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1913. Maximum Product Difference Between Two Pairs
Description
The product difference between two pairs (a, b) and (c, d) is defined as (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.
Example 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 <= 1041 <= nums[i] <= 104
Solutions
-
class Solution { public int maxProductDifference(int[] nums) { Arrays.sort(nums); int n = nums.length; return nums[n - 1] * nums[n - 2] - nums[0] * nums[1]; } } -
class Solution { public: int maxProductDifference(vector<int>& nums) { sort(nums.begin(), nums.end()); int n = nums.size(); return nums[n - 1] * nums[n - 2] - nums[0] * nums[1]; } }; -
class Solution: def maxProductDifference(self, nums: List[int]) -> int: nums.sort() return nums[-1] * nums[-2] - nums[0] * nums[1] -
func maxProductDifference(nums []int) int { sort.Ints(nums) n := len(nums) return nums[n-1]*nums[n-2] - nums[0]*nums[1] } -
/** * @param {number[]} nums * @return {number} */ var maxProductDifference = function (nums) { nums.sort((a, b) => a - b); let n = nums.length; let ans = nums[n - 1] * nums[n - 2] - nums[0] * nums[1]; return ans; };