Array Partition I LeetCode

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].

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Solution:

 

	/**
	Array Partition I
	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].
	*/
	import java.util.*;
	public class ArrayPartitionI{
	/**
	-sort the array
	-take the min of every two numbers and accumulate the sum
	O(nlogn)
	*/
	public static int arrayPairSum(int[] nums) {
	//base case
	if(nums==null || nums.length==0){
	return 0;
	}
	if(nums.length%2!=0){
	return 0;
	}
	Arrays.sort(nums);
	//now find the sum of the pairs
	int sum = 0;
	for(int i=0;i<nums.length; i+=2){
	sum+=Math.min(nums[i], nums[i+1]);
	}
	return sum;
	}
	public static void main(String args[]){
	int[] nums = new int[]{1,4,3,2};
	System.out.println("sum for array "+Arrays.toString(nums)+" is: "+arrayPairSum(nums));
	nums = new int[]{1,1};
	System.out.println("sum for array "+Arrays.toString(nums)+" is: "+arrayPairSum(nums));
	}
	}

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