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

• Last Updated : 31 May, 2021

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: 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`
Output:

`No`

Performance Analysis:

• Time Complexity O(N)

• Auxiliary Space: O(123)

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