Leetcode 561. Array Partition I, Java解法

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:

1. n is a positive integer, which is in the range of [1, 10000].
2. All the integers in the array will be in the range of [-10000, 10000].

Approach #1 Mysolution

 

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

Complexity Analysis

• Time complexity : O(nlog(n)). Sorting takes O(nlog(n))time. We iterate over the array only once.

• Space complexity : O(1)Constant extra space is used.

Let me try to prove the algorithm...

1. Assume in each pair i, bi >= ai.
2. Denote Sm = min(a1, b1) + min(a2, b2) + ... + min(an, bn). The biggest Sm is the answer of this problem. Given 1, Sm = a1 + a2 + ... + an.
3. Denote Sa = a1 + b1 + a2 + b2 + ... + an + bn. Sa is constant for a given input.
4. Denote di = |ai - bi|. Given 1, di = bi - ai. Denote Sd = d1 + d2 + ... + dn.
5. 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.
6. 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's not intuitive enough, see attached picture. Case 1 has the smallest Sd.

Approach #2 Using Extra Array

public class Solution {
    public int arrayPairSum(int[] nums) {
        int[] arr = new int[20001];
        int lim = 10000;
        for (int num: nums)
            arr[num + lim]++;
        int d = 0, sum = 0;
        for (int i = -10000; i <= 10000; i++) {
            sum += (arr[i + lim] + 1 - d) / 2 * i;
            d = (2 + arr[i + lim] - d) % 2;
        }
        return sum;
    }
} 

 While traversing the hashmap, we determine the correct number of times each element needs to be considered as discussed above. Note that the flag dd and the sumsum remains unchanged if the current element of the hashmap doesn't exist in the array.

Complexity Analysis

• Time complexity : O(n) The whole hashmap arr of size n is traversed only once.

• Space complexity : O(n) A hashmap arrarr of size n is used.