README.md

1464. Maximum Product of Two Elements in an Array (Easy)

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

Companies: Amazon, Cisco, Samsung

Related Topics:
Array, Sorting, Heap (Priority Queue)

Hints:

• Use brute force: two loops to select i and j, then select the maximum value of (nums[i]-1)*(nums[j]-1).

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int N = nums.size(), ans = 0;
        for (int i = 0; i < N; ++i) {
            for (int j = i + 1; j < N; ++j) ans = max(ans, (nums[i] - 1) * (nums[j] - 1)) ;
        }
        return ans;
    }
};

Solution 2. Heap

We just need find the greatest two elements.

// OJ: https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int maxProduct(vector<int>& nums) {
        make_heap(begin(nums), end(nums));
        pop_heap(begin(nums), end(nums));
        pop_heap(begin(nums), end(nums) - 1);
        return (nums.back() - 1) * (*(nums.end() - 2) - 1);
    }
};

Solution 3. Two pass

// OJ: https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxProduct(vector<int>& nums) {
        auto it = max_element(begin(nums), end(nums));
        swap(*it, nums[0]);
        it = max_element(begin(nums) + 1, end(nums));
        return (nums[0] - 1) * (*it - 1);
    }
};

Solution 4. One pass

// OJ: https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int a = 0, b = 0;
        for (int n : nums) {
            if (n >= a) {
                b = a;
                a = n;
            } else if (n > b) b = n;
        }
        return (a - 1) * (b - 1);
    }
};