-
Notifications
You must be signed in to change notification settings - Fork 4.9k
/
MaximumProductOfTwoElementsInAnArray.cpp
51 lines (49 loc) · 1.48 KB
/
MaximumProductOfTwoElementsInAnArray.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
// Source : https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/
// Author : Hao Chen
// Date : 2020-10-02
/*****************************************************************************************************
*
* Given the array of integers nums, you will choose two different indices i and j of that array.
* Return the maximum value of (nums[i]-1)*(nums[j]-1).
*
* Example 1:
*
* Input: nums = [3,4,5,2]
* Output: 12
* Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum
* value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12.
*
* Example 2:
*
* Input: nums = [1,5,4,5]
* Output: 16
* Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of
* (5-1)*(5-1) = 16.
*
* Example 3:
*
* Input: nums = [3,7]
* Output: 12
*
* Constraints:
*
* 2 <= nums.length <= 500
* 1 <= nums[i] <= 10^3
******************************************************************************************************/
class Solution {
public:
int maxProduct(vector<int>& nums) {
int max1=max(nums[0], nums[1]);
int max2=min(nums[0], nums[1]);
for (int i=2; i< nums.size(); i++) {
if (nums[i] > max1) {
max2 = max1;
max1 = nums[i];
continue;
}else if (nums[i] > max2) {
max2 = nums[i];
}
}
return (max1-1)*(max2-1);
}
};