1913. Maximum Product Difference Between Two Pairs LeetCode Solution
In this guide, you will get 1913. Maximum Product Difference Between Two Pairs LeetCode Solution with the best time and space complexity. The solution to Maximum Product Difference Between Two Pairs problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Maximum Product Difference Between Two Pairs
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.
1913. Maximum Product Difference Between Two Pairs LeetCode Solution in C++
class Solution {
public:
int maxProductDifference(vector<int>& nums) {
int max1 = INT_MIN;
int max2 = INT_MIN;
int min1 = INT_MAX;
int min2 = INT_MAX;
for (const int num : nums) {
if (num > max1) {
max2 = max1;
max1 = num;
} else if (num > max2) {
max2 = num;
}
if (num < min1) {
min2 = min1;
min1 = num;
} else if (num < min2) {
min2 = num;
}
}
return max1 * max2 - min1 * min2;
}
}; /* code provided by PROGIEZ */
1913. Maximum Product Difference Between Two Pairs LeetCode Solution in Java
class Solution {
public int maxProductDifference(int[] nums) {
int max1 = Integer.MIN_VALUE;
int max2 = Integer.MIN_VALUE;
int min1 = Integer.MAX_VALUE;
int min2 = Integer.MAX_VALUE;
for (final int num : nums) {
if (num > max1) {
max2 = max1;
max1 = num;
} else if (num > max2) {
max2 = num;
}
if (num < min1) {
min2 = min1;
min1 = num;
} else if (num < min2) {
min2 = num;
}
}
return max1 * max2 - min1 * min2;
}
} // code provided by PROGIEZ
1913. Maximum Product Difference Between Two Pairs LeetCode Solution in Python
class Solution:
def maxProductDifference(self, nums: list[int]) -> int:
max1 = -math.inf
max2 = -math.inf
min1 = math.inf
min2 = math.inf
for num in nums:
if num > max1:
max2 = max1
max1 = num
elif num > max2:
max2 = num
if num < min1:
min2 = min1
min1 = num
elif num < min2:
min2 = num
return max1 * max2 - min1 * min2 # code by PROGIEZ
