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Minimum cost for constructing the subsequence of length K from given string S

Last Updated : 06 Feb, 2023
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Given a string S consisting of N lowercase English alphabets, and an integer K and, an array cost[] of size 26 denoting the cost of each lowercase English alphabet, the task is to find the minimum cost to construct a subsequence of length K from the characters of the string S.

Examples:

Input: S = "aabcbc", K = 3, cost[] = {2, 1, 3, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9}
Output: 4
Explanation:
One way to construct a string of size K(= 3) is:
Form the string "abb" taking two 'b's with a cost of (2*1 = 2), and one 'a' with a cost of (1*2 = 2).
Therefore, the total cost to construct the string "abb" is (2 + 2 = 4), which is the minimum possible.

Input: S = "aaaca", K = 1, cost[] = {2, 1, 3, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9}
Output: 2

Approach: The given problem can be solved by sorting the array cost[] in increasing order and include the first K characters having the minimum cost that are present in the given string S. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

C++ 
// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Custom comparator to sort according
// to the second element
bool comparator(pair<char, int> p1,
                pair<char, int> p2)
{
    return p1.second < p2.second;
}

// Function to find the minimum cost
// to construct a subsequence of the
// length K
int minimumCost(string S, int N,
                int K, int cost[])
{
    // Stores the minimum cost
    int minCost = 0;

    // Stores the pair of character
    // and the cost of that character
    vector<pair<char, int> > V;

    // Stores the frequency of each
    // character
    unordered_map<char, int> mp;

    // Iterate in the range [0, N-1]
    for (int i = 0; i < N; i++)
        mp[S[i]]++;

    // Iterate in the range [0, 25]
    for (int i = 0; i < 26; i++) {
        V.push_back({ char('a' + i), cost[i] });
    }

    // Sort the vector of pairs V
    // wrt the second element
    sort(V.begin(), V.end(), comparator);

    // Iterate in the range [0, 25]
    for (int i = 0; i < 26; i++) {

        // Stores the frequency of the
        // current char in the string
        int count = mp[char('a' + i)];

        // If count is less than
        // or equal to K
        if (count >= K) {

            // Update the value of
            // minCost
            minCost += V[i].second * K;
            break;
        }
        else if (count < K) {

            // Update the value of
            // minCost
            minCost += V[i].second * count;

            // Update the value
            // of K
            K = K - count;
        }
    }

    // Print the value of minCost
    return minCost;
}

// Driver Code
int main()
{
    string S = "aabcbc";
    int K = 3;
    int cost[26] = { 2, 1, 3, 9, 9, 9, 9,
                     9, 9, 9, 9, 9, 9, 9,
                     9, 9, 9, 9, 9, 9, 9,
                     9, 9, 9, 9, 9 };
    int N = S.length();
    cout << minimumCost(S, N, K, cost);

    return 0;
}
Java 
// Java program for the above approach

import java.util.*;

public class Main {
    
    
    static class Pair {
        char first;
        int second;

        Pair(char first, int second) {
            this.first = first;
            this.second = second;
        }
    }
    
    

    // Custom comparator to sort according
    // to the second element
    public static Comparator<Pair> comparator = new Comparator<Pair>() {
        @Override
        public int compare(Pair p1, Pair p2) {
            return p1.second - p2.second;
        }
    };

    public static int minimumCost(String S, int N, int K, int[] cost) {
        // Stores the minimum cost
        int minCost = 0;

        // Stores the pair of character
        // and the cost of that character
        List<Pair> V = new ArrayList<>();

        // Stores the frequency of each
        // character
        Map<Character, Integer> mp = new HashMap<>();

        // Iterate in the range [0, N-1]
        for (int i = 0; i < N; i++) {
            if (mp.containsKey(S.charAt(i))) {
                mp.put(S.charAt(i), mp.get(S.charAt(i)) + 1);
            } else {
                mp.put(S.charAt(i), 1);
            }
        }

        // Iterate in the range [0, 25]
        for (int i = 0; i < 26; i++) {
            if(mp.containsKey((char)('a' + i))){
                V.add(new Pair((char)('a' + i), cost[i]));
            }
        }

        // Sort the list of pairs V
        // wrt the second element
        Collections.sort(V, comparator);

        // Iterate in the range [0, 25]
        for (Pair p : V) {

            // Stores the frequency of the
            // current char in the string
            int count = mp.get(p.first);

            // If count is less than
            // or equal to K
            if (count >= K) {

                // Update the value of
                // minCost
                minCost += p.second * K;
                break;
            } else if (count < K) {

                // Update the value of
                // minCost
                minCost += p.second * count;

                // Update the value
                // of K
                K = K - count;
            }
        }

        // Return the value of minCost
        return minCost;
    }
    
    // Driver Code
    public static void main(String[] args) {
        
        String S = "aabcbc";
        int K = 3;
        int[] cost = { 2, 1, 3, 9, 9, 9, 9,
                       9, 9, 9, 9, 9, 9, 9,
                       9, 9, 9, 9, 9, 9, 9,
                       9, 9, 9, 9, 9 };
        int N = S.length();
        System.out.println(minimumCost(S, N, K, cost));
    }

    
}
Python3 
# Python3 program for the above approach

# Function to find the minimum cost
# to construct a subsequence of the
# length K
def minimumCost(S, N, K, cost):
    
    # Stores the minimum cost
    minCost = 0

    # Stores the pair of character
    # and the cost of that character
    V = []

    # Stores the frequency of each
    # character
    mp = {}

    # Iterate in the range [0, N-1]
    for i in range(N):
        if (S[i] in mp):
            mp[S[i]] += 1
        else:
            mp[S[i]] = 1

    # Iterate in the range [0, 25]
    for i in range(26):
        V.append([chr(ord('a') + i), cost[i]])

    # Sort the array of pairs V
    # with the second element
    while (1):
        flag = 0

        for i in range(len(V) - 1):
            if (V[i][1] > V[i + 1][1]):
                temp = V[i]
                V[i] = V[i + 1]
                V[i + 1] = temp
                flag = 1

        if (flag == 0):
            break

    # Iterate in the range [0, 25]
    for i in range(26):

        # Stores the frequency of the
        # current char in the string
        count = mp[chr(ord('a') + i)]

        # If count is less than
        # or equal to K
        if (count >= K):

            # Update the value of
            # minCost
            minCost += V[i][1] * K
            break

        elif (count < K):

            # Update the value of
            # minCost
            minCost += V[i][1] * count

            # Update the value
            # of K
            K = K - count

    # Print the value of minCost
    return minCost

# Driver Code
S = "aabcbc"
K = 3
cost = [ 2, 1, 3, 9, 9, 9, 9,
         9, 9, 9, 9, 9, 9, 9,
         9, 9, 9, 9, 9, 9, 9,
         9, 9, 9, 9, 9 ]
N = len(S)

print(minimumCost(S, N, K, cost))

# This code is contributed by _saurabh_jaiswal
C# 
// C# implementation of the approach

using System;
using System.Collections.Generic;

class GFG {
  public class Pair {
    public char first;
    public int second;

    public Pair(char first, int second)
    {
      this.first = first;
      this.second = second;
    }
  }

  public static int minimumCost(string S, int N, int K,
                                int[] cost)
  {
    // Stores the minimum cost
    int minCost = 0;

    // Stores the pair of character
    // and the cost of that character
    List<Pair> V = new List<Pair>();

    // Stores the frequency of each
    // character
    Dictionary<char, int> mp
      = new Dictionary<char, int>();

    // Iterate in the range [0, N-1]
    for (int i = 0; i < N; i++) {
      if (mp.ContainsKey(S[i])) {
        mp[S[i]] = mp[S[i]] + 1;
      }
      else {
        mp.Add(S[i], 1);
      }
    }

    // Iterate in the range [0, 25]
    for (int i = 0; i < 26; i++) {
      if (mp.ContainsKey((char)('a' + i))) {
        V.Add(new Pair((char)('a' + i), cost[i]));
      }
    }

    // Sort the list of pairs V
    // wrt the second element
    V.Sort((x, y) => x.second.CompareTo(y.second));

    // Iterate in the range [0, 25]
    foreach(Pair p in V)
    {
      // Stores the frequency of the
      // current char in the string
      int count = mp[p.first];

      // If count is less than
      // or equal to K
      if (count >= K) {
        // Update the value of
        // minCost
        minCost += p.second * K;
        break;
      }
      else if (count < K) {
        // Update the value of
        // minCost
        minCost += p.second * count;

        // Update the value
        // of K
        K = K - count;
      }
    }

    // Return the value of minCost
    return minCost;
  }

  // Driver code
  static void Main(string[] args)
  {
    string S = "aabcbc";
    int K = 3;
    int[] cost
      = { 2, 1, 3, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9,
         9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9 };
    int N = S.Length;
    Console.WriteLine(minimumCost(S, N, K, cost));
  }
}

// This code is contributed by ishankhandelwals.
JavaScript 
<script>
// Javascript program for the above approach


// Function to find the minimum cost
// to construct a subsequence of the
// length K
function minimumCost(S, N, K, cost) {
    // Stores the minimum cost
    let minCost = 0;

    // Stores the pair of character
    // and the cost of that character
    let V = [];

    // Stores the frequency of each
    // character
    let mp = new Map();

    // Iterate in the range [0, N-1]
    for (let i = 0; i < N; i++) {
        if (mp.has(S[i])) {
            mp.set(S[i], mp.get(S[i]) + 1)
        } else {
            mp.set(S[i], 1)
        }
    }

    // Iterate in the range [0, 25]
    for (let i = 0; i < 26; i++) {
        V.push([String.fromCharCode('a'.charCodeAt(0) + i), cost[i]]);
    }

    // Sort the array of pairs V
    // with the second element

    while (1) {
        let flag = 0;

        for (let i = 0; i < V.length - 1; i++) {
            if (V[i][1] > V[i + 1][1]) {
                let temp = V[i];
                V[i] = V[i + 1];
                V[i + 1] = temp;
                flag = 1
            }
        }
        if(flag == 0){
            break
        }
    }


    // Iterate in the range [0, 25]
    for (let i = 0; i < 26; i++) {

        // Stores the frequency of the
        // current char in the string
        let count = mp.get(String.fromCharCode('a'.charCodeAt(0) + i));

        // If count is less than
        // or equal to K
        if (count >= K) {

            // Update the value of
            // minCost
            minCost += V[i][1] * K;
            break;
        }
        else if (count < K) {

            // Update the value of
            // minCost
            minCost += V[i][1] * count;

            // Update the value
            // of K
            K = K - count;
        }
    }

    // Print the value of minCost
    return minCost;
}

// Driver Code

let S = "aabcbc";
let K = 3;
let cost = [2, 1, 3, 9, 9, 9, 9,
    9, 9, 9, 9, 9, 9, 9,
    9, 9, 9, 9, 9, 9, 9,
    9, 9, 9, 9, 9];
let N = S.length;
document.write(minimumCost(S, N, K, cost));

// This code is contributed by gfgking.
</script>

Output
4

Time Complexity: O(N)
Auxiliary Space: O(1)


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