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 ` |

## Python 3

`# Python 3 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 ` |

**Output:**

geeksgeeks

This code has a time complexity of 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 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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