561. Array Partition
Description
Given an integer array nums
of 2n
integers, group these integers into n
pairs (a1, b1), (a2, b2), ..., (an, bn)
such that the sum of min(ai, bi)
for all i
is maximized. Return the maximized sum.
Example 1:
Input: nums = [1,4,3,2] Output: 4 Explanation: All possible pairings (ignoring the ordering of elements) are: 1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3 2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3 3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4 So the maximum possible sum is 4.
Example 2:
Input: nums = [6,2,6,5,1,2] Output: 9 Explanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.
Constraints:
1 <= n <= 104
nums.length == 2 * n
-104 <= nums[i] <= 104
Solutions
Solution 1: Sorting
For a pair of numbers $(a, b)$, we can assume $a \leq b$, then $\min(a, b) = a$. In order to make the sum as large as possible, the $b$ we choose should be as close to $a$ as possible, so as to retain a larger number.
Therefore, we can sort the array $nums$, then divide every two adjacent numbers into a group, and add the first number of each group.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array $nums$.
1 2 3 4 |
|
1 2 3 4 5 6 7 8 9 10 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 |
|
1 2 3 4 5 6 |
|
1 2 3 4 5 6 7 8 |
|