# 628.Maximum-Product-of-Three-Numbers

## 628. Maximum Product of Three Numbers

## 题目地址

<https://leetcode.com/problems/maximum-product-of-three-numbers/>

## 题目描述

```
Given an integer array, find three numbers whose product is maximum and output the maximum product.

Example 1:
Input: [1,2,3]
Output: 6

Example 2:
Input: [1,2,3,4]
Output: 24

Note:
The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].
Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer.
```

## 代码

### Approach 1: Brute Force

Time complexity : O(n^3)

### Approach 2: Using Sorting

```java
public class Solution {
  public int maximumProduct(int[] nums) {
    Arrays.sort(nums);

    // it could also be possible that two negative numbers lying at the left extreme 
    return Math.max(nums[0] * nums[1] * nums[nums.length - 1], nums[nums.length - 1] * nums[nums.length - 2] * nums[nums.length - 3]);
  }
}
```

### Approach #3 Single Scan

**Complexity Analysis**

* Time complexity : O(n). Only one iteration over the nums array of length n is required.
* Space complexity : O(1). Constant extra space is used.

```java
class Solution {
  public int maximumProduct(int[] nums) {
    int min1 = Integer.MAX_VALUE;
    int min2 = Integer.MAX_VALUE;
    int max1 = Integer.MIN_VALUE;
    int max2 = Integer.MIN_VALUE;
    int max3 = Integer.MIN_VALUE;

    for (int n: nums) {
      if (n <= min1) {
        min2 = min1;
        min1 = n;
      } else if (n <= min2) {
        min2 = n;
      } 

      if (n >= max1) {
        max3 = max2;
        max2 = max1;
        max1 = n;
      } else if (n >= max2) {
        max3 = max2;
        max2 = n;
      } else if (n >= max3) {
        max3 = n;
      }
    }

    return Math.max(min1 * min2 * max1, max1 * max2 * max3);
  }
}
```


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