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Construct a string that has exactly K subsequences from given string

Last Updated : 18 May, 2021
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Given a string str and an integer K, the task is to find a string S such that it has exactly K subsequences of given string str. 
Examples:  

Input: str = "gfg", K = 10 
Output: gggggffg 
Explanation: 
There are 10 possible subsequence of the given string "gggggffg". They are: 
1. gggggffg 
2. gggggffg 
3. gggggffg 
4. gggggffg 
5. gggggffg 
6. gggggffg 
7. gggggffg 
8. gggggffg 
9. gggggffg 
10. gggggffg.
Input: str = "code", K = 20 
Output: cccccoodde 
Explanation: 
There are 20 possible subsequence of the string "cccccoodde". 
 

Approach:
To solve the problem mentioned above we have to follow the steps given below:  

  • The idea is to find the prime factors of K and store the prime factors(say factors).
  • Create an empty array count of the size of the given string to store the count of each character in the resultant string s. Initialize the array with 1.
  • Now pop elements from the list factors and multiply to each position of array in a cyclic way until the list becomes empty. Finally, we have the count of each character of str in the array.
  • Iterate in the array count[] and append the number of characters for each character ch to the resultant string s.

Below is the implementation of the above approach:
 

C++ 
// C++ program for the above approach
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Function that computes the string s
void printSubsequenceString(string str,
                            long long k)
{
    // Length of the given string str
    int n = str.size();
    int i;

    // List that stores all the prime
    // factors of given k
    vector<long long> factors;

    // Find the prime factors
    for (long long i = 2;
         i <= sqrt(k); i++) {

        while (k % i == 0) {
            factors.push_back(i);
            k /= i;
        }
    }
    if (k > 1)
        factors.push_back(k);

    // Initialize the count of each
    // character position as 1
    vector<long long> count(n, 1);

    int index = 0;

    // Loop until the list
    // becomes empty
    while (factors.size() > 0) {

        // Increase the character
        // count by multiplying it
        // with the prime factor
        count[index++] *= factors.back();
        factors.pop_back();

        // If we reach end then again
        // start from beginning
        if (index == n)
            index = 0;
    }

    // Store the output
    string s;

    for (i = 0; i < n; i++) {
        while (count[i]-- > 0) {
            s += str[i];
        }
    }

    // Print the string
    cout << s;
}

// Driver code
int main()
{
    // Given String
    string str = "code";

    long long k = 20;

    // Function Call
    printSubsequenceString(str, k);
    return 0;
}
Java 
// Java program for the above approach 
import java.util.*;
class GFG{ 

// Function that computes the String s 
static void printSubsequenceString(String str, 
                                      int k) 
{ 
    // Length of the given String str 
    int n = str.length(); 
    int i; 

    // List that stores all the prime 
    // factors of given k 
    Vector<Integer> factors = new Vector<Integer>(); 

    // Find the prime factors 
    for (i = 2; i <= Math.sqrt(k); i++) 
    { 
        while (k % i == 0) 
        { 
            factors.add(i); 
            k /= i; 
        } 
    } 
    if (k > 1) 
        factors.add(k); 

    // Initialize the count of each 
    // character position as 1 
    int []count = new int[n]; 
    Arrays.fill(count, 1);
    int index = 0; 

    // Loop until the list 
    // becomes empty 
    while (factors.size() > 0) 
    { 

        // Increase the character 
        // count by multiplying it 
        // with the prime factor 
        count[index++] *= factors.get(factors.size() - 1); 
        factors.remove(factors.get(factors.size() - 1)); 

        // If we reach end then again 
        // start from beginning 
        if (index == n) 
            index = 0; 
    } 

    // Store the output 
    String s = ""; 

    for (i = 0; i < n; i++) 
    { 
        while (count[i]-- > 0) 
        { 
            s += str.charAt(i); 
        } 
    } 

    // Print the String 
    System.out.print(s); 
} 

// Driver code 
public static void main(String[] args) 
{ 
    // Given String 
    String str = "code"; 

    int k = 20; 

    // Function Call 
    printSubsequenceString(str, k); 
}
} 

// This code is contributed by sapnasingh4991
Python3 
# Python3 program for 
# the above approach
import math

# Function that computes 
# the string s
def printSubsequenceString(st, k):
    # Length of the given 
    # string str
    n = len(st)

    # List that stores 
    # all the prime
    # factors of given k
    factors = []
  
    # Find the prime factors
    sqt = (int(math.sqrt(k)))
    for i in range (2, sqt + 1):

        while (k % i == 0):
            factors.append(i)
            k //= i
   
    if (k > 1):
        factors.append(k)

    # Initialize the count of each
    # character position as 1
    count = [1] * n

    index = 0

    # Loop until the list
    # becomes empty
    while (len(factors) > 0):

        # Increase the character
        # count by multiplying it
        # with the prime factor
        count[index] *= factors[-1]
        factors.pop()
        index += 1

        # If we reach end then again
        # start from beginning
        if (index == n):
            index = 0
            
    # store output
    s = ""
    for i in range (n):
        while (count[i] > 0):
            s += st[i]
            count[i] -= 1  
       
    # Print the string
    print (s)

# Driver code
if __name__ == "__main__":
  
    # Given String
    st = "code"

    k = 20

    # Function Call
    printSubsequenceString(st, k)
    
# This code is contributed by Chitranayal
C# 
// C# program for the above approach 
using System;
using System.Collections.Generic;

class GFG{ 

// Function that computes the String s 
static void printSubsequenceString(String str, 
                                      int k) 
{ 
    // Length of the given String str 
    int n = str.Length; 
    int i; 

    // List that stores all the prime 
    // factors of given k 
    List<int> factors = new List<int>(); 

    // Find the prime factors 
    for (i = 2; i <= Math.Sqrt(k); i++) 
    { 
        while (k % i == 0) 
        { 
            factors.Add(i); 
            k /= i; 
        } 
    } 
    if (k > 1) 
        factors.Add(k); 

    // Initialize the count of each 
    // character position as 1 
    int []count = new int[n]; 
    for (i = 0; i < n; i++) 
        count[i] = 1;
    int index = 0; 

    // Loop until the list 
    // becomes empty 
    while (factors.Count > 0) 
    { 

        // Increase the character 
        // count by multiplying it 
        // with the prime factor 
        count[index++] *= factors[factors.Count - 1]; 
        factors.Remove(factors[factors.Count - 1]); 

        // If we reach end then again 
        // start from beginning 
        if (index == n) 
            index = 0; 
    } 

    // Store the output 
    String s = ""; 

    for (i = 0; i < n; i++) 
    { 
        while (count[i]-- > 0) 
        { 
            s += str[i]; 
        } 
    } 

    // Print the String 
    Console.Write(s); 
} 

// Driver code 
public static void Main(String[] args) 
{ 
    // Given String 
    String str = "code"; 

    int k = 20; 

    // Function Call 
    printSubsequenceString(str, k); 
}
} 

// This code is contributed by sapnasingh4991
JavaScript 
<script>

// Javascript program for the above approach

// Function that computes the string s
function printSubsequenceString(str, k)
{
    // Length of the given string str
    let n = str.length;
    let i;

    // List that stores all the prime
    // factors of given k
    let factors = new Array();

    // Find the prime factors
    for (let i = 2;
        i <= Math.sqrt(k); i++) {

        while (k % i == 0) {
            factors.push(i);
            k /= i;
        }
    }
    if (k > 1)
        factors.push(k);

    // Initialize the count of each
    // character position as 1
    let count = new Array(n).fill(1);

    let index = 0;

    // Loop until the list
    // becomes empty
    while (factors.length > 0) {

        // Increase the character
        // count by multiplying it
        // with the prime factor
        count[index++] *= factors[factors.length - 1];
        factors.pop();

        // If we reach end then again
        // start from beginning
        if (index == n)
            index = 0;
    }

    // Store the output
    let s = new String();

    for (i = 0; i < n; i++) {
        while (count[i]-- > 0) {
            s += str[i];
        }
    }

    // Print the string
    document.write(s);
}

// Driver code

// Given String
let str = "code";

let k = 20;

// Function Call
printSubsequenceString(str, k);

// This code is contributed by _saurabh_jaiswal

</script>

Output: 
cccccoodde

 

Time Complexity: O(N*log2(log2(N))) 
Auxiliary Space: O(K)


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