561 Array Partition I

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最新推荐文章于 2022-01-31 22:01:47 发布

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分类专栏： LeetCode

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本文介绍了一种算法，用于将给定的整数数组通过最优配对求得最大最小值之和。该算法首先对数组进行排序，然后依次选取相邻元素中的较小者进行累加，最终得到的最大和即为所求。通过对不同配对方案的分析，证明了此方法的有效性和正确性。

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题目

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.

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).

Code

public class Solution {
    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        int result = 0;
        for (int i = 0; i < nums.length; i += 2) {
            result += nums[i];
        }
        return result;
    }
}

Prove

Let me try to prove the algorithm…

Assume in each pair i, bi >= ai.
Denote Sm = min(a1, b1) + min(a2,b2) + … + min(an, bn). The biggest Sm is the answer of thisproblem. Given 1, Sm = a1 + a2 + … + an.
Denote Sa = a1 + b1+ a2 + b2 + … + an + bn. Sa is constant for a given input. Denote di = |ai - bi|. Given 1, di = bi - ai.
Denote Sd =d1 + d2 + … + dn. So Sa = a1 + a1 + d1 + a2 + a2 + d2 + … +an + an + dn = 2Sm + Sd => Sm = (Sa - Sd) / 2. To get the max Sm, given Sa is constant, we need to make Sd as small as possible. So this problem becomes finding pairs in an array that makes sum of di (distance between ai and bi) as small as possible. Apparently, sum of these distances of adjacent elements is the smallest. If that’snot intuitive enough, see attached picture. Case 1 has the smallest Sd.