Convert to a string that is repetition of a substring of k length

06 Jul, 2022
Given a string, find if it is possible to convert it to a string that is the repetition of a substring with k characters. To convert, we can replace one substring of length k starting at index i (zero-based indexing) such that i is divisible by K, with k characters.
Examples:  
Input: str = "bdac",  k = 2
Output: True
We can either replace "bd" with "ac" or 
"ac" with "bd".

Input: str = "abcbedabcabc",  k = 3
Output: True
Replace "bed" with "abc" so that the 
whole string becomes repetition of "abc".

Input: str = "bcacc", k = 3
Output: False
k doesn't divide string length i.e. 5%3 != 0

Input: str = "bcacbcac", k = 2
Output: False

Input: str = "bcdbcdabcedcbcd", k = 3
Output: False
This can be used in compression. If we have a string where the complete string is repetition except one substring, then we can use this algorithm to compress the string.
One observation is, length of string must be a multiple of k as we can replace only one substring. 
The idea is to declare a map mp which maps strings of length k to an integer denoting its count. So, if there are only two different sub-strings of length k in the map container and the count of one of the sub-string is 1 then the answer is true. Otherwise, answer is false. 
Implementation:
C++
Java
Python3
C#
Javascript
// C++ program to check if a string can be converted to// a string that has repeated substrings of length k.#include<bits/stdc++.h>using namespace std; // Returns true if str can be converted to a string// with k repeated substrings after replacing k// characters.bool checkString(string str, long k){    // Length of string must be a multiple of k    int n = str.length();    if (n%k != 0)        return false;     // Map to store strings of length k and their counts    unordered_map<string, int> mp;    for (int i=0; i<n; i+=k)        mp[str.substr(i, k)]++;     // If string is already a repetition of k substrings,    // return true.    if (mp.size() == 1)        return true;     // If number of distinct substrings is not 2, then    // not possible to replace a string.    if (mp.size() != 2)        return false;     // One of the two distinct must appear exactly once.    // Either the first entry appears once, or it appears    // n/k-1 times to make other substring appear once.    if ((mp.begin()->second == (n/k - 1)) ||                    mp.begin()->second == 1)       return true;     return false;} // Driver codeint main(){    checkString("abababcd", 2)? cout << "Yes" :                                cout << "No";    return 0;}
// Java program to check if a string// can be converted to a string that has// repeated substrings of length k.import java.util.HashMap;import java.util.Iterator; class GFG{     // Returns true if str can be converted    // to a string with k repeated substrings    // after replacing k characters.    static boolean checkString(String str, int k)    {         // Length of string must be        // a multiple of k        int n = str.length();        if (n % k != 0)            return false;         // Map to store strings of        // length k and their counts        HashMap<String, Integer> mp = new HashMap<>();        try        {            for (int i = 0; i < n; i += k)                mp.put(str.substring(i, k),                mp.get(str.substring(i, k)) == null ? 1 :                mp.get(str.substring(i, k)) + 1);        } catch (Exception e) {    }         // If string is already a repetition        // of k substrings, return true.        if (mp.size() == 1)            return true;         // If number of distinct substrings is not 2,        // then not possible to replace a string.        if (mp.size() != 2)            return false;         HashMap.Entry<String,                Integer> entry = mp.entrySet().iterator().next();         // One of the two distinct must appear        // exactly once. Either the first entry        // appears once, or it appears n/k-1 times        // to make other substring appear once.        if (entry.getValue() == (n / k - 1) ||            entry.getValue() == 1)            return true;         return false;    }     // Driver code    public static void main(String[] args)    {        if (checkString("abababcd", 2))            System.out.println("Yes");        else            System.out.println("No");    }} // This code is contributed by// sanjeev2552
# Python3 program to check if a can be converted to# a that has repeated subs of length k. # Returns True if S can be converted to a# with k repeated subs after replacing k# characters.def check( S, k):         # Length of must be a multiple of k    n = len(S)         if (n % k != 0):        return False     # Map to store s of length k and their counts    mp =  {}    for i in range(0, n, k):        mp[S[i:k]] = mp.get(S[i:k], 0) + 1     # If is already a repetition of k subs,    # return True.    if (len(mp) == 1):        return True     # If number of distinct subs is not 2, then    # not possible to replace a .    if (len(mp) != 2):        return False     # One of the two distinct must appear exactly once.    # Either the first entry appears once, or it appears    # n/k-1 times to make other sub appear once.    for i in mp:        if i == (n//k - 1) or mp[i] == 1:            return True     return False # Driver code if check("abababcd", 2):    print("Yes")else:    print("No")     # This code is contributed by mohit kumar 29   
// C# program to check if a string// can be converted to a string that has// repeated substrings of length k.using System;using System.Collections.Generic; class GFG{     // Returns true if str can be converted    // to a string with k repeated substrings    // after replacing k characters.    static bool checkString(String str, int k)    {         // Length of string must be        // a multiple of k        int n = str.Length;        if (n % k != 0)            return false;         // Map to store strings of        // length k and their counts        Dictionary<String,                   int> mp = new Dictionary<String,                                            int>();         for (int i = 0; i < n; i += k)        {            if(!mp.ContainsKey(str.Substring(i, k)))                mp.Add(str.Substring(i, k), 1);            else                mp[str.Substring(i, k)] = mp[str.Substring(i, k)] + 1;        }         // If string is already a repetition        // of k substrings, return true.        if (mp.Count == 1)            return true;         // If number of distinct substrings is not 2,        // then not possible to replace a string.        if (mp.Count != 2)            return false;         foreach(KeyValuePair<String, int> entry in mp)        {             // One of the two distinct must appear            // exactly once. Either the first entry            // appears once, or it appears n/k-1 times            // to make other substring appear once.            if (entry.Value == (n / k - 1) ||                entry.Value == 1)                return true;        }        return false;    }     // Driver code    public static void Main(String[] args)    {        if (checkString("abababcd", 2))            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }} // This code is contributed by Princi Singh
<script>// Javascript program to check if a string// can be converted to a string that has// repeated substrings of length k.         // Returns true if str can be converted    // to a string with k repeated substrings    // after replacing k characters.    function checkString(str,k)    {        // Length of string must be        // a multiple of k        let n = str.length;        if (n % k != 0)            return false;           // Map to store strings of        // length k and their counts        let mp = new Map();        for (let i = 0; i < n; i += k)                {            if(mp.has(str.substring(i, i+k)))                mp.set(str.substring(i, i+k),mp.get(str.substring(i, i+k)) + 1);            else                mp.set(str.substring(i, i+k),1);        }               // If string is already a repetition        // of k substrings, return true.        if (mp.size == 1)            return true;           // If number of distinct substrings is not 2,        // then not possible to replace a string.        if (mp.size != 2)        {               return false;          }                    // One of the two distinct must appear        // exactly once. Either the first entry        // appears once, or it appears n/k-1 times        // to make other substring appear once.        for (let [key, value] of mp.entries())        {                         if(value == (Math.floor(n/k) - 1) || value == 1)                return true;        }           return false;    }         // Driver code    if (checkString("abababcd", 2))        document.write("Yes");    else        document.write("No");              // This code is contributed by unknown2108</script>
Yes
Time complexity : O(n) Auxiliary Space : O(n)
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mohit kumar 29
sanjeev2552
princi singh
unknown2108
surindertarika1234
saurabh1990aror
youmailmahibagi
sangramjagaxcka
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