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1913. Maximum Product Difference Between Two Pairs

題目​

完整題目

The product difference between two pairs (a, b) and (c, d) is defined as (a∗b)−(c∗d)(a * b) - (c * d).

For example, the product difference between (5, 6) and (2, 7) is (5 6) - (2 7) = 16. Given an integer array nums, choose four distinct indices w, x, y, and z such that the product difference between pairs (nums[w], nums[x]) and (nums[y], nums[z]) is maximized.

Return the maximum such product difference.

Example 1

Input: nums = [5,6,2,7,4]
Output: 34
Explanation: We can choose indices 1 and 3 for the first pair (6, 7) and indices 2 and 4 for the second pair (2, 4).
The product difference is (6 * 7) - (2 * 4) = 34.

Exmaple 2

Input: nums = [4,2,5,9,7,4,8]
Output: 64
Explanation: We can choose indices 3 and 6 for the first pair (9, 8) and indices 1 and 5 for the second pair (2, 4).
The product difference is (9 * 8) - (2 * 4) = 64.

Constraints

  • 4 <= nums.length <= 10410^4
  • 1 <= nums[i] <= 10410^4

題目難易度​

Easy

解題想法​

題目是 (a∗b)(a * b) 和 (c∗d)(c * d) 相減，所以換句話來說就是要讓 (a∗b)(a * b) 越大，(c∗d)(c * d)越小。那我們只需要先把 nums 排序，a 和 b 抓最大的兩個數字，c 和 d 抓最小的兩個數字。

初試​

Runtime: 76ms, 62.25%

Memory Usage: 45.10MB, 38.87%

  • 時間複雜度：O(n∗logn)O(n * logn)
  • 空間複雜度：O(1)O(1)
/**
 * @param {number[]} nums
 * @return {number}
 */
var maxProductDifference = function (nums) {
  const a = nums.sort((a, b) => b - a);
  return a[0] * a[1] - a[nums.length - 1] * a[nums.length - 2];
};