carloscn
commented
1 year ago Analysis
pub fn max_product_difference(nums: Vec<i32>) -> i32
{
if nums.len() < 4 {
return 0;
}
let mut ret:i32 = 0;
let (mut first_max, mut sec_max, mut first_min, mut sec_min) =
(i32::MIN, i32::MIN + 1, i32::MAX, i32::MAX - 1);
for i in 0..nums.len() {
if nums[i] > first_max {
sec_max = first_max;
first_max = nums[i];
} else if nums[i] <= first_max && nums[i] > sec_max {
sec_max = nums[i];
}
if nums[i] < first_min {
sec_min = first_min;
first_min = nums[i];
} else if nums[i] >= first_min && nums[i] < sec_min {
sec_min = nums[i];
}
}
ret = (first_max * sec_max) - (first_min * sec_min);
return ret;
}
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 <= 104 1 <= nums[i] <= 104