A Product Array Puzzle

Given an array arr[] of n integers, construct a Product Array prod[] (of same size) such that prod[i] is equal to the product of all the elements of arr[] except arr[i]. Solve it without 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

Naive Solution:

Approach: 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, initilize 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.

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
/* Function to print product array 
for a given array arr[] of size n */
void productArray(int arr[], int n)
{
  
    // Base case
    if (n == 1) {
        cout << 0;
        return;
    }
    /* Allocate memory for temporary 
arrays left[] and right[] */
    int* left = new int[sizeof(int) * n];
    int* right = new int[sizeof(int) * n];
  
    /* Allocate memory for the product array */
    int* prod = new int[sizeof(int) * n];
  
    int i, j;
  
    /* Left most element of left 
array is always 1 */
    left[0] = 1;
  
    /* Rightmost most element of right 
array is always 1 */
    right[n - 1] = 1;
  
    /* Construct the left array */
    for (i = 1; i < n; i++)
        left[i] = arr[i - 1] * left[i - 1];
  
    /* Construct the right array */
    for (j = n - 2; j >= 0; j--)
        right[j] = arr[j + 1] * right[j + 1];
  
    /* Construct the product array using 
        left[] and right[] */
    for (i = 0; i < n; i++)
        prod[i] = left[i] * right[i];
  
    /* print the constructed prod array */
    for (i = 0; i < n; i++)
        cout << prod[i] << " ";
  
    return;
}
  
/* Driver code*/
int main()
{
    int arr[] = { 10, 3, 5, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "The product array is: \n";
    productArray(arr, n);
}
  
// This is code is contributed by rathbhupendra

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C

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#include <stdio.h>
#include <stdlib.h>
  
/* Function to print product array 
for a given array arr[] of size n */
void productArray(int arr[], int n)
{
  
    // Base case
    if (n == 1) {
        printf("0");
        return;
    }
  
    /* Allocate memory for temporary 
arrays left[] and right[] */
    int* left = (int*)malloc(
        sizeof(int) * n);
    int* right = (int*)malloc(
        sizeof(int) * n);
  
    /* Allocate memory for the product array */
    int* prod = (int*)malloc(
        sizeof(int) * n);
  
    int i, j;
  
    /* Left most element of left array 
is always 1 */
    left[0] = 1;
  
    /* Rightmost most element of right 
array is always 1 */
    right[n - 1] = 1;
  
    /* Construct the left array */
    for (i = 1; i < n; i++)
        left[i] = arr[i - 1] * left[i - 1];
  
    /* Construct the right array */
    for (j = n - 2; j >= 0; j--)
        right[j] = arr[j + 1] * right[j + 1];
  
    /* Construct the product array using 
    left[] and right[] */
    for (i = 0; i < n; i++)
        prod[i] = left[i] * right[i];
  
    /* print the constructed prod array */
    for (i = 0; i < n; i++)
        printf("%d ", prod[i]);
  
    return;
}
  
/* Driver program to test above functions */
int main()
{
    int arr[] = { 10, 3, 5, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("The product array is: \n");
    productArray(arr, n);
    getchar();
}

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Java

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class ProductArray {
    /* Function to print product array 
    for a given array arr[] of size n */
    void productArray(int arr[], int n)
    {
  
        // Base case
        if (n == 1) {
            System.out.print(0);
            return;
        }
        // Initialize memory to all arrays
        int left[] = new int[n];
        int right[] = new int[n];
        int prod[] = new int[n];
  
        int i, j;
  
        /* Left most element of left array 
is always 1 */
        left[0] = 1;
  
        /* Rightmost most element of right 
array is always 1 */
        right[n - 1] = 1;
  
        /* Construct the left array */
        for (i = 1; i < n; i++)
            left[i] = arr[i - 1] * left[i - 1];
  
        /* Construct the right array */
        for (j = n - 2; j >= 0; j--)
            right[j] = arr[j + 1] * right[j + 1];
  
        /* Construct the product array using 
        left[] and right[] */
        for (i = 0; i < n; i++)
            prod[i] = left[i] * right[i];
  
        /* print the constructed prod array */
        for (i = 0; i < n; i++)
            System.out.print(prod[i] + " ");
  
        return;
    }
  
    /* Driver program to test the aboe function */
    public static void main(String[] args)
    {
        ProductArray pa = new ProductArray();
        int arr[] = { 10, 3, 5, 6, 2 };
        int n = arr.length;
        System.out.println("The product array is : ");
        pa.productArray(arr, n);
    }
}
  
// This code has been contributed by Mayank Jaiswal

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Python3

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# Python implementation of the above approach 
  
# Function to print product array for a given array 
# arr[] of size n 
  
  
def productArray(arr, n): 
  
    # Base case
    if(n == 1):
        print(0)
        return
          
    # Allocate memory for temporary arrays left[] and right[] 
    left = [0]*
    right = [0]*
  
    # Allocate memory for the product array 
    prod = [0]*
  
    # Left most element of left array is always 1 
    left[0] = 1
  
    # Rightmost most element of right array is always 1 
    right[n - 1] = 1
  
    # Construct the left array 
    for i in range(1, n): 
        left[i] = arr[i - 1] * left[i - 1
  
    # Construct the right array 
    for j in range(n-2, -1, -1): 
        right[j] = arr[j + 1] * right[j + 1
  
    # Construct the product array using 
    # left[] and right[] 
    for i in range(n): 
        prod[i] = left[i] * right[i] 
  
    # print the constructed prod array 
    for i in range(n): 
        print(prod[i], end =' '
  
  
# Driver code 
arr = [10, 3, 5, 6, 2
n = len(arr) 
print("The product array is:"
productArray(arr, n) 
  
# This code is contributed by ankush_953 

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C#

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using System;
  
class GFG {
  
    /* Function to print product array 
    for a given array arr[] of size n */
    static void productArray(int[] arr, int n)
    {
  
        // Base case
        if (n == 1) {
            Console.Write(0);
            return;
        }
        // Initialize memory to all arrays
        int[] left = new int[n];
        int[] right = new int[n];
        int[] prod = new int[n];
  
        int i, j;
  
        /* Left most element of left array 
        is always 1 */
        left[0] = 1;
  
        /* Rightmost most element of right 
        array is always 1 */
        right[n - 1] = 1;
  
        /* Construct the left array */
        for (i = 1; i < n; i++)
            left[i] = arr[i - 1] * left[i - 1];
  
        /* Construct the right array */
        for (j = n - 2; j >= 0; j--)
            right[j] = arr[j + 1] * right[j + 1];
  
        /* Construct the product array using 
        left[] and right[] */
        for (i = 0; i < n; i++)
            prod[i] = left[i] * right[i];
  
        /* print the constructed prod array */
        for (i = 0; i < n; i++)
            Console.Write(prod[i] + " ");
  
        return;
    }
  
    /* Driver program to test the aboe function */
    public static void Main()
    {
        int[] arr = { 10, 3, 5, 6, 2 };
        int n = arr.Length;
        Console.Write("The product array is :\n");
  
        productArray(arr, n);
    }
}
  
// This code is contributed by nitin mittal.

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PHP

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<?php 
// Function to print product 
// array for a given array 
// arr[] of size n 
function productArray($arr, $n
  
    // Base case
    if($n == 1) {
        echo "0";
        return;
    }
    // Initialize memory 
    // to all arrays 
    $left = array(); 
    $right = array(); 
    $prod = array(); 
  
    $i; $j
  
    // Left most element of 
    // left array is always 1 
    $left[0] = 1; 
  
    // Rightmost most element of 
    // right array is always 1 
    $right[$n - 1] = 1; 
  
    // Construct the left array 
    for ($i = 1; $i < $n; $i++) 
        $left[$i] = $arr[$i - 1] * 
                    $left[$i - 1]; 
  
    // Construct the right array 
    for ($j = $n - 2; $j >= 0; $j--) 
        $right[$j] = $arr[$j + 1] * 
                    $right[$j + 1]; 
  
    // Construct the product array 
    // using left[] and right[] 
    for ($i = 0; $i < $n; $i++) 
        $prod[$i] = $left[$i] * 
                    $right[$i]; 
  
    // print the constructed prod array 
    for ($i = 0; $i < $n; $i++) 
        echo $prod[$i], " "
  
    return
  
// Driver Code 
$arr = array(10, 3, 5, 6, 2); 
$n = count($arr); 
echo "The product array is : \n"
productArray($arr, $n); 
  
// This code has been contributed by anuj_67. 
?> 

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Output :

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).
  • Space Complexity: 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 solution:

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 initilize 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. initilize 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.

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
/* Function to print product array 
for a given array arr[] of size n */
void productArray(int arr[], int n)
{
  
    // Base case
    if (n == 1) {
        cout << 0;
        return;
    }
  
    int i, temp = 1;
  
    /* Allocate memory for the product array */
    int* prod = new int[(sizeof(int) * n)];
  
    /* Initialize the product array as 1 */
    memset(prod, 1, n);
  
    /* In this loop, temp variable contains product of 
       elements on left side excluding arr[i] */
    for (i = 0; i < n; i++) {
        prod[i] = temp;
        temp *= arr[i];
    }
  
    /* Initialize temp to 1 
    for product on right side */
    temp = 1;
  
    /* In this loop, temp variable contains product of 
       elements on right side excluding arr[i] */
    for (i = n - 1; i >= 0; i--) {
        prod[i] *= temp;
        temp *= arr[i];
    }
  
    /* print the constructed prod array */
    for (i = 0; i < n; i++)
        cout << prod[i] << " ";
  
    return;
}
  
// Driver Code
int main()
{
    int arr[] = { 10, 3, 5, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "The product array is: \n";
    productArray(arr, n);
}
  
// This code is contributed by rathbhupendra

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Java

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class ProductArray {
    void productArray(int arr[], int n)
    {
  
        // Base case
        if (n == 1) {
            System.out.print("0");
            return;
        }
  
        int i, temp = 1;
  
        /* Allocate memory for the product array */
        int prod[] = new int[n];
  
        /* Initialize the product array as 1 */
        for (int j = 0; j < n; j++)
            prod[j] = 1;
  
        /* In this loop, temp variable contains product of
           elements on left side excluding arr[i] */
        for (i = 0; i < n; i++) {
            prod[i] = temp;
            temp *= arr[i];
        }
  
        /* Initialize temp to 1 for product on right side */
        temp = 1;
  
        /* In this loop, temp variable contains product of
           elements on right side excluding arr[i] */
        for (i = n - 1; i >= 0; i--) {
            prod[i] *= temp;
            temp *= arr[i];
        }
  
        /* print the constructed prod array */
        for (i = 0; i < n; i++)
            System.out.print(prod[i] + " ");
  
        return;
    }
  
    /* Driver program to test above functions */
    public static void main(String[] args)
    {
        ProductArray pa = new ProductArray();
        int arr[] = { 10, 3, 5, 6, 2 };
        int n = arr.length;
        System.out.println("The product array is : ");
        pa.productArray(arr, n);
    }
}
  
// This code has been contributed by Mayank Jaiswal

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Python3

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# Python3 program for A Product Array Puzzle
def productArray(arr, n):
  
    # Base case
    if n == 1:
        print(0)
        return
  
    i, temp = 1, 1
  
    # Allocate memory for the product array 
    prod = [1 for i in range(n)]
  
    # Initialize the product array as 1 
  
    # In this loop, temp variable contains product of
    # elements on left side excluding arr[i] 
    for i in range(n):
        prod[i] = temp
        temp *= arr[i]
  
    # Initialize temp to 1 for product on right side 
    temp = 1
  
    # In this loop, temp variable contains product of
    # elements on right side excluding arr[i] 
    for i in range(n - 1, -1, -1):
        prod[i] *= temp
        temp *= arr[i]
  
    # Print the constructed prod array 
    for i in range(n):
        print(prod[i], end = " ")
  
    return
  
# Driver Code
arr = [10, 3, 5, 6, 2]
n = len(arr)
print("The product array is: n")
productArray(arr, n)
  
# This code is contributed by mohit kumar

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C#

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using System;
  
class GFG {
  
    static void productArray(int[] arr, int n)
    {
  
        // Base case
        if (n == 1) {
            Console.Write(0);
            return;
        }
        int i, temp = 1;
  
        /* Allocate memory for the product
        array */
        int[] prod = new int[n];
  
        /* Initialize the product array as 1 */
        for (int j = 0; j < n; j++)
            prod[j] = 1;
  
        /* In this loop, temp variable contains
        product of elements on left side
        excluding arr[i] */
        for (i = 0; i < n; i++) {
            prod[i] = temp;
            temp *= arr[i];
        }
  
        /* Initialize temp to 1 for product on 
        right side */
        temp = 1;
  
        /* In this loop, temp variable contains
        product of elements on right side 
        excluding arr[i] */
        for (i = n - 1; i >= 0; i--) {
            prod[i] *= temp;
            temp *= arr[i];
        }
  
        /* print the constructed prod array */
        for (i = 0; i < n; i++)
            Console.Write(prod[i] + " ");
  
        return;
    }
  
    /* Driver program to test above functions */
    public static void Main()
    {
        int[] arr = { 10, 3, 5, 6, 2 };
        int n = arr.Length;
        Console.WriteLine("The product array is : ");
  
        productArray(arr, n);
    }
}
  
// This code is contributed by nitin mittal.

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PHP

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<?php
// PHP program for 
// A Product Array Puzzle
      
function productArray($arr, $n
    {
  
        // Base case
        if ($n == 1) {
            echo "0";
            return;
        }
        $i; $temp = 1;
          
        /* Allocate memory for 
           the productarray */
        $prod = array();
  
        /* Initialize the product 
           array as 1 */
        for( $j = 0; $j < $n; $j++)
            $prod[$j] = 1;
  
        /* In this loop, temp 
           variable contains
           product of elements
           on left side
           excluding arr[i] */
        for ($i = 0; $i < $n; $i++) 
        {
            $prod[$i] = $temp;
            $temp *= $arr[$i];
        }
  
        /* Initialize temp to 1 
           for product on right
           side */
        $temp = 1;
  
        /* In this loop, temp 
           variable contains
           product of elements 
           on right side 
           excluding arr[i] */
        for ($i = $n - 1; $i >= 0; $i--) 
        {
            $prod[$i] *= $temp;
            $temp *= $arr[$i];
        }
  
        /* print the constructed
           prod array */
        for ($i = 0; $i < $n; $i++)
            echo $prod[$i], " ";
  
        return;
    }
  
        // Driver Code    
        $arr = array(10, 3, 5, 6, 2);
        $n = count($arr);
        echo "The product array is : \n";
        productArray($arr, $n);
      
// This code is contributed by anuj_67.
?>

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Output :

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.
  • Space Complexity: 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.

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