Given string str with uppercase characters and an integer K, the task is to find the length of the longest subsequence such that the frequency of the first K alphabet is the same.
Examples:
Input: str = “ACAABCCAB”, K=3
Output: 6
Explanation: One of the possible subsequences is “ACABCB”.Input: str = “ACAABCCAB”, K=4
Output: 0
Explanation: Since the string does not contain ‘D’, no such subsequence can be obtained.
Approach:
Traverse the string and find the least frequent of the first K alphabets. Once found, (frequency of that element) * K gives the desired result.
Below is the implementation of the above approach:
C++
// C++ program to find the longest // subsequence with first K // alphabets having same frequency #include <bits/stdc++.h> using namespace std; // Function to return the // length of the longest // subsequence with first K // alphabets having same frequency int lengthOfSubsequence(string str, int K) { // Map to store frequency // of all characters in // the string map<char,int> mp; for (char ch : str) { mp[ch]++; } // Variable to store the // frequency of the least // frequent of first K // alphabets int minimum = mp['A']; for (int i = 1; i < K; i++) { minimum = min(minimum, mp[(char)(i + 'A')]); } return minimum * K; } int main() { string str = "ACAABCCAB"; int K = 3; cout << lengthOfSubsequence(str, K); return 0; } |
Java
// Java program to find the longest // subsequence with first K alphabets // having same frequency import java.util.*; class GFG{ // Function to return the // length of the longest // subsequence with first K // alphabets having same frequency static int lengthOfSubsequence(String str, int K) { // Map to store frequency // of all characters in // the string Map<Character, Integer> mp = new HashMap<>(); for(char ch : str.toCharArray()) { mp.put(ch, mp.getOrDefault(ch, 0) + 1); } // Variable to store the // frequency of the least // frequent of first K // alphabets int minimum = mp.get('A'); for(int i = 1; i < K; i++) { minimum = Math.min(minimum, mp.get((char)(i + 'A'))); } return minimum * K; } // Driver code public static void main(String[] args) { String str = "ACAABCCAB"; int K = 3; System.out.println(lengthOfSubsequence(str, K)); } } // This code is contributed by offbeat |
Python3
# Python3 program to find the longest # subsequence with first K alphabets # having same frequency from collections import defaultdict # Function to return the # length of the longest # subsequence with first K # alphabets having same frequency def lengthOfSubsequence(st, K): # Map to store frequency # of all characters in # the string mp = defaultdict(int) for ch in st: mp[ch] += 1 # Variable to store the # frequency of the least # frequent of first K # alphabets minimum = mp['A'] for i in range(1, K): minimum = min(minimum, mp[chr(i + ord('A'))]) return (minimum * K) # Driver code if __name__ == "__main__": st = "ACAABCCAB" K = 3 print(lengthOfSubsequence(st, K)) # This code is contributed by chitranayal |
C#
// C# program to find the longest// subsequence with first K alphabets// having same frequencyusing System;using System.Collections.Generic;class GFG{// Function to return the// length of the longest// subsequence with first K// alphabets having same frequencystatic int lengthOfSubsequence(string str, int K){ // Store frequency // of all characters in // the string Dictionary<char, int> mp = new Dictionary<char, int>(); foreach(char ch in str.ToCharArray()) { mp[ch] = mp.GetValueOrDefault(ch, 0) + 1; } // Variable to store the frequency // of the least frequent of first K // alphabets int minimum = mp['A']; for(int i = 1; i < K; i++) { minimum = Math.Min(minimum, mp[(char)(i + 'A')]); } return minimum * K;}// Driver codepublic static void Main(string[] args){ string str = "ACAABCCAB"; int K = 3; Console.Write(lengthOfSubsequence(str, K));}}// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript program to find the longest // subsequence with first K // alphabets having same frequency // Function to return the // length of the longest // subsequence with first K // alphabets having same frequency function lengthOfSubsequence(str, K) { // Map to store frequency // of all characters in // the string var mp = new Map(); str.split('').forEach(ch => { if(mp.has(ch)) mp.set(ch, mp.get(ch)+1) else mp.set(ch, 1) }); // Variable to store the // frequency of the least // frequent of first K // alphabets var minimum = mp.get('A'); for (var i = 1; i < K; i++) { minimum = Math.min(minimum, mp.get(String.fromCharCode(i + 'A'.charCodeAt(0)))); } return minimum * K; } var str = "ACAABCCAB"; var K = 3; document.write( lengthOfSubsequence(str, K)); </script> |
Output:
6
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