1+ /*
2+ This code uses prefix and postfix product to evaluate answer.
3+ We just need to traverse the array twice, once to the left and once to the right.
4+ Then answer of ith place can be calculated using constant time.
5+ 6+ Time Complexity : O(n)
7+ Space Complexity : O(n)
8+ */
9+ 10+ 11+ 12+ class Solution {
13+ public:
14+ vector<int> productExceptSelf(vector<int>& nums) {
15+ int n = nums.size(); //Variable for size of the array
16+ //pre[] stores product of all numbers to the left of ith element
17+ //post[] stores product of all numbers to the right of ith element
18+ int pre[n],post[n];
19+ 20+ //loop to assign values to pre[]
21+ int mul=1;
22+ for(int i=0; i<n; i++){
23+ mul*=nums[i];
24+ pre[i]=mul;
25+ }
26+ 27+ //loop to assign values to post[]
28+ mul=1;
29+ for(int i=n-1; i>=0; i--){
30+ mul*=nums[i];
31+ post[i]=mul;
32+ }
33+ 34+ //declare a vector to return
35+ vector <int> out;
36+ 37+ //first element of out is just going to be product of all elements except first one
38+ out.push_back(post[1]);
39+ 40+ //value of out[i] = product of all elements except ith element
41+ //which is nothing but pre[i-1]*[post[i+1]]
42+ for(int i=1; i<n-1; i++){
43+ int p=i-1;
44+ int s=i+1;
45+ out.push_back(pre[p]*post[s]);
46+ }
47+ 48+ //last element of out is just going to be product of all elements except last one
49+ out.push_back(pre[n-2]);
50+ 51+ //return the vector
52+ return out;
53+ }
54+ };
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