ArrayPartitionI.java

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package array;

import java.util.*;

/**

* Created by gouthamvidyapradhan on 14/08/2019 Given an array of 2n integers, your task is to group

* these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of

* min(ai, bi) for all i from 1 to n as large as possible.

Example 1: Input: [1,4,3,2]

Output: 4 Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

* Note: n is a positive integer, which is in the range of [1, 10000]. All the integers in the array

* will be in the range of [-10000, 10000].

Solution: O(n log n) General idea is to pair the smallest with the next smallest value inorder

* to get the max sum of minimum.

public class ArrayPartitionI {

public static void main(String[] args) {

int[] A = {1, 2, 3, 4};

System.out.println(new ArrayPartitionI().arrayPairSum(A));

}

public int arrayPairSum(int[] nums) {

Arrays.sort(nums);

int sum = 0;

for (int i = 1; i < nums.length; i += 2) {

sum += Math.min(nums[i - 1], nums[i]);

}

return sum;

}