-
Notifications
You must be signed in to change notification settings - Fork 4.9k
/
MinimizeMaximumPairSumInArray.cpp
53 lines (51 loc) · 1.65 KB
/
MinimizeMaximumPairSumInArray.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
// Source : https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/
// Author : Hao Chen
// Date : 2021-11-12
/*****************************************************************************************************
*
* The pair sum of a pair (a,b) is equal to a + b. The maximum pair sum is the largest pair sum in a
* list of pairs.
*
* For example, if we have pairs (1,5), (2,3), and (4,4), the maximum pair sum would be
* max(1+5, 2+3, 4+4) = max(6, 5, 8) = 8.
*
* Given an array nums of even length n, pair up the elements of nums into n / 2 pairs such that:
*
* Each element of nums is in exactly one pair, and
* The maximum pair sum is minimized.
*
* Return the minimized maximum pair sum after optimally pairing up the elements.
*
* Example 1:
*
* Input: nums = [3,5,2,3]
* Output: 7
* Explanation: The elements can be paired up into pairs (3,3) and (5,2).
* The maximum pair sum is max(3+3, 5+2) = max(6, 7) = 7.
*
* Example 2:
*
* Input: nums = [3,5,4,2,4,6]
* Output: 8
* Explanation: The elements can be paired up into pairs (3,5), (4,4), and (6,2).
* The maximum pair sum is max(3+5, 4+4, 6+2) = max(8, 8, 8) = 8.
*
* Constraints:
*
* n == nums.length
* 2 <= n <= 10^5
* n is even.
* 1 <= nums[i] <= 10^5
******************************************************************************************************/
class Solution {
public:
int minPairSum(vector<int>& nums) {
sort(nums.begin(), nums.end());
int i = 0, j = nums.size() - 1;
int m = 0;
for(int i = 0, j = nums.size() - 1; i < j; i++,j--) {
m = max(m, nums[i] + nums[j]);
}
return m;
}
};