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

• Last Updated : 07 Aug, 2020

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[][2], ``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][1]``                        ``- arr[i][0];`` ` `        ``// 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][0];``        ``else``            ``res += arr[temp[i].second][1];``    ``}`` ` `    ``// Return the final Sum``    ``return` `res;``}`` ` `// Driver Code``int` `main()``{``    ``// Given array of pairs``    ``int` `arr[4][2] = { { 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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