# Minimum sum obtained by choosing N number from given N pairs

Given an array arr[] of N pairs of integers (A, B) where N is even, the task is to find the minimum sum of choosing N elements such that value A and B from all the pairs are chosen exactly (N/2) times.

Examples:

Input: N = 4, arr[][] = { {7, 20}, {300, 50}, {30, 200}, {30, 20} }
Output: 107
Explanation:
Choose value-A from 1st pair = 7.
Choose value-B from 2nd pair = 50.
Choose value-A from 3rd pair = 30.
Choose value-B from 4th pair = 20.
The minimum sum is 7 + 50 + 30 + 20 = 107.

Input: N = 4, arr[][] = { {10, 20}, {400, 50}, {30, 200}, {30, 20} }
Output: 110
Explanation:
Choose value-A from 1st pair = 10.
Choose value-B from 2nd pair = 50.
Choose value-A from 3rd pair = 30.
Choose value-B from 4th pair = 20.
The minimum sum is 10 + 50 + 30 + 20 = 110.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem can be solved using Greedy Approach. Below are the steps:

1. For each pair (A, B) in the given array, store the value of (B – A) with the corresponding index in temporary array(say temp[]). The value (B – A) actually defines how much cost is minimized if A is chosen over B for each element.
2. The objective is to minimize the total cost. Hence, sort the array temp[] in decreasing order.
3. Pick the first N/2 elements from the array temp[] by choosing A as first N/2 elements will have the maximum sum when A is chosen over B.
4. For remaining N/2 elements choose B as the sum of values can be minimized.

Below is the implementation of the above approach:

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to choose the elements A ` `// and B such the sum of all elements ` `// is minimum ` `int` `minSum(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Create an array of pair to ` `    ``// store Savings and index ` `    ``pair<``int``, ``int``> temp[n]; ` ` `  `    ``// Traverse the given array of pairs ` `    ``for` `(``int` `i = 0; i < 2 * n; i++) { ` ` `  `        ``// Sum minimized when A ` `        ``// is chosen over B for ` `        ``// i-th element. ` `        ``temp[i].first = arr[i] ` `                        ``- arr[i]; ` ` `  `        ``// Store index for the ` `        ``// future reference. ` `        ``temp[i].second = i; ` `    ``} ` ` `  `    ``// Sort savings array in ` `    ``// non-increasing order. ` `    ``sort(temp, temp + 2 * n, ` `         ``greater >()); ` ` `  `    ``// Storing result ` `    ``int` `res = 0; ` ` `  `    ``for` `(``int` `i = 0; i < 2 * n; i++) { ` ` `  `        ``// First n elements choose ` `        ``// A and rest choose B ` `        ``if` `(i < n) ` `            ``res += arr[temp[i].second]; ` `        ``else` `            ``res += arr[temp[i].second]; ` `    ``} ` ` `  `    ``// Return the final Sum ` `    ``return` `res; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given array of pairs ` `    ``int` `arr = { { 7, 20 }, ` `                      ``{ 300, 50 }, ` `                      ``{ 30, 200 }, ` `                      ``{ 30, 20 } }; ` ` `  `    ``// Function Call ` `    ``cout << minSum(arr, 2); ` `} `

Output:

```107
```

Time Complexity: O(N*log(N))
Auxiliary Space Complexity: O(N)

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