**Special Product List - Amazon Top Interview Questions**

### Problem Statement :

Given a list of integers nums, return a new list such that each element at index i of the new list is the product of all the numbers in the original list except the one at i. Do this without using division. Constraints 2 ≤ n ≤ 100,000 where n is the length of nums Example 1 Input nums = [1, 2, 3, 4, 5] Output [120, 60, 40, 30, 24] Explanation 120 = 2 * 3 * 4 * 5, 60 = 1 * 3 * 4 * 5, and so on. Example 2 Input nums = [3, 2, 1] Output [2, 3, 6]

### Solution :

` ````
Solution in C++ :
vector<int> solve(vector<int>& nums) {
int n = nums.size(), p = 1;
vector<int> res(n, 1);
for (int i = 0; i < n; i++) {
res[i] *= p;
p *= nums[i];
}
p = 1;
for (int i = n - 1; i >= 0; i--) {
res[i] *= p;
p *= nums[i];
}
return res;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int[] solve(int[] nums) {
int n = nums.length;
int[] prefix = new int[n + 1];
int[] suffix = new int[n + 1];
Arrays.fill(prefix, 1);
Arrays.fill(suffix, 1);
for (int i = 1; i <= n; i++) {
prefix[i] = prefix[i - 1] * nums[i - 1];
}
for (int i = n - 1; i >= 0; i--) {
suffix[i] = suffix[i + 1] * nums[i];
}
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = prefix[i] * suffix[i + 1];
}
return res;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
prod, n = 1, len(nums)
result = [1] * n
for i in range(n):
result[i] = prod
prod *= nums[i]
prod = 1
for i in range(n - 1, -1, -1):
result[i] *= prod
prod *= nums[i]
return result
```

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