# 238. Product of Array Except Self
Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
Example:
Input: [1,2,3,4]
Output: [24,12,8,6]
Note: Please solve it without division and in O(n).
Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)
# Solution
Approach 1: O(n) time and O(n) space. Two passes -- once from the left, and once from the right. Can save space by using only 1 array, and updating on it.
Approach 2: O(n) time and O(1) space except the output.
# Code (Python)
Approach 1:
def productExceptSelf(self, nums: List[int]) -> List[int]:
prod_from_left = []
for num in nums:
prod_from_left.append(prod_from_left[-1] * num if prod_from_left else num)
prod_from_right = 1
for i in range(len(nums) - 1, -1, -1):
if i != 0:
prod_from_left[i] = prod_from_left[i-1] * prod_from_right
prod_from_right *= nums[i]
else:
prod_from_left[i] = prod_from_right
return prod_from_left
Approach 2:
# Code (C++)
Approach 1:
// O(n) time and O(n) space.
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
vector<int> res;
int n = nums.size();
if (n == 0)
return res;
vector<int> left = vector<int>(n, 1);
vector<int> right = vector<int>(n, 1);
for (int i = 1; i < n; ++i)
{
left[i] *= left[i-1] * nums[i-1];
}
for (int i = n - 2; i >= 0; --i)
{
right[i] *= right[i+1] * nums[i+1];
}
for (int i = 0; i < n; ++i)
{
res.push_back(left[i] * right[i]);
}
return res;
}
};
Approach 2:
// O(n) time and O(1) space except the output.
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> res = vector<int>(n, 1);
if (n == 0)
return res;
for (int i = 0; i < n; ++i)
{
if (i - 1 >= 0)
res[i] *= res[i-1] * nums[i-1];
}
int prev = 1;
for (int i = n - 1; i >= 0; --i)
{
if (i + 1 < n)
res[i] *= prev;
prev *= nums[i];
}
return res;
}
};