Product of Array except itself

Given an array arr[] of n integers, construct a Product Array prod[] (of the same size) such that prod[i] is equal to the product of all the elements of arr[] except arr[i]. 

Note: Solve it without the division operator in O(n) time.

Example : 

Input: arr[]  = {10, 3, 5, 6, 2}
Output: prod[]  = {180, 600, 360, 300, 900}
3 * 5 * 6 * 2 product of other array 
elements except 10 is 180
10 * 5 * 6 * 2 product of other array 
elements except 3 is 600
10 * 3 * 6 * 2 product of other array 
elements except 5 is 360
10 * 3 * 5 * 2 product of other array 
elements except 6 is 300
10 * 3 * 6 * 5 product of other array 
elements except 2 is 900


Input: arr[]  = {1, 2, 3, 4, 5}
Output: prod[]  = {120, 60, 40, 30, 24 }
2 * 3 * 4 * 5  product of other array 
elements except 1 is 120
1 * 3 * 4 * 5  product of other array 
elements except 2 is 60
1 * 2 * 4 * 5  product of other array 
elements except 3 is 40
1 * 2 * 3 * 5  product of other array 
elements except 4 is 30
1 * 2 * 3 * 4  product of other array 
elements except 5 is 24

A naive approach: Store the product of all the elements is variable and then iterate the array and add product/current_index_value in a new array. and then return this new array. But as per the problem statement this is not a valid approach:

An approach using prefix and suffix multiplication: Create two extra space, i.e. two extra arrays to store the product of all the array elements from start, up to that index and another array to store the product of all the array elements from the end of the array to that index. 
To get the product excluding that index, multiply the prefix product up to index i-1 with the suffix product up to index i+1.

Algorithm: 

1. Create two array prefix and suffix of length n, i.e length of the original array, initialize prefix[0] = 1 and suffix[n-1] = 1 and also another array to store the product.
2. Traverse the array from second index to end.
3. For every index i update prefix[i] as prefix[i] = prefix[i-1] * array[i-1], i.e store the product upto i-1 index from the start of array.
4. Traverse the array from second last index to start.
5. For every index i update suffix[i] as suffix[i] = suffix[i+1] * array[i+1], i.e store the product upto i+1 index from the end of array
6. Traverse the array from start to end.
7. For every index i the output will be prefix[i] * suffix[i], the product of the array element except that element.

The product array is: 
180 600 360 300 900 

 Complexity Analysis: 

• Time Complexity: O(n). 
  The array needs to be traversed three times, so the time complexity is O(n).
• Auxiliary Space: O(n). 
  Two extra arrays and one array to store the output is needed so the space complexity is O(n)

Note: The above method can be optimized to work in space complexity O(1). Thanks to Dileep for suggesting the below solution.

Efficient Approach: In the previous solution, two extra arrays were created to store the prefix and suffix, in this solution store the prefix and suffix product in the output array (or product array) itself. Thus reducing the space required.

Algorithm: 

1. Create an array product and initialize its value to 1 and a variable temp = 1.
2. Traverse the array from start to end.
3. For every index i update product[i] as product[i] = temp and temp = temp * array[i], i.e store the product upto i-1 index from the start of array.
4. initialize temp = 1 and traverse the array from last index to start.
5. For every index i update product[i] as product[i] = product[i] * temp and temp = temp * array[i], i.e multiply with the product upto i+1 index from the end of array.
6. Print the product array.

The product array is: 
180 600 360 300 900 

Complexity Analysis: 

• Time Complexity: O(n). 
  The original array needs to be traversed only once, so the time complexity is constant.
• Auxiliary Space: O(n). 
  Even though the extra arrays are removed, the space complexity remains O(n), as the product array is still needed.

A product array puzzle | Set 2 (O(1) Space)
Related Problem: 
Construct an Array from XOR of all elements of array except element at same index
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.