Reorder the given string to form a K-concatenated string

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Given a string S and an integer K. The task is to form a string T such that the string T is a reordering of the string S in a way that it is a K-Concatenated-String. A string is said to be a K-Concatenated-String if it contains exactly K copies of some string.
For example, the string “geekgeek” is a 2-Concatenated-String formed by concatenating 2 copies of the string “geek”.
Note: Multiple answers are possible.
Examples:

Input : s = “gkeekgee”, k=2
Output: geekgeek
eegkeegk is another possible K-Concatenated-String
Input: s = “abcd”, k=2
Output: Not Possible

Approach: To find a valid ordering that is a K-Concatenated-string, iterate through the entire string and maintain a frequency array for the characters to hold the number of times each character occurs in a string. Since, in a K-Concatenated-string, the number of times a character occurs should be divisible by K. If any character is found not following this, then the string cant be ordered in any way to represent a K-Concatenated-string, else there can be exactly (Frequency of i^th character / K) copies of the i^th character in a single copy of the k-Concatenated-string. Such single copies when repeated K times form a valid K-Concatenated-string.
Below is the implementation of the above approach:

C++

// C++ program to form a
// K-Concatenated-String from a given string
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the k-concatenated string
string kStringGenerate(string str, int k)
{
 
    // maintain a frequency table
    // for lowercase letters
    int frequency[26] = { 0 };
 
    int len = str.length();
 
    for (int i = 0; i < len; i++) {
 
        // for each character increment its frequency
        // in the frequency array
        frequency[str[i] - 'a']++;
    }
 
    // stores a single copy of a string,
    // which on repeating forms a k-string
    string single_copy = "";
 
    // iterate over frequency array
    for (int i = 0; i < 26; i++) {
 
        // if the character occurs in the string,
        // check if it is divisible by k,
        // if not divisible then k-string cant be formed
        if (frequency[i] != 0) {
 
            if ((frequency[i] % k) != 0) {
 
                string ans = "Not Possible";
                return ans;
            }
            else {
 
                // ith character occurs (frequency[i]/k) times in a
                // single copy of k-string
                int total_occurrences = (frequency[i] / k);
 
                for (int j = 0; j < total_occurrences; j++) {
                    single_copy += char(i + 97);
                }
            }
        }
    }
 
    string kString;
 
    // append the single copy formed k times
    for (int i = 0; i < k; i++) {
        kString += single_copy;
    }
 
    return kString;
}
 
// Driver Code
int main()
{
    string str = "gkeekgee";
    int K = 2;
 
    string kString = kStringGenerate(str, K);
    cout << kString;
    return 0;
}

Java

// Java program to form a
// K-Concatenated-String 
// from a given string
import java.io.*;
import java.util.*;
import java.lang.*;
 
class GFG
{
     
// Function to print 
// the k-concatenated string
static String kStringGenerate(String str,
                              int k)
{
 
    // maintain a frequency table
    // for lowercase letters
    int frequency[] = new int[26];
    Arrays.fill(frequency,0);
    int len = str.length();
 
    for (int i = 0; i < len; i++) 
    {
 
        // for each character 
        // increment its frequency
        // in the frequency array
        frequency[str.charAt(i) - 'a']++;
    }
 
    // stores a single copy 
    // of a string, which on 
    // repeating forms a k-string
    String single_copy = "";
 
    // iterate over 
    // frequency array
    for (int i = 0; i < 26; i++) 
    {
 
        // if the character occurs 
        // in the string, check if 
        // it is divisible by k,
        // if not divisible then
        // k-string cant be formed
        if (frequency[i] != 0) 
        {
 
            if ((frequency[i] % k) != 0)
            {
                String ans = "Not Possible";
                return ans;
            }
            else
            {
 
                // ith character occurs 
                // (frequency[i]/k) times in 
                // a single copy of k-string
                int total_occurrences = (frequency[i] / k);
 
                for (int j = 0; 
                         j < total_occurrences; j++) 
                {
                    single_copy += (char)(i + 97);
                }
            }
        }
    }
 
    String kString = "";
 
    // append the single
    // copy formed k times
    for (int i = 0; i < k; i++) 
    {
        kString += single_copy;
    }
 
    return kString;
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "gkeekgee";
    int K = 2;
     
    String kString = kStringGenerate(str, K);
    System.out.print(kString);
}
}

Python3

# Python 3 program to form a
# K-Concatenated-String from a given string
 
# Function to print the k-concatenated string
 
 
def kStringGenerate(st, k):
 
    # maintain a frequency table
    # for lowercase letters
    frequency = [0] * 26
 
    length = len(st)
 
    for i in range(length):
 
        # for each character increment its frequency
        # in the frequency array
        frequency[ord(st[i]) - ord('a')] += 1
 
    # stores a single copy of a string,
    # which on repeating forms a k-string
    single_copy = ""
 
    # iterate over frequency array
    for i in range(26):
 
        # if the character occurs in the string,
        # check if it is divisible by k,
        # if not divisible then k-string cant be formed
        if (frequency[i] != 0):
 
            if ((frequency[i] % k) != 0):
 
                ans = "Not Possible"
                return ans
 
            else:
 
                # ith character occurs (frequency[i]/k) times in a
                # single copy of k-string
                total_occurrences = (frequency[i] // k)
 
                for j in range(total_occurrences):
                    single_copy += chr(i + 97)
 
    kString = ""
 
    # append the single copy formed k times
    for i in range(k):
        kString += single_copy
 
    return kString
 
 
# Driver Code
if __name__ == "__main__":
 
    st = "gkeekgee"
    K = 2
 
    kString = kStringGenerate(st, K)
    print(kString)
 
    # This code is contributed by ukasp.

C#

// C# program to form a
// K-Concatenated-String 
// from a given string
using System;
 
class GFG
{
     
// Function to print 
// the k-concatenated string
static String kStringGenerate(String str,
                            int k)
{
 
    // maintain a frequency table
    // for lowercase letters
    int []frequency = new int[26];
    int len = str.Length;
 
    for (int i = 0; i < len; i++) 
    {
 
        // for each character 
        // increment its frequency
        // in the frequency array
        frequency[str[i]- 'a']++;
    }
 
    // stores a single copy 
    // of a string, which on 
    // repeating forms a k-string
    String single_copy = "";
 
    // iterate over 
    // frequency array
    for (int i = 0; i < 26; i++) 
    {
 
        // if the character occurs 
        // in the string, check if 
        // it is divisible by k,
        // if not divisible then
        // k-string cant be formed
        if (frequency[i] != 0) 
        {
 
            if ((frequency[i] % k) != 0)
            {
                String ans = "Not Possible";
                return ans;
            }
            else
            {
 
                // ith character occurs 
                // (frequency[i]/k) times in 
                // a single copy of k-string
                int total_occurrences = (frequency[i] / k);
 
                for (int j = 0; 
                        j < total_occurrences; j++) 
                {
                    single_copy += (char)(i + 97);
                }
            }
        }
    }
 
    String kString = "";
 
    // append the single
    // copy formed k times
    for (int i = 0; i < k; i++) 
    {
        kString += single_copy;
    }
 
    return kString;
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "gkeekgee";
    int K = 2;
     
    String kString = kStringGenerate(str, K);
    Console.Write(kString);
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
      // JavaScript program to form a
      // K-Concatenated-String
      // from a given string
       
      // Function to print
      // the k-concatenated string
      function kStringGenerate(str, k) {
        // maintain a frequency table
        // for lowercase letters
        var frequency = new Array(26).fill(0);
        var len = str.length;
 
        for (var i = 0; i < len; i++) {
          // for each character
          // increment its frequency
          // in the frequency array
          frequency[str[i].charCodeAt(0) - "a".charCodeAt(0)]++;
        }
 
        // stores a single copy
        // of a string, which on
        // repeating forms a k-string
        var single_copy = "";
 
        // iterate over
        // frequency array
        for (var i = 0; i < 26; i++) {
          // if the character occurs
          // in the string, check if
          // it is divisible by k,
          // if not divisible then
          // k-string cant be formed
          if (frequency[i] != 0) {
            if (frequency[i] % k != 0) {
              var ans = "Not Possible";
              return ans;
            } else {
              // ith character occurs
              // (frequency[i]/k) times in
              // a single copy of k-string
              var total_occurrences = parseInt(frequency[i] / k);
 
              for (var j = 0; j < total_occurrences; j++) {
                single_copy += String.fromCharCode(i + 97);
              }
            }
          }
        }
 
        var kString = "";
 
        // append the single
        // copy formed k times
        for (var i = 0; i < k; i++) {
          kString += single_copy;
        }
 
        return kString;
      }
 
      // Driver Code
      var str = "gkeekgee";
      var K = 2;
 
      var kString = kStringGenerate(str, K);
      document.write(kString);
       
    </script>

Output:

eegkeegk

Time Complexity: O(N), Here N is the length of the string.

Auxiliary Space: O(N), Here N is the length of the string.

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