Given a string ‘s’ and an integer k, find other string ‘t’ such that ‘t’ is the largest subsequence of given string ‘s’ and each character of ‘t’ must occur at least k times in string s.
Examples :
Input : s = "neveropen"
k = 2
Output : neveropenneveropen
Input : s = "baaabaacba"
k = 3
Output : baaabaaba
A simple solution is to generate all subsequences. For every subsequence, check if it has all characters at least k times. Find the longest such subsequence. The time complexity of this approach is exponential.
Efficient Approach We can take another array to keep the record of the count of each character from string s, if any character occurred more than or equal to k times, then we simply print it.
Implementation:
CPP
// CPP Program to find the subsequence// with each character occurring at least// k times in string str#include <iostream>using namespace std;#define MAX_CHAR 26// Function to find the subsequencevoid findSubsequence(string str, int k){ // Taking an extra array to keep // record for character count in s int a[MAX_CHAR] = { 0 }; // Counting occurrences of all // characters in str[] for (int i = 0; i < str.size(); i++) a[str[i] - 'a']++; // Printing characters with count // >= k in same order as they appear // in str. for (int i = 0; i < str.size(); i++) if (a[str[i] - 'a'] >= k) cout << str[i]; }// Driver codeint main(){ int k = 2; findSubsequence("neveropen", k); return 0;} |
Java
// Java Program to find the subsequence// with each character occurring at least// k times in string sclass GFG { static final int MAX_CHAR = 26; // Function to find the subsequence static void findSubsequence(String str, int k) { // Taking an extra array to keep // record for character count in s int a[] = new int[MAX_CHAR]; // Counting occurrences of all // characters in str[] for (int i = 0; i < str.length(); i++) a[str.charAt(i) - 'a']++; // Printing characters with count // >= k in same order as they appear // in str. for (int i = 0; i < str.length(); i++) if (a[str.charAt(i) - 'a'] >= k) System.out.print(str.charAt(i)); } // Driver code public static void main(String[] args) { int k = 2; findSubsequence("neveropen", k); }}// This code is contributed by Anant Agarwal. |
Python3
# Python Program to find the subsequence# with each character occurring at least# k times in string sMAX_CHAR = 26# Function to find the subsequencedef findSubsequence(stri, k): # Taking an extra array to keep # record for character count in s a = [0] * MAX_CHAR; # Counting occurrences of all # characters in str[] for i in range(len(stri)): a[ord(stri[i]) - ord('a')] += 1 # Printing characters with count # >= k in same order as they appear # in str. for i in range(len(stri)): if a[ord(stri[i]) - ord('a')] >= k: print(stri[i],end='') # Driver codek = 2findSubsequence("neveropen", k)# This code is contributed by Shubham Rana |
C#
// C# Program to find the subsequence// with each character occurring at // least k times in string susing System;class GFG { static int MAX_CHAR = 26; // Function to find the subsequence static void findSubsequence(string str, int k) { // Taking an extra array to keep // record for character count in s int []a = new int[MAX_CHAR]; // Counting occurrences of all // characters in str[] for (int i = 0; i < str.Length; i++) a[str[i] - 'a']++; // Printing characters with count // >= k in same order as they appear // in str. for (int i = 0; i < str.Length; i++) if (a[str[i] - 'a'] >= k) Console.Write(str[i]); } // Driver code public static void Main() { int k = 2; findSubsequence("neveropen", k); }}// This code is contributed by vt_m. |
PHP
<?php// PHP Program to find the // subsequence with each // character occurring at // least k times in string s// Function to find the subsequencefunction findSubsequence($str, $k) { // Taking an extra array to keep // record for character count in s $a = array(1024); for($i = 0; $i < 26; $i++) $a[$i] = 0; // Counting occurrences of all // characters in str[] for ($i = 0; $i < strlen($str); $i++) { $temp = ord($str[$i]) - ord('a'); $a[$temp] += 1; } // Printing characters with // count >= k in same order // as they appear in str. for ($i = 0; $i < strlen($str); $i++) if ($a[ord($str[$i]) - ord('a')] >= $k) echo $str[$i];}// Driver code$k = 2;findSubsequence("neveropen", $k); // This code is contributed by Sam007?> |
Javascript
<script>// Javascript Program to find the subsequence// with each character occurring at least// k times in string svar MAX_CHAR = 26;// Function to find the subsequencefunction findSubsequence(str, k){ // Taking an extra array to keep // record for character count in s var a = Array(MAX_CHAR).fill(0); // Counting occurrences of all // characters in str[] for (var i = 0; i < str.length; i++) a[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++; // Printing characters with count // >= k in same order as they appear // in str. for (var i = 0; i < str.length; i++) if (a[str[i].charCodeAt(0) - 'a'.charCodeAt(0)] >= k) document.write( str[i]); }// Driver codevar k = 2;findSubsequence("neveropen", k);</script> |
Output
neveropenneveropen
Time Complexity: O(n), where n is the size of the given string.
Auxiliary Space: O(1)
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