Find the length of the longest subsequence with first K alphabets having same frequency

Given a 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 first K alphabets is 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++

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// 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;
}

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Python3

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# 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

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Output:

6

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Improved By : chitranayal