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.
Approach: This problem can be solved using Greedy Approach. Below are the steps:
- 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.
- The objective is to minimize the total cost. Hence, sort the array temp[] in decreasing order.
- 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.
- 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 <bits/stdc++.h> 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<pair<int, int> >()); // 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); } |
107
Time Complexity: O(N*log(N))
Auxiliary Space Complexity: O(N)
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