Problem

Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a₁, b₁), (a₂, b₂), …, (a_n, b_n) which makes sum of min(a_i, b_i) 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].

Solution 1: Sorting

Time complexity: O(nlogn)

Space complexity: O(1)

// Author: Huahua

// Running time: 52 ms

class Solution {

public:

int arrayPairSum(vector<int>& nums) {

std::sort(nums.begin(), nums.end());

int ans = 0;

for(int i = 0; i < nums.size(); i += 2)

ans += nums[i];

return ans;

}

};

Solution 2: HashTable

Time complexity: O(n + max(nums) – min(nums))

Space complexity: O(max(nums) – min(nums))

// Author: Huahua

// Running time: 39 ms

class Solution {

public:

int arrayPairSum(vector<int>& nums) {

const int max_val = *max_element(begin(nums), end(nums));

const int min_val = *min_element(begin(nums), end(nums));

const int offset = -min_val;

vector<int> c(max_val - min_val + 1);

for (int num : nums)

++c[num + offset];

int ans = 0;

int index = 0;

int n = min_val;

while (n <= max_val) {

if (c[n + offset] == 0) {

++n;

continue;

}

ans += (++index & 1) ? n : 0;

--c[n + offset];

}

return ans;

}

};

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花花酱 LeetCode 561. Array Partition I

Problem

Solution 1: Sorting

Solution 2: HashTable

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