628. Maximum Product of Three Numbers LeetCode Solution

In this guide, you will get 628. Maximum Product of Three Numbers LeetCode Solution with the best time and space complexity. The solution to Maximum Product of Three Numbers 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
Complexity Analysis
Maximum Product of Three Numbers solution in C++
Maximum Product of Three Numbers solution in Java
Maximum Product of Three Numbers solution in Python
Additional Resources

628. Maximum Product of Three Numbers LeetCode Solution image

Problem Statement of Maximum Product of Three Numbers

Given an integer array nums, find three numbers whose product is maximum and return the maximum product.

Example 1:
Input: nums = [1,2,3]
Output: 6
Example 2:
Input: nums = [1,2,3,4]
Output: 24
Example 3:
Input: nums = [-1,-2,-3]
Output: -6

Constraints:

3 <= nums.length <= 104
-1000 <= nums[i] <= 1000

Complexity Analysis

Time Complexity: O(\texttt{sort})
Space Complexity: O(\texttt{sort})

628. Maximum Product of Three Numbers LeetCode Solution in C++

class Solution {
 public:
  int maximumProduct(vector<int>& nums) {
    const int n = nums.size();
    ranges::sort(nums);
    return max(nums[n - 1] * nums[0] * nums[1],
               nums[n - 1] * nums[n - 2] * nums[n - 3]);
  }
};
/* code provided by PROGIEZ */

628. Maximum Product of Three Numbers LeetCode Solution in Java

class Solution {
  public int maximumProduct(int[] nums) {
    final int n = nums.length;
    Arrays.sort(nums);
    return Math.max(nums[n - 1] * nums[0] * nums[1], nums[n - 1] * nums[n - 2] * nums[n - 3]);
  }
}
// code provided by PROGIEZ

628. Maximum Product of Three Numbers LeetCode Solution in Python

class Solution:
  def maximumProduct(self, nums: list[int]) -> int:
    nums.sort()
    return max(nums[-1] * nums[0] * nums[1],
               nums[-1] * nums[-2] * nums[-3])
 # code by PROGIEZ