# Longest subsequence where every character appears at-least k times

• Difficulty Level : Basic
• Last Updated : 18 Jul, 2022

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
Explaination: 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. 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.

Implementation:

## C++

 `// C++ program to Find longest subsequence where``//  every character appears at-least k times``#include``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(s, k):   ` `    ``n ``=` `len``(s)                 `` ` `    ``# Count frequencies of all characters``    ``freq ``=` `[``0``]``*``MAX_CHARS                ``    ``for` `i ``in` `range``(n): ``        ``freq[``ord``(s[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``(s[i]) ``-` `ord``(``'a'``)] >``=` `k):              ``            ``print``(s[i],end``=``"")`` ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``s ``=` `"geeksforgeeks"``    ``k ``=` `2``    ``longestSubseqWithK(s, 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

 ``

Output

`geeksgeeks`

Time complexity: O(n), where n is the size of the string.
Auxiliary Space: O(n), where n is the length of the string. This is because when a string is passed to any function it is passed by value and creates a copy of itself in the stack.

Method 3 (Efficient way – Using HashMap) :

## C++14

 `// C++ program to Find longest subsequence where every``// character appears at-least k times``#include ``using` `namespace` `std;` `void` `longestSubseqWithK(string str, ``int` `k)``{``    ``int` `n = str.size();``    ``map<``char``, ``int``> mp;` `    ``// Count frequencies of all characters``    ``for` `(``int` `i = 0; i < n; i++) {``        ``char` `t = str[i];``        ``mp[t]++;``    ``}` `    ``// 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` `t = str[i];``        ``if` `(mp[t] >= k) {``            ``cout << t;``        ``}``    ``}``}` `// Driver code``int` `main()``{``    ``string str = ``"geeksforgeeks"``;``    ``int` `k = 2;``    ``longestSubseqWithK(str, k);` `    ``return` `0;``}` `// This code is contributed by rakeshsahni`

## 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 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`

## Python3

 `# Python3 program to Find longest subsequence where every``# character appears at-least k times``def` `longestSubseqWithK(``Str``, k):``  ``n ``=` `len``(``Str``)``  ``hm ``=` `{}` `  ``# Count frequencies of all characters``  ``for` `i ``in` `range``(n):``    ``c ``=` `Str``[i]``    ``if``(c ``in` `hm):``        ``hm ``+``=` `1``    ` `    ``else``:``        ``hm ``=` `1` `  ``# Traverse given string again and print``  ``# all those characters whose frequency``  ``# is more than or equal to k.``  ``for` `i ``in` `range``(n):``    ``c ``=` `Str``[i]``    ``if` `(hm >``=` `k):``      ``print``(c,end``=``"")``    ` `# Driver code``Str` `=` `"geeksforgeeks"``k ``=` `2``longestSubseqWithK(``Str``, k)` `# This code is contributed by shinjanpatra`

## Javascript

 `

Output

`geeksgeeks`

Time complexity: O(n*log(n)). This is because the time complexity of inserting an element in a map is O(log(n)).
Auxiliary Space: O(n), where n is the length of the string. This is because when the string is passed to any function it is passed by value and creates a copy of itself in the stack.

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.

My Personal Notes arrow_drop_up