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Longest subsequence where every character appears at-least k times

  • Difficulty Level : Basic
  • Last Updated : 30 Jul, 2021
Geek Week
 

Given a string and a number k, find the longest subsequence of a string where every character appears at-least k times.

Examples: 

Input : str = "geeksforgeeks"
         k = 2
Output : geeksgeeks
Every character in the output
subsequence appears at-least 2
times.

Input : str = "aabbaabacabb"
          k = 5
Output : aabbaabaabb

Method 1 (Brute force) 
We generate all subsequences. For every subsequence count distinct characters in it and find the longest subsequence where every character appears at-least k times.

Method 2 (Efficient way) 
1. Find the frequency of the string and store it in an integer array of size 26 representing the alphabets. 
2. After finding the frequency iterate the string character by character and if the frequency of that character is greater than or equal to the required number of repetitions then print that character then and there only. 

C++




// C++ program to Find longest subsequence where
//  every character appears at-least k times
#include<bits/stdc++.h>
using namespace std;
 
const int MAX_CHARS = 26;
 
void longestSubseqWithK(string str, int k)   
{
    int n = str.size();                  
 
    // Count frequencies of all characters
    int freq[MAX_CHARS] = {0};                   
    for (int i = 0 ; i < n; i++)   
        freq[str[i] - 'a']++;             
     
    // Traverse given string again and print
    // all those characters whose frequency
    // is more than or equal to k.
    for (int i = 0 ; i < n ; i++)  
        if (freq[str[i] - 'a'] >= k)              
            cout << str[i];     
}
 
// Driver code
int main() {
    string str = "geeksforgeeks";
    int k = 2;
    longestSubseqWithK(str, k);      
    return 0;
}

Java




// Java program to Find longest subsequence where
//  every character appears at-least k times
 
class GFG {
 
    static final int MAX_CHARS = 26;
 
    static void longestSubseqWithK(String str, int k) {
        int n = str.length();
 
        // Count frequencies of all characters
        int freq[] = new int[MAX_CHARS];
        for (int i = 0; i < n; i++) {
            freq[str.charAt(i) - 'a']++;
        }
 
        // Traverse given string again and print
        // all those characters whose frequency
        // is more than or equal to k.
        for (int i = 0; i < n; i++) {
            if (freq[str.charAt(i) - 'a'] >= k) {
                System.out.print(str.charAt(i));
            }
        }
    }
 
// Driver code
    static public void main(String[] args) {
        String str = "geeksforgeeks";
        int k = 2;
        longestSubseqWithK(str, k);
 
    }
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program to Find longest subsequence where
#  every character appears at-least k times
  
MAX_CHARS = 26
  
def longestSubseqWithK(str, k):   
 
    n = len(str)                 
  
    # Count frequencies of all characters
    freq = [0]*MAX_CHARS                
    for i in range(n): 
        freq[ord(str[i]) - ord('a')]+=1             
      
    # Traverse given string again and print
    # all those characters whose frequency
    # is more than or equal to k.
    for i in range(n ):
        if (freq[ord(str[i]) - ord('a')] >= k):              
            print(str[i],end="")
  
# Driver code
if __name__ == "__main__":
     
    str = "geeksforgeeks"
    k = 2
    longestSubseqWithK(str, k)

C#




     
// C# program to Find longest subsequence where
//  every character appears at-least k times
using System;
public class GFG {
  
    static readonly int MAX_CHARS = 26;
  
    static void longestSubseqWithK(String str, int k) {
        int n = str.Length;
  
        // Count frequencies of all characters
        int []freq = new int[MAX_CHARS];
        for (int i = 0; i < n; i++) {
            freq[str[i]- 'a']++;
        }
  
        // Traverse given string again and print
        // all those characters whose frequency
        // is more than or equal to k.
        for (int i = 0; i < n; i++) {
            if (freq[str[i] - 'a'] >= k) {
                Console.Write(str[i]);
            }
        }
    }
  
// Driver code
    static public void Main() {
        String str = "geeksforgeeks";
        int k = 2;
        longestSubseqWithK(str, k);
  
    }
}
  
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript program to find longest subsequence
// where every character appears at-least k times
let MAX_CHARS = 26;
 
function longestSubseqWithK(str, k)
{
    let n = str.length;
 
    // Count frequencies of all characters
    let freq = new Array(MAX_CHARS);
    for(let i = 0; i < freq.length; i++)
    {
        freq[i] = 0
    }
     
    for(let i = 0; i < n; i++)
    {
        freq[str[i].charCodeAt(0) -
                'a'.charCodeAt(0)]++;
    }
 
    // Traverse given string again and print
    // all those characters whose frequency
    // is more than or equal to k.
    for(let i = 0; i < n; i++)
    {
        if (freq[str[i].charCodeAt(0) -
                    'a'.charCodeAt(0)] >= k)
        {
            document.write(str[i]);
        }
    }
}
 
// Driver code
let str = "geeksforgeeks";
let k = 2;
 
longestSubseqWithK(str, k);
 
// This code is contributed by avanitrachhadiya2155
     
</script>
Output



geeksgeeks

This code has a time complexity of O(n) where n is the size of the string. 

Method 3 (Efficient way – Using HashMap) 

Java




/*package whatever //do not write package name here */
// Java program to Find longest subsequence where every
// character appears at-least k times
 
import java.io.*;
import java.util.HashMap;
 
class GFG {
 
    static void longestSubseqWithK(String str, int k)
    {
        int n = str.length();
        HashMap<Character, Integer> hm = new HashMap<>();
 
        // Count frequencies of all characters
        for (int i = 0; i < n; i++) {
            char c = str.charAt(i);
            if (hm.containsKey(c))
                hm.put(c, hm.get(c) + 1);
            else
                hm.put(c, 1);
        }
 
        // Traverse given string again and print
        // all those characters whose frequency
        // is more than or equal to k.
        for (int i = 0; i < n; i++) {
            char c = str.charAt(i);
            if (hm.get(c) >= k) {
                System.out.print(c);
            }
        }
    }
 
    // Driver code
    static public void main(String[] args)
    {
        String str = "geeksforgeeks";
        int k = 2;
        longestSubseqWithK(str, k);
    }
}
 
// This code is contributed by Chandan-Bedi
Output
geeksgeeks

Time complexity: O(n) where n is the size of the string. 

This article is contributed by Mohak Agrawal. 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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