Maximize the happiness of the groups on the Trip

A trip to mystical land is going to be organized in ByteLand, the city of Bytes. Unfortunately, there are limited seats say A and there are N number of groups of people. Every group can have old person o, child c, man m and woman w. The organizing committee wants to maximize the happiness value of the trip. Happiness value of the trip is the sum of the happiness value of all the groups that are going. A group will go for the trip if every member can get a seat (Breaking a group is not a good thing).

  1. The happiness of child c = 4
  2. The happiness of woman w = 3
  3. The happiness of man m = 2
  4. The happiness of the old person o = 1

The happiness of group G, H(G) = (sum of happiness of people in it) * (number of people in the group).
The happiness of the group (‘coow’) = (4 + 1 + 1 + 3) * 4 = 36.

Given the groups and the total seating capacity, the task is to maximize the happiness and print the maximized happiness of the groups going on the trip.



Examples:

Input: groups[] = {“mmo”, “oo”, “cmw”, “cc”, “c”}, A = 5
Output: 43
Pick these groups [‘cmw’, ‘cc’] to get the maximum profit of (4 + 2 + 3) * 3 + (4 + 4) * 2 = 43

Input: groups[] = {“ccc”, “oo”, “cm”, “mm”, “wwo”}, A = 10
Output: 77

Approach: The problem can be considered as a slight modification of the 0-1 knapsack problem. The total seats available can be considered as the size of the knapsack. The happiness of each group can be considered as the profit of each item and the number of people in each group can be considered as the weight of each item. Now similar to the dynamic programming approach for 0-1 knapsack problem apply dynamic programming here to get the maximum happiness.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the maximized happiness
int MaxHappiness(int A, int N, vector<string> v)
{
    string str;
  
    // Two arrays similar to
    // 0 1 knapsack problem
    int val[N], wt[N], c = 0;
    for (int i = 0; i < N; i++) {
        str = v[i];
  
        // To store the happiness
        // of the current group
        c = 0;
        for (int j = 0; str[j]; j++) {
  
            // Current person is a child
            if (str[j] == 'c')
                c += 4;
  
            // Woman
            else if (str[j] == 'w')
                c += 3;
  
            // Man
            else if (str[j] == 'm')
                c += 2;
  
            // Old person
            else
                c++;
        }
  
        // Group's happiness is the sum of happiness
        // of the people in the group multiplied by
        // the number of people
        c *= str.length();
        val[i] = c;
        wt[i] = str.length();
    }
  
    // Solution using 0 1 knapsack
    int k[N + 1][A + 1];
    for (int i = 0; i <= N; i++) {
        for (int w = 0; w <= A; w++) {
            if (i == 0 || w == 0)
                k[i][w] = 0;
            else if (wt[i - 1] <= w)
                k[i][w] = max(val[i - 1]
                                  + k[i - 1][w - wt[i - 1]],
                              k[i - 1][w]);
            else
                k[i][w] = k[i - 1][w];
        }
    }
    return k[N][A];
}
  
// Driver code
int main()
{
  
    // Number of seats
    int A = 5;
  
    // Groups
    vector<string> v = { "mmo", "oo", "cmw", "cc", "c" };
    int N = v.size();
    cout << MaxHappiness(A, N, v);
  
    return 0;
}

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Java

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// Java implementation of the approach 
class GFG 
{
    // Function to return the maximized happiness
    static int maxHappiness(int A, int N, String[] v) 
    {
        String str;
  
        // Two arrays similar to
        // 0 1 knapsack prolem
        int[] val = new int[N];
        int[] wt = new int[N];
        int c = 0;
        for (int i = 0; i < N; i++) 
        {
            str = v[i];
  
            // To store the happiness
            // of the current goup
            c = 0;
            for (int j = 0; j < str.length(); j++) 
            {
                // Current person is a child
                if (str.charAt(j) == 'c')
                    c += 4;
  
                // Woman
                else if (str.charAt(j) == 'w')
                    c += 3;
  
                // Man
                else if (str.charAt(j) == 'm')
                    c += 2;
  
                // Old Person
                else
                    c++;
            }
  
            // Group's happiness is the sum of happiness
            // of the people in the group multiplie
            // the number of people
            c *= str.length();
            val[i] = c;
            wt[i] = str.length();
        }
  
        // Solution using 0 1 knapsack
        int[][] k = new int[N + 1][A + 1];
        for (int i = 0; i <= N; i++) 
        {
            for (int w = 0; w <= A; w++) 
            {
                if (i == 0 || w == 0)
                    k[i][w] = 0;
                else if (wt[i - 1] <= w) 
                {
                    k[i][w] = Math.max(val[i - 1]+ k[i - 1][w - wt[i - 1]], k[i-1][w]);
                
                else 
                {
                    k[i][w] = k[i - 1][w];
                }
            }
        }
        return k[N][A];
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        // Number of seats
        int A = 5;
  
        // Groups
        String[] v = { "mmo", "oo", "cmw", "cc", "c" };
        int N = v.length;
        System.out.println(maxHappiness(A, N, v));
    }
}
  
// This code is contributed by Vivek Kumar Singh

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Python3

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# Python3 implementation of the approach 
import numpy as np
  
# Function to return the maximized happiness 
def MaxHappiness(A, N, v) :
      
    # Two arrays similar to 
    # 0 1 knapsack problem 
    val = [0] * N; wt = [0] * N; c = 0
      
    for i in range(N) :
        string = v[i]; 
  
        # To store the happiness 
        # of the current group 
        c = 0
        for j in range(len(string)) :
  
            # Current person is a child 
            if (string[j] == 'c') :
                c += 4
  
            # Woman 
            elif (string[j] == 'w') :
                c += 3
  
            # Man 
            elif (string[j] == 'm') :
                c += 2
  
            # Old person 
            else :
                c += 1
  
        # Group's happiness is the sum of happiness 
        # of the people in the group multiplied by 
        # the number of people 
        c *= len(string); 
          
        val[i] = c;
          
        wt[i] = len(string); 
  
    # Solution using 0 1 knapsack 
    k = np.zeros((N + 1, A + 1))
      
    for i in range(N + 1) :
          
        for w in range(A + 1) :
            if (i == 0 or w == 0) :
                k[i][w] = 0
            elif (wt[i - 1] <= w) :
                k[i][w] = max(val[i - 1] + 
                                k[i - 1][w - wt[i - 1]], 
                                k[i - 1][w]); 
            else :
                k[i][w] = k[i - 1][w]; 
                  
    return k[N][A]; 
  
# Driver code 
if __name__ == "__main__"
  
    # Number of seats 
    A = 5
  
    # Groups 
    v = [ "mmo", "oo", "cmw", "cc", "c" ]; 
      
    N = len(v); 
    print(MaxHappiness(A, N, v)); 
      
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach 
using System;
  
class GFG 
{
    // Function to return the maximized happiness
    static int maxHappiness(int A, int N,
                              String[] v) 
    {
        String str;
  
        // Two arrays similar to
        // 0 1 knapsack prolem
        int[] val = new int[N];
        int[] wt = new int[N];
        int c = 0;
        for (int i = 0; i < N; i++) 
        {
            str = v[i];
  
            // To store the happiness
            // of the current goup
            c = 0;
            for (int j = 0; j < str.Length; j++) 
            {
                // Current person is a child
                if (str[j] == 'c')
                    c += 4;
  
                // Woman
                else if (str[j] == 'w')
                    c += 3;
  
                // Man
                else if (str[j] == 'm')
                    c += 2;
  
                // Old Person
                else
                    c++;
            }
  
            // Group's happiness is the sum of happiness
            // of the people in the group multiplie
            // the number of people
            c *= str.Length;
            val[i] = c;
            wt[i] = str.Length;
        }
  
        // Solution using 0 1 knapsack
        int[ , ] k = new int[N + 1, A + 1];
        for (int i = 0; i <= N; i++) 
        {
            for (int w = 0; w <= A; w++) 
            {
                if (i == 0 || w == 0)
                    k[i, w] = 0;
                else if (wt[i - 1] <= w) 
                {
                    k[i, w] = Math.Max(val[i - 1]+ 
                                       k[i - 1, w - wt[i - 1]],
                                       k[i - 1, w]);
                
                else
                {
                    k[i, w] = k[i - 1, w];
                }
            }
        }
        return k[N, A];
    }
  
    // Driver code
    public static void Main() 
    {
        // Number of seats
        int A = 5;
  
        // Groups
        String[] v = { "mmo", "oo", "cmw", "cc", "c" };
        int N = v.Length;
        Console.WriteLine(maxHappiness(A, N, v));
    }
}
  
// This code is contributed by Mohit kumar 29

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Output:

43


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