Problem

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:

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

Idea:

Find the top 3 numbers t1, t2, t3, and bottom 2 numbers, b1, b2.

If all numbers are positives,  answer must be t1 * t2 * t3.

Since the number can go negative, the answer must be either t1*t2*t3 or b1 * b2 * t1, if b1 and b2 are both negatives.

ex. nums: [5, 1, -6, 3, -1]

t1, t2, t3: 5, 3, 1

b1, b2: -6, -1

t1 * t2 * t3 = 15

t1 * b1 * b2 = 30

Solution 1: Manual Tracking

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua
// Running time: 28 ms
class Solution {
public:
  int maximumProduct(vector<int>& nums) {
    int max1 = INT_MIN;
    int max2 = INT_MIN;
    int max3 = INT_MIN;
    int min1 = INT_MAX;
    int min2 = INT_MAX;

    for (const int num: nums) {
      if (num > max1) {
          max3 = max2;
          max2 = max1;
          max1 = num;
      } else if (num > max2) {
          max3 = max2;
          max2 = num;
      } else if (num > max3) {
          max3=num;
      }

      if (num < min1) {
          min2 = min1;
          min1 = num;
      } else if (num < min2) {
          min2=num;
      }
    }

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

Solution 2: Sorting

Time complexity: O(nlogn)

Space complexity: O(1)

// Author: Huahua
// Running time: 48 ms
class Solution {
public:
  int maximumProduct(vector<int>& nums) {
    int n = nums.size();
    sort(nums.rbegin(), nums.rend());
    return max(nums[0] * nums[1] * nums[2], nums[0] * nums[n - 1] * nums[n - 2]);
  }
};

Solution 3: Two Heaps (Priority Queues)

Time complexity: O(nlog3)

Space complexity: O(2 + 3)

// Author: Huahua
// Running time: 40 ms
class Solution {
public:
  int maximumProduct(vector<int>& nums) {
    priority_queue<int, vector<int>, less<int>> min_q; // max_heap
    priority_queue<int, vector<int>, greater<int>> max_q; // min_heap
    
    for (int num : nums) {
      max_q.push(num);
      if (max_q.size() > 3) max_q.pop();
      min_q.push(num);
      if (min_q.size() > 2) min_q.pop();
    }
    
    int max3 = max_q.top(); max_q.pop();
    int max2 = max_q.top(); max_q.pop();
    int max1 = max_q.top(); max_q.pop();
    int min2 = min_q.top(); min_q.pop();
    int min1 = min_q.top(); min_q.pop();
    
    return max(max1 * max2 * max3, max1 * min1 * min2); 
  }
};

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