Interview Question Array Partition I

By Fatih KABAKCI

Problem #561 - Array Partition I

Description: 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.
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]..

Difficulty: Easy

Solution in Java

package com.fatihkabakci.Easy.ArrayPartitionI;

import java.util.Arrays;

/**
 * @author fkabakci 
 * Problem Description: 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.
 * 
 * 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:
 * 1. Sort the array to discriminate later on easily the elements that are closest to the max element
 *    in their groups.
 *    
 *    For example, lets consider above the case that is 1 2 3 4
 *    
 *    3 is the closest one to 4.
 *    1 is the closest one to 2.
 *    
 *    1 + 3 = 4 is largest value that is among the sum of the min pairs.
 */

public class Solution {
	public int arrayPairSum(int[] nums) {
		Arrays.sort(nums);
		
        int n = nums.length, sum = 0;
        for(int i = n - 1; i > 0; i-=2) {
        	sum += Math.min(nums[i], nums[i - 1]);
        }
        return sum;
    }
}