https://leetcode.com/problems/array-partition-i
https://discuss.leetcode.com/topic/87206/java-solution-sorting-and-rough-proof-of-algorithm
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].
https://discuss.leetcode.com/topic/87206/java-solution-sorting-and-rough-proof-of-algorithm
The algorithm is first sort the input array and then the sum of 1st, 3rd, 5th..., is the answer.
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;
}
}
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 biggestSm
is the answer of this problem. Given1
,Sm = a1 + a2 + ... + an
. - Denote
Sa = a1 + b1 + a2 + b2 + ... + an + bn
.Sa
is constant for a given input. - Denote
di = |ai - bi|
. Given1
,di = bi - ai
. DenoteSd = d1 + d2 + ... + dn
. - So
Sa = a1 + a1 + d1 + a2 + a2 + d2 + ... + an + an + di = 2Sm + Sd
=>Sm = (Sa - Sd) / 2
. To get the maxSm
, givenSa
is constant, we need to makeSd
as small as possible. - So this problem becomes finding pairs in an array that makes sum of
di
(distance betweenai
andbi
) as small as possible. Apparently, sum of these distances of adjacent elements is the smallest. If that's not intuitive enough, see attached picture. Case 1 has the smallestSd
.