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

  • Difficulty Level : Medium
  • Last Updated : 26 Jul, 2021

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 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. 
 

C++




// 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 code
int main()
{
    checkString("abababcd", 2)? cout << "Yes" :
                                cout << "No";
    return 0;
}

Java




// 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




# 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#




// 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

Javascript




<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>

Output:



Yes

This article is contributed by Himanshu Gupta(Bagri). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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