Approach: Sort, compare two candidates

def maximumProduct(nums):
    a = sorted(nums)
    return max(a[-1]*a[-2]*a[-3], a[0]*a[1]*a[-1])

Complexity: O(n log n) time, O(1) space.

Full Solution — all languages

class Solution:
    def maximumProduct(self, nums):
        a = sorted(nums)
        return max(a[-1] * a[-2] * a[-3], a[0] * a[1] * a[-1])

import java.util.*;
class Solution {
    public int maximumProduct(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        return Math.max(nums[n-1]*nums[n-2]*nums[n-3], nums[0]*nums[1]*nums[n-1]);
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int maximumProduct(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        return max(nums[n-1]*nums[n-2]*nums[n-3], nums[0]*nums[1]*nums[n-1]);
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */ var maximumProduct = function(nums) {
    const a = nums.slice().sort((x, y) => x - y);
    const n = a.length;
    return Math.max(a[n-1]*a[n-2]*a[n-3], a[0]*a[1]*a[n-1]);
};

Maximum Product of Three Numbers

Example 1

Example 2

Constraints