# 628. Maximum Product of Three Numbers

## Description

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

## Solutions

• class Solution {
public int maximumProduct(int[] nums) {
final int inf = 1 << 30;
int mi1 = inf, mi2 = inf;
int mx1 = -inf, mx2 = -inf, mx3 = -inf;
for (int x : nums) {
if (x < mi1) {
mi2 = mi1;
mi1 = x;
} else if (x < mi2) {
mi2 = x;
}
if (x > mx1) {
mx3 = mx2;
mx2 = mx1;
mx1 = x;
} else if (x > mx2) {
mx3 = mx2;
mx2 = x;
} else if (x > mx3) {
mx3 = x;
}
}
return Math.max(mi1 * mi2 * mx1, mx1 * mx2 * mx3);
}
}

• class Solution {
public:
int maximumProduct(vector<int>& nums) {
const int inf = 1 << 30;
int mi1 = inf, mi2 = inf;
int mx1 = -inf, mx2 = -inf, mx3 = -inf;
for (int x : nums) {
if (x < mi1) {
mi2 = mi1;
mi1 = x;
} else if (x < mi2) {
mi2 = x;
}
if (x > mx1) {
mx3 = mx2;
mx2 = mx1;
mx1 = x;
} else if (x > mx2) {
mx3 = mx2;
mx2 = x;
} else if (x > mx3) {
mx3 = x;
}
}
return max(mi1 * mi2 * mx1, mx1 * mx2 * mx3);
}
};

• class Solution:
def maximumProduct(self, nums: List[int]) -> int:
top3 = nlargest(3, nums)
bottom2 = nlargest(2, nums, key=lambda x: -x)
return max(top3[0] * top3[1] * top3[2], top3[0] * bottom2[0] * bottom2[1])


• func maximumProduct(nums []int) int {
const inf = 1 << 30
mi1, mi2 := inf, inf
mx1, mx2, mx3 := -inf, -inf, -inf
for _, x := range nums {
if x < mi1 {
mi1, mi2 = x, mi1
} else if x < mi2 {
mi2 = x
}
if x > mx1 {
mx1, mx2, mx3 = x, mx1, mx2
} else if x > mx2 {
mx2, mx3 = x, mx2
} else if x > mx3 {
mx3 = x
}
}
return max(mi1*mi2*mx1, mx1*mx2*mx3)
}

• function maximumProduct(nums: number[]): number {
const inf = 1 << 30;
let mi1 = inf,
mi2 = inf;
let mx1 = -inf,
mx2 = -inf,
mx3 = -inf;
for (const x of nums) {
if (x < mi1) {
mi2 = mi1;
mi1 = x;
} else if (x < mi2) {
mi2 = x;
}
if (x > mx1) {
mx3 = mx2;
mx2 = mx1;
mx1 = x;
} else if (x > mx2) {
mx3 = mx2;
mx2 = x;
} else if (x > mx3) {
mx3 = x;
}
}
return Math.max(mi1 * mi2 * mx1, mx1 * mx2 * mx3);
}