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 <= 10530 <= nums[i] <= 30answer[i] 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.)
https://leetcode.com/problems/product-of-array-except-self/description/
Using O(n) time and space complexity
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> preProd(n,1), postProd(n,1);
preProd[0] = nums[0];
postProd[n-1] = nums[n-1];
for(int i=1;i<n;i++) {
preProd[i] = preProd[i-1] * nums[i];
}
for(int i=n-2;i>=0;i--) {
postProd[i] = postProd[i+1] * nums[i];
}
vector<int> res(n,1);
for(int i=0;i<n;i++) {
int a = 1, b = 1;
if(i-1>=0) {
a = preProd[i-1];
}
if(i+1<n) {
b = postProd[i+1];
}
res[i] = a*b;
}
return res;
}
};
Using O(n) time and O(1) space complexity
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> prod(n);
int cur = 1;
for(int i = 0; i < n; i++){
prod[i] = cur;
cur *= nums[i];
}
cur = 1;
for(int i = n - 1; i >= 0; i--){
prod[i] *= cur;
cur *= nums[i];
}
return prod;
}
};