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# Check if any permutation of string is a K times repeated string

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 `   `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;` `    `  `    ``// 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

 `// 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

 `# 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#

 `// 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``; ` `     `  `    ``// 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`

## Javascript

 ``

Output:

`No`

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.

1. 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.
2. 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 ` `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;` `}`

Output :

`Yes`

Time Complexity : O(n^2)

Auxiliary Space : O(1)

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