Table of Contents
Maximum Product of Three Numbers – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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
C++ Maximum Product of Three Numbers LeetCode Solution
class Solution {
public:
int maximumProduct(vector<int>& nums) {
int f[2][4], i, j;
f[0][0] = f[1][0] = 1;
for (j = 3; j > 0; --j) {
f[0][j] = INT_MIN;
f[1][j] = INT_MAX;
}
for (i = 0; i < nums.size(); ++i) {
for (j = 3; j > 0; --j) {
if (f[0][j - 1] == INT_MIN) {
continue;
}
f[0][j] = max(f[0][j], max(f[0][j - 1] * nums[i], f[1][j - 1] * nums[i]));
f[1][j] = min(f[1][j], min(f[0][j - 1] * nums[i], f[1][j - 1] * nums[i]));
}
}
return f[0][3];
}
};
Java Maximum Product of Three Numbers LeetCode Solution
public int maximumProduct(int[] nums) {
int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE, min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
for (int n : nums) {
if (n > max1) {
max3 = max2;
max2 = max1;
max1 = n;
} else if (n > max2) {
max3 = max2;
max2 = n;
} else if (n > max3) {
max3 = n;
}
if (n < min1) {
min2 = min1;
min1 = n;
} else if (n < min2) {
min2 = n;
}
}
return Math.max(max1*max2*max3, max1*min1*min2);
}
Python 3 Maximum Product of Three Numbers LeetCode Solution
def maximumProduct(self, nums):
return max(nums) * max(a * b for a, b in [heapq.nsmallest(2, nums), heapq.nlargest(3, nums)[1:]])
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