Maximum Product of Two Elements in an Array

Samsung

2020-05-31

1464. Maximum Product of Two Elements in an Array

Question:

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

Solution:

Intuition: find the max two numbers.

Time: O(n)
Space: O(1)

class Solution {
    public int maxProduct(int[] nums) {
        int n = Integer.MIN_VALUE;;
        int m = Integer.MIN_VALUE;;
        for (int i : nums) {
            if (i >= n) {
                m = n;
                n = i;
            } else if (i >= m) {
                m = i;
            }
        }
        return (--n) * (--m);
    }
}

Using array sort.

Time: O(nlogn)
Space: O(1)

class Solution {
    public int maxProduct(int[] nums) {
        Arrays.sort(nums);
        return (--nums[nums.length - 1]) * (--nums[nums.length - 2]);
    }
}

Using ProrityQueue.

Time: O(nlogn)
Space: O(n)

class Solution {
    public int maxProduct(int[] nums) {
        PriorityQueue<Integer> pq = new PriorityQueue<>(nums.length, new Comparator<Integer>(){
            @Override
            public int compare(Integer a, Integer b) {
                return b - a;
            }
        });
        for (int i : nums) {
            pq.add(i);
        }
        return (pq.poll() - 1) * (pq.poll() - 1);
    }
}