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

Given a string, find if it is possible to convert it to a string that is repetition of 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 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 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 count of one of the sub-string is 1 then answer is true. Otherwise answer is false.

C++

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// 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 coverted 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 repition 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;
}

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Java

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// 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 coverted
    // 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 repition 
        // 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

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Python3

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# Python3 program to check if a can be converted to
# a that has repeated subs of length k.
  
# Returns True if S can be coverted 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 repition 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    

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C#

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// 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 coverted
    // 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 repition 
        // 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

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Output:

Yes

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