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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0501_0600.s0561_array_partition_i;

// #Easy #Array #Sorting #Greedy #Counting_Sort
// #2022_08_03_Time_14_ms_(84.99%)_Space_44.2_MB_(95.29%)

import java.util.Arrays;

/**
 * 561 - Array Partition I\.
 *
 * Easy
 *
 * Given an integer array `nums` of `2n` integers, group these integers into `n` pairs (a₁, b₁), (a₂, b₂), ..., (a_n, b_n) such that the sum of min(a_i, b_i) 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
**/
public class Solution {
    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        int sum = 0;
        for (int i = 0; i < nums.length - 1; i = i + 2) {
            sum += Math.min(nums[i], nums[i + 1]);
        }
        return sum;
    }
}