https://leetcode.com/problems/product-of-array-except-self
Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. You must write an algorithm that runs in O(n) time and without using the division operation.
Example 1: Input: nums = [1,2,3,4] Output: [24,12,8,6] Example 2: Input: nums = [-1,1,0,-3,3] Output: [0,0,9,0,0]
Constraints:
2 <= nums.length <= 105
-30 <= nums[i] <= 30
The product of any prefix or suffix of nums is **guaranteed** to fit in a **32-bit** integer.
Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)
Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. You must write an algorithm that runs in O(n) time and without using the division operation.
Example 1: Input: nums = [1,2,3,4] Output: [24,12,8,6]
- code
class Solution {
public int[] productExceptSelf(int[] nums) {
int n = nums.length;
int[] res = new int[n];
int product = 1;
for (int i = 0; i < n; i++)
res[i] = 1;
for (int i = 0; i < n; i++){
res[i] = product;
product *= nums[i];
}
product = 1;
for (int i = n - 1; i > -1; i--){
res[i] *= product;
product *= nums[i];
}
return res;
}
}
- code
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
res = [1] * len(nums)
# left to right, for each i, put product of left array to i
product = 1
for i in range(len(nums)-1):
product *= nums[i]
res[i+1] = product
# right to left
product = 1
for i in range(len(nums)-1, 0, -1):
product *= nums[i]
res[i-1] *= product
return res
- code
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
res = [1] * len(nums)
# left to right, for each i, put product of left array to i
product = 1
for i in range(len(nums)):
res[i] = product
product *= nums[i]
# right to left
product = 1
for i in range(len(nums)-1, -1, -1):
res[i] *= product
product *= nums[i]
return res