ArrayPartition1.java

/*
* 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].
* */

public class ArrayPartition1 {
    public static void main (String args[]){
        int[] nums = {1,4,3,2};
        System.out.println(arrayPairSum(nums));
    }
    public static int arrayPairSum(int[] nums) {
        int[] exist = new int[20001];
        for (int i = 0; i < nums.length; i++) {
            exist[nums[i] + 10000]++;
        }
        int sum = 0;
        boolean odd = true;
        for (int i = 0; i < exist.length; i++) {
            while (exist[i] > 0) {
                if (odd) {
                    sum += i - 10000;
                }
                odd = !odd;
                exist[i]--;
            }
        }
        return sum;
    }
}