Check if any permutation of string is a K times repeated string
Last Updated :
20 Oct, 2023
Given a string S and an integer K, the task is to check that if any permutation of the string can be formed by K times repeating any other string.
Examples:
Input: S = "abba", K = 2
Output: Yes
Explanation:
Permutations of given string -
{"aabb", "abab", "abba", "baab", "baba", "bbaa"}
As "abab" is repeating string of "ab"+"ab" = "abab", which is also permutation of string.
Input: S = "abcabd", K = 2
Output: No
Explanation:
There is no such repeating string in all permutations of the given string.
Approach 1: The idea is to find the frequency of each character of the string and check that the frequency of the character is a multiple of the given integer K. If the frequency of all characters of the string is divisible by K, then there is a string which is a permutation of the given string and also a K times repeated string.
Below is the implementation of the above approach:
C++
// C++ implementation to check that
// the permutation of the given string
// is K times repeated string
#include <bits/stdc++.h>
using namespace std;
// Function to check that permutation
// of the given string is a
// K times repeating String
bool repeatingString(string s,
int n, int k)
{
// if length of string is
// not divisible by K
if (n % k != 0) {
return false;
}
// Frequency Array
int frequency[123];
// Initially frequency of each
// character is 0
for (int i = 0; i < 123; i++) {
frequency[i] = 0;
}
// Computing the frequency of
// each character in the string
for (int i = 0; i < n; i++) {
frequency[s[i]]++;
}
int repeat = n / k;
// Loop to check that frequency of
// every character of the string
// is divisible by K
for (int i = 0; i < 123; i++) {
if (frequency[i] % repeat != 0) {
return false;
}
}
return true;
}
// Driver Code
int main()
{
string s = "abcdcba";
int n = s.size();
int k = 3;
if (repeatingString(s, n, k)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
return 0;
}
// Java implementation to check that
// the permutation of the given String
// is K times repeated String
class GFG{
// Function to check that permutation
// of the given String is a
// K times repeating String
static boolean repeatingString(String s,
int n, int k)
{
// if length of String is
// not divisible by K
if (n % k != 0) {
return false;
}
// Frequency Array
int []frequency = new int[123];
// Initially frequency of each
// character is 0
for (int i = 0; i < 123; i++) {
frequency[i] = 0;
}
// Computing the frequency of
// each character in the String
for (int i = 0; i < n; i++) {
frequency[s.charAt(i)]++;
}
int repeat = n / k;
// Loop to check that frequency of
// every character of the String
// is divisible by K
for (int i = 0; i < 123; i++) {
if (frequency[i] % repeat != 0) {
return false;
}
}
return true;
}
// Driver Code
public static void main(String[] args)
{
String s = "abcdcba";
int n = s.length();
int k = 3;
if (repeatingString(s, n, k)) {
System.out.print("Yes" +"\n");
}
else {
System.out.print("No" +"\n");
}
}
}
// This code is contributed by PrinciRaj1992
# Python3 implementation to check that
# the permutation of the given string
# is K times repeated string
# Function to check that permutation
# of the given string is a
# K times repeating String
def repeatingString(s, n, k):
# If length of string is
# not divisible by K
if (n % k != 0):
return False
# Frequency Array
frequency = [0 for i in range(123)]
# Initially frequency of each
# character is 0
for i in range(123):
frequency[i] = 0
# Computing the frequency of
# each character in the string
for i in range(n):
frequency[s[i]] += 1
repeat = n // k
# Loop to check that frequency of
# every character of the string
# is divisible by K
for i in range(123):
if (frequency[i] % repeat != 0):
return False
return True
# Driver Code
if __name__ == '__main__':
s = "abcdcba"
n = len(s)
k = 3
if (repeatingString(s, n, k)):
print("Yes")
else:
print("No")
# This code is contributed by Samarth
// C# implementation to check that
// the permutation of the given String
// is K times repeated String
using System;
class GFG{
// Function to check that permutation
// of the given String is a
// K times repeating String
static bool repeatingString(String s,
int n, int k)
{
// if length of String is
// not divisible by K
if (n % k != 0) {
return false;
}
// Frequency Array
int []frequency = new int[123];
// Initially frequency of each
// character is 0
for (int i = 0; i < 123; i++) {
frequency[i] = 0;
}
// Computing the frequency of
// each character in the String
for (int i = 0; i < n; i++) {
frequency[s[i]]++;
}
int repeat = n / k;
// Loop to check that frequency of
// every character of the String
// is divisible by K
for (int i = 0; i < 123; i++) {
if (frequency[i] % repeat != 0) {
return false;
}
}
return true;
}
// Driver Code
public static void Main(String[] args)
{
String s = "abcdcba";
int n = s.Length;
int k = 3;
if (repeatingString(s, n, k)) {
Console.Write("Yes" +"\n");
}
else {
Console.Write("No" +"\n");
}
}
}
// This code is contributed by Rajput-Ji
<script>
// JavaScript implementation to check that
// the permutation of the given string
// is K times repeated string
// Function to check that permutation
// of the given string is a
// K times repeating String
function repeatingString( s, n, k)
{
// if length of string is
// not divisible by K
if (n % k != 0) {
return false;
}
// Frequency Array
var frequency = new Array(123);
// Initially frequency of each
// character is 0
for (let i = 0; i < 123; i++) {
frequency[i] = 0;
}
// Computing the frequency of
// each character in the string
for (let i = 0; i < n; i++) {
frequency[s[i]]++;
}
var repeat = n / k;
// Loop to check that frequency of
// every character of the string
// is divisible by K
for (let i = 0; i < 123; i++) {
if (frequency[i] % repeat != 0) {
return false;
}
}
return true;
}
// Driver Code
var s = "abcdcba";
var n = s.length;
var k = 3;
if (repeatingString(s, n, k)) {
console.log("Yes");
}
else {
console.log("No" );
}
// This code is contributed by ukasp.
</script>
Performance Analysis:
Time Complexity O(N)
Auxiliary Space: O(1)
Approach 2 :
We can check for all possible substrings of the given string and then check if any of these substrings can be repeated K times to form a permutation of the original string.
- For each substring of length l = n/K, we can check if all characters in the substring have the same frequency (count) in the original string S. If this condition is not satisfied, we move to the next substring.
- We repeat step 1 for all substrings of length l in the original string S. If we find any substring which satisfies the condition, we return "Yes", else we return "No".
Below is the code for above approach :
C++
#include <bits/stdc++.h>
using namespace std;
// Function to check if a repeating substring of given length
// exists in the given string
bool isRepeatingSubstring(string s, int len)
{
int n = s.length();
// Check if given substring length is greater than string length
if(len >= n) {
return false;
}
// Map to store the frequency of characters
unordered_map<char, int> freq;
// Find the frequency of characters in first len characters of the string
for(int i = 0; i < len; i++) {
freq[s[i]]++;
}
// Check if the substring consisting of first len characters is repeating
bool repeating = true;
for(auto it : freq) {
if(it.second != n/len) {
repeating = false;
break;
}
}
// If the substring is not repeating, check for other substrings
if(!repeating) {
for(int i = len; i < n; i++) {
// Update the frequency map
freq[s[i-len]]--;
freq[s[i]]++;
// Check if the substring consisting of previous len characters is repeating
if(freq[s[i-len]] == 0) {
freq.erase(s[i-len]);
}
if(freq.size() == 1 && freq.begin()->second == n/len) {
return true;
}
}
}
return false;
}
// Driver code
int main() {
string S = "abba";
int K = 2;
if(isRepeatingSubstring(S, K)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
return 0;
}
//code in Java for the above approach
import java.util.HashMap;
import java.util.Map;
public class Main {
// Function to check if a repeating substring of given length
// exists in the given string
public static boolean isRepeatingSubstring(String s, int len) {
int n = s.length();
// Check if given substring length is greater than string length
if (len >= n) {
return false;
}
// Map to store the frequency of characters
Map<Character, Integer> freq = new HashMap<>();
// Find the frequency of characters in the first len characters of the string
for (int i = 0; i < len; i++) {
char ch = s.charAt(i);
freq.put(ch, freq.getOrDefault(ch, 0) + 1);
}
// Check if the substring consisting of the first len characters is repeating
boolean repeating = true;
for (Map.Entry<Character, Integer> entry : freq.entrySet()) {
if (entry.getValue() != n / len) {
repeating = false;
break;
}
}
// If the substring is not repeating, check for other substrings
if (!repeating) {
for (int i = len; i < n; i++) {
char prevChar = s.charAt(i - len);
char currentChar = s.charAt(i);
// Update the frequency map
freq.put(prevChar, freq.get(prevChar) - 1);
if (freq.get(prevChar) == 0) {
freq.remove(prevChar);
}
freq.put(currentChar, freq.getOrDefault(currentChar, 0) + 1);
// Check if the substring consisting of the previous len characters is repeating
if (freq.size() == 1 && freq.values().iterator().next() == n / len) {
return true;
}
}
}
return false;
}
// Driver code
public static void main(String[] args) {
String S = "abba";
int K = 2;
if (isRepeatingSubstring(S, K)) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}
# Function to check if a repeating substring of a given length
# exists in the given string
def isRepeatingSubstring(s, length):
n = len(s)
# Check if the given substring length is greater than or equal to the string length
if length >= n:
return False
# Dictionary to store the frequency of characters
freq = {}
# Find the frequency of characters in the first 'length' characters of the string
for i in range(length):
if s[i] in freq:
freq[s[i]] += 1
else:
freq[s[i]] = 1
# Check if the substring consisting of the first 'length' characters is repeating
repeating = True
for count in freq.values():
if count != n // length:
repeating = False
break
# If the substring is not repeating, check for other substrings
if not repeating:
for i in range(length, n):
# Update the frequency dictionary
freq[s[i - length]] -= 1
if freq[s[i - length]] == 0:
del freq[s[i - length]]
if s[i] in freq:
freq[s[i]] += 1
else:
freq[s[i]] = 1
# Check if the substring consisting of the previous 'length' characters is repeating
if len(freq) == 1 and list(freq.values())[0] == n // length:
return True
return False
# Driver code
S = "abba"
K = 2
if isRepeatingSubstring(S, K):
print("Yes")
else:
print("No")
using System;
using System.Collections.Generic;
class GFG
{
static bool IsRepeatingSubstring(string s, int len)
{
int n = s.Length;
// Check if given substring length is greater than string length
if (len >= n)
{
return false;
}
// Dictionary to store the frequency of characters
Dictionary<char, int> freq = new Dictionary<char, int>();
// Find the frequency of characters in the
// first len characters of the string
for (int i = 0; i < len; i++)
{
if (freq.ContainsKey(s[i]))
{
freq[s[i]]++;
}
else
{
freq[s[i]] = 1;
}
}
// Check if the substring consisting of
// the first len characters is repeating
bool repeating = true;
foreach (var kvp in freq)
{
if (kvp.Value != n / len)
{
repeating = false;
break;
}
}
// If the substring is not repeating
// check for other substrings
if (!repeating)
{
for (int i = len; i < n; i++)
{
// Update the frequency dictionary
freq[s[i - len]]--;
if (freq[s[i - len]] == 0)
{
freq.Remove(s[i - len]);
}
if (freq.ContainsKey(s[i]))
{
freq[s[i]]++;
}
else
{
freq[s[i]] = 1;
}
// Check if the substring consisting of the previous len characters is repeating
if (freq.Count == 1 && freq.ContainsValue(n / len))
{
return true;
}
}
}
return false;
}
static void Main(string[] args)
{
string S = "abba";
int K = 2;
if (IsRepeatingSubstring(S, K))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
function isRepeatingSubstring(s, len) {
const n = s.length;
// Check if given substring length is
// greater than string length
if (len >= n) {
return false;
}
// Map to store the frequency of characters
const freq = new Map();
// Find the frequency of characters in the first
// len characters of the string
for (let i = 0; i < len; i++) {
const char = s[i];
freq.set(char, (freq.get(char) || 0) + 1);
}
// Check if the substring consisting of the
// first len characters is repeating
let repeating = true;
for (const [char, count] of freq) {
if (count !== n / len) {
repeating = false;
break;
}
}
// If the substring is not repeating
// check for other substrings
if (!repeating) {
for (let i = len; i < n; i++) {
// Update the frequency map
const charToRemove = s[i - len];
const charToAdd = s[i];
freq.set(charToRemove, freq.get(charToRemove) - 1);
if (freq.get(charToRemove) === 0) {
freq.delete(charToRemove);
}
freq.set(charToAdd, (freq.get(charToAdd) || 0) + 1);
// Check if the substring consisting of the previous
// len characters is repeating
if (freq.size === 1 && [...freq.values()][0] === n / len) {
return true;
}
}
}
return false;
}
// Driver code
const S = "abba";
const K = 2;
if (isRepeatingSubstring(S, K)) {
console.log("Yes");
} else {
console.log("No");
}
Output :
Yes
Time Complexity : O(n^2)
Auxiliary Space : O(1)
Similar Reads
String which when repeated exactly K times gives a permutation of S Given an integer K and a string str of lowercase English characters, the task is to find a string s such that when s is repeated exactly K times, it gives a permutation of S. If no such string exists, print -1. Examples: Input: str = "aabb", k = 2 Output: ab "ab" when repeated 2 times gives "abab" w
12 min read
Time complexity of all permutations of a string Time complexity of printing all permutations of a string. void perm(String str){ perm(str, ""); } void perm(String str, String prefix){ if(str.length() == 0){ System.out.println(prefix); } else{ for(int i = 0; i < str.length(); i++){ String rem = str.substring(0, i) + str.substring(i + 1); perm(r
3 min read
Print all Unique permutations of a given string. Given a string that may contain duplicates, the task is find all unique permutations of given string in any order.Examples: Input: "ABC"Output: ["ABC", "ACB", "BAC", "BCA", "CAB", "CBA"]Explanation: Given string ABC has 6 unique permutations as "ABC", "ACB", "BAC", "BCA", "CAB" and "CBA".Input: "AAA
12 min read
Check if a binary string contains all permutations of length k Given a binary string and k, to check whether it's contains all permutations of length k or not. Examples: Input : Binary string 11001 k : 2 Output : Yes 11001 contains all possibilities of binary sequences with k = 2, 00, 01, 10, 11 Input : Binary string: 1001 k : 2 Output: No 1001 does not contain
13 min read
Check if Permutation of Pattern is Substring 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: trueExplanation: "eek" is a permutation of "eke" which exists in "geeks".Input: txt = "programming", pat = "rain"Output: fa
12 min read