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 :



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                        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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