time Complexity: O(n*MAX_CHAR), where n is the length of txt and MAX_CHAR = 26 as txt and pat have lowercase letters only.
Auxiliary Space: O(1)