Product array puzzle

Problem Description

Given an array of integers A, find and return the product array of the same size where the ith element of the product array will be equal to the product of all the elements divided by the ith element of the array.

Note: It is always possible to form the product array with integer (32 bit) values. Solve it without using the division operator.

Problem Constraints

2 <= length of the array <= 1000
1 <= A[i] <= 10

Input Format

The only argument given is the integer array A.

Output Format

Return the product array.

Example Input

Input 1:
A = [1, 2, 3, 4, 5]

Input 2:
A = [5, 1, 10, 1]

Example Output

Output 1:
[120, 60, 40, 30, 24]

Output 2:
[10, 50, 5, 50]

Output

Java

public class ProductArrayPuzzle {
    public int[] productArray(int[] A) {
        int n = A.length;
        int[] leftProducts = new int[n];
        int[] rightProducts = new int[n];
        int[] productArray = new int[n];
 
        leftProducts[0] = 1;
        rightProducts[n - 1] = 1;
 
        // Calculating products of elements to the left
        for (int i = 1; i < n; i++) {
            leftProducts[i] = A[i - 1] * leftProducts[i - 1];
        }
 
        // Calculating products of elements to the right
        for (int i = n - 2; i >= 0; i--) {
            rightProducts[i] = A[i + 1] * rightProducts[i + 1];
        }
 
        // Calculate the product array
        for (int i = 0; i < n; i++) {
            productArray[i] = leftProducts[i] * rightProducts[i];
        }
 
        return productArray;
    }
}

Python

def product_array(A):
    n = len(A)
    left_products = [0] * n
    right_products = [0] * n
    product_array = [0] * n
 
    left_products[0] = 1
    right_products[n - 1] = 1
 
    # Calculating products of elements to the left
    for i in range(1, n):
        left_products[i] = A[i - 1] * left_products[i - 1]
 
    # Calculating products of elements to the right
    for i in range(n - 2, -1, -1):
        right_products[i] = A[i + 1] * right_products[i + 1]
 
    # Calculate the product array
    for i in range(n):
        product_array[i] = left_products[i] * right_products[i]
 
    return product_array

JavaScript

function productArray(A) {
    const n = A.length;
    const leftProducts = new Array(n).fill(0);
    const rightProducts = new Array(n).fill(0);
    const productArray = new Array(n).fill(0);
 
    leftProducts[0] = 1;
    rightProducts[n - 1] = 1;
 
    // Calculating products of elements to the left
    for (let i = 1; i < n; i++) {
        leftProducts[i] = A[i - 1] * leftProducts[i - 1];
    }
 
    // Calculating products of elements to the right
    for (let i = n - 2; i >= 0; i--) {
        rightProducts[i] = A[i + 1] * rightProducts[i + 1];
    }
 
    // Calculate the product array
    for (let i = 0; i < n; i++) {
        productArray[i] = leftProducts[i] * rightProducts[i];
    }
 
    return productArray;
}

In Place Prefix Sum Range Sum Query