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.

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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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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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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