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# C++ Program to find out the maximum rated parts set

Suppose, there is a manufacturer that makes specific parts for a particular product. The manufacturer has n different variations of the parts, and the parts have a specific rating on three criteria. The ratings of the n products are given in the array 'ratings' where each element is of the format (A, B, C) where A, B, and C are different rating criteria of the product. Now, an OEM wants to buy m parts for each product they make from the manufacturer of the parts. The OEM chooses the parts satisfying the below conditions −

Two or more units of the same part should not be bought.

Choose the set of parts such that the value V is maximized, where V = |total rating of criteria A| + |total rating of criteria B| + |total rating of criteria C|.

We have to find the maximum possible value of V from the parts that the OEM chooses.

So, if the input is like n = 6, m = 4, ratings = {{2, 3, 5}, {3, 5, 2}, {4, 8, 5}, {1, 5, 3}, {7, 2, 7}, {4, 3, 6}}, then the output will be 56.

If the OEM chooses parts 1, 3, 5, and 6, then the total ratings for each category are −

Category A = 2 + 4 + 7 + 4 = 17 Category B = 3 + 8 + 2 + 3 = 16. Category C = 5 + 5 + 7 + 6 = 23 The total value of V is 17 + 16 + 23 = 56.

To solve this, we will follow these steps −

N := 100 Define an array arr of size: 9 x N. Define an array ans. for initialize i := 0, when i < n, update (increase i by 1), do: a := first value of ratings[i] b := second value of ratings[i] c := third value of ratings[i] arr[1, i] := a + b + c arr[2, i] := a - b - c arr[3, i] := a + b - c arr[4, i] := a - b + c arr[5, i] := -a + b + c arr[6, i] := -a - b - c arr[7, i] := -a + b - c arr[8, i] := -a - b + c for initialize i := 1, when i <= 8, update (increase i by 1), do: sort the array arr[i] for initialize i := 1, when i <= 8, update (increase i by 1), do: reverse the array arr[i] if m is the same as 0, then: V := 0 Otherwise for initialize j := 1, when j <= 8, update (increase j by 1), do: k := 0 for initialize i := 0, when i < m, update (increase i by 1), do: k := k + arr[j, i] V := maximum of V and k return V

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h> using namespace std; const int INF = 1e9; const int modval = (int) 1e9 + 7; #define N 100 int solve(int n, int m, vector<tuple<int, int, int>> ratings) { int V, arr[9][N] ; vector<int> ans ; for(int i = 0 ; i < n ; i++) { int a, b, c; tie(a, b, c) = ratings[i]; arr[1][i] = a + b + c ; arr[2][i] = a - b - c ; arr[3][i] = a + b - c ; arr[4][i] = a - b + c ; arr[5][i] = -a + b + c ; arr[6][i] = -a - b - c ; arr[7][i] = -a + b - c ; arr[8][i] = -a - b + c ; } for(int i = 1 ; i <= 8 ; i++) sort(arr[i] , arr[i] + n) ; for(int i = 1 ; i <= 8 ; i++) reverse(arr[i] , arr[i] + n) ; if (m == 0) V = 0 ; else { for (int j = 1; j <= 8; j++) { int k = 0; for (int i = 0; i < m; i++) k += arr[j][i]; V = max(V, k); } } return V; } int main() { int n = 6, m = 4; vector<tuple<int, int, int>> ratings = {{2, 3, 5}, {3, 5, 2}, {4, 8, 5}, {1, 5, 3}, {7, 2, 7}, {4, 3, 6}}; cout<< solve(n, m, ratings); return 0; }

## Input

6, 4, {{2, 3, 5}, {3, 5, 2}, {4, 8, 5}, {1, 5, 3}, {7, 2, 7}, {4, 3,6}}

## Output

56

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