A product array puzzle | Set 2 (O(1) Space)

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 and in O(n).

Example:

Input: arr[] = {10, 3, 5, 6, 2}
Output: prod[] = {180, 600, 360, 300, 900}
The elements of output array are 
{3*5*6*2, 10*5*6*2, 10*3*6*2, 
10*3*5*2, 10*3*5*6}
Input: arr[] = {1, 2, 1, 3, 4}
Output: prod[] = {24, 12, 24, 8, 6}
The elements of output array are 
{3*4*1*2, 1*1*3*4, 4*3*2*1, 
1*1*4*2, 1*1*3*2}

There is already a discussed O(n) approach in A product array puzzle | set 1. The previous approach uses extra O(n) space for constructing product array.

Solution 1: Using log property. 

Approach: In this post, a better approach has been discussed which uses log property to find the product of all elements of the array except at a particular index. This approach uses no extra space.

Use property of log to multiply large numbers

x = a * b * c * d
log(x) = log(a * b * c * d)
log(x) = log(a) + log(b) + log(c) + log(d)
x = antilog(log(a) + log(b) + log(c) + log(d))

So the idea is simple,  
Traverse the array and find the sum of log of all the elements,   

log(a[0]) + log(a[1]) + 
.. + log(a[n-1])

Then again traverse through the array and find the product using this formula.  

antilog((log(a[0]) + log(a[1]) +
 .. + log(a[n-1])) - log(a[i]))

This equals to product of all the elements except a[i], i.e antilog(sum- log(a[i])). 

Implementation:

C++

// C++ program for product array puzzle
// with O(n) time and O(1) space.
#include <bits/stdc++.h>
using namespace std;

// epsilon value to maintain precision
#define EPS 1e-9

void productPuzzle(int a[], int n)
{
    // to hold sum of all values
    long double sum = 0;
    for (int i = 0; i < n; i++)
        sum += (long double)log10(a[i]);

    // output product for each index
    // antilog to find original product value
    for (int i = 0; i < n; i++)
        cout << (int)(EPS + pow((long double)10.00, sum - log10(a[i]))) << " ";
}

// Driver code
int main()
{
    int a[] = { 10, 3, 5, 6, 2 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << "The product array is: \n";
    productPuzzle(a, n);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program for product array puzzle
// with O(n) time and O(1) space.
#include <stdio.h>
#include <math.h>

// epsilon value to maintain precision
#define EPS 1e-9

void productPuzzle(int a[], int n)
{
    // to hold sum of all values
    long double sum = 0;
    for (int i = 0; i < n; i++)
        sum += (long double)log10(a[i]);

    // output product for each index
    // antilog to find original product value
    for (int i = 0; i < n; i++)
        printf("%d ",(int)(EPS + pow((long double)10.00, sum - log10(a[i]))));
}

// Driver code
int main()
{
    int a[] = { 10, 3, 5, 6, 2 };
    int n = sizeof(a) / sizeof(a[0]);
    printf("The product array is: \n");
    productPuzzle(a, n);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program for product array puzzle
// with O(n) time and O(1) space.
public class Array_puzzle_2 {

    // epsilon value to maintain precision
    static final double EPS = 1e-9;

    static void productPuzzle(int a[], int n)
    {
        // to hold sum of all values
        double sum = 0;
        for (int i = 0; i < n; i++)
            sum += Math.log10(a[i]);

        // output product for each index
        // anti log to find original product value
        for (int i = 0; i < n; i++)
            System.out.print(
                (int)(EPS
                      + Math.pow(
                            10.00, sum
                                       - Math.log10(a[i])))
                + " ");
    }

    // Driver code
    public static void main(String args[])
    {
        int a[] = { 10, 3, 5, 6, 2 };
        int n = a.length;
        System.out.println("The product array is: ");
        productPuzzle(a, n);
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Python program for product array puzzle
# with O(n) time and O(1) space.

import math

# epsilon value to maintain precision
EPS = 1e-9

def productPuzzle(a, n):

    # to hold sum of all values
    sum = 0
    for i in range(n):
        sum += math.log10(a[i])

    # output product for each index
    # antilog to find original product value
    for i in range(n):
        print (int((EPS + pow(10.00, sum - math.log10(a[i])))),end=" ")

    return

# Driver code
a = [10, 3, 5, 6, 2 ]
n = len(a)
print ("The product array is: ")
productPuzzle(a, n)

# This code is contributed by Sachin Bisht

C#

// C# program for product
// array puzzle with O(n)
// time and O(1) space.
using System;
class GFG {

    // epsilon value to
    // maintain precision
    static double EPS = 1e-9;

    static void productPuzzle(int[] a,
                              int n)
    {
        // to hold sum of all values
        double sum = 0;
        for (int i = 0; i < n; i++)
            sum += Math.Log10(a[i]);

        // output product for each
        // index anti log to find
        // original product value
        for (int i = 0; i < n; i++)
            Console.Write((int)(EPS + Math.Pow(10.00, sum - Math.Log10(a[i]))) + " ");
    }

    // Driver code
    public static void Main()
    {
        int[] a = { 10, 3, 5, 6, 2 };
        int n = a.Length;
        Console.WriteLine("The product array is: ");
        productPuzzle(a, n);
    }
}

// This code is contributed by mits

Javascript

<script>

// javascript program for product array puzzle
// with O(n) time and O(1) space.

// epsilon value to maintain precision
var EPS = 1e-9;

function productPuzzle(a , n)
{
    // to hold sum of all values
    var sum = 0;
    for (var i = 0; i < n; i++)
        sum += Math.log10(a[i]);

    // output product for each index
    // anti log to find original product value
    for (var i = 0; i < n; i++)
        document.write(
            parseInt((EPS
                  + Math.pow(
                        10.00, sum
                                   - Math.log10(a[i]))))
            + " ");
}

// Driver code
var a = [ 10, 3, 5, 6, 2 ];
var n = a.length;
document.write("The product array is: ");
productPuzzle(a, n);

// This code is contributed by 29AjayKumar
</script>

PHP

<?php
// PHP program for product array puzzle
// with O(n) time and O(1) space.

// epsilon value to maintain precision
$EPS=1e-9;

function productPuzzle($a, $n)
{
    global $EPS;
    // to hold sum of all values
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        $sum += (double)log10($a[$i]);

    // output product for each index
    // antilog to find original product value
    for ($i = 0; $i < $n; $i++)
        echo (int)($EPS + pow((double)10.00, $sum - log10($a[$i])))." ";
}

// Driver code

    $a = array(10, 3, 5, 6, 2 );
    $n = count($a);
    echo "The product array is: \n";
    productPuzzle($a, $n);

// This code is contributed by mits
?>

Output

The product array is: 
180 600 360 300 900

Complexity Analysis:

• Time Complexity: O(n). 
  Only two traversals of the array is required.
• Space Complexity: O(1). 
  No extra space is required.

Alternate Approach: Here’s another approach to solve the above problem by the use of pow() function, does not use division and works in O(n) time. 
Traverse the array and find the product of all the elements in the array. Store the product in a variable. 

Then again traverse the array and find the product of all the elements except that number by using the formula (product * pow(a[i], -1)) 

Implementation:

C++

// C++ program for product array puzzle
// with O(n) time and O(1) space.
#include <bits/stdc++.h>
using namespace std;

// Solve function which prints the answer
void solve(int arr[], int n)
{

    // Initialize a variable to store the
    // total product of the array elements
    int prod = 1;
    for (int i = 0; i < n; i++)
        prod *= arr[i];

    // we know x/y mathematically is same
    // as x*(y to power -1)
    for (int i = 0; i < n; i++) {
        cout << (int)(prod * pow(arr[i], -1)) << ' ';
    }
}

// Driver Code
int main()
{
    int arr[] = { 10, 3, 5, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    solve(arr, n);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C#

// C# program for product array puzzle
// with O(n) time and O(1) space.
using System;

class GFG {

public
    class ArrayPuzzle {

        static void solve(int[] arr, int n)
        {
            // Initialize a variable to store the
            // total product of the array elements
            int prod = 1;
            for (int i = 0; i < n; i++)
                prod *= arr[i];

            // we know x/y mathematically is same
            // as x*(y to power -1)
            for (int i = 0; i < n; i++)
                Console.Write(
                    (int)prod * Math.Pow(arr[i], -1) + " ");
        }

        // Driver code
        static public void Main()
        {
            int[] arr = { 10, 3, 5, 6, 2 };
            int n = arr.Length;
            solve(arr, n);
        }
    }
}
// This code is contributed by shivanisinghss2110

C

// C program for product array puzzle
// with O(n) time and O(1) space.
#include <math.h>
#include <stdio.h>

// Solve function which prints the answer
void solve(int arr[], int n)
{

    // Initialize a variable to store the
    // total product of the array elements
    int prod = 1;
    for (int i = 0; i < n; i++)
        prod *= arr[i];

    // we know x/y mathematically is same
    // as x*(y to power -1)
    for (int i = 0; i < n; i++)
        printf("%d ", (int)(prod * pow(arr[i], -1)));
}

// Driver Code
int main()
{
    int arr[] = { 10, 3, 5, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    solve(arr, n);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program for product array puzzle
// with O(n) time and O(1) space.
public class ArrayPuzzle {

    static void solve(int arr[], int n)
    {
        // Initialize a variable to store the
        // total product of the array elements
        int prod = 1;
        for (int i = 0; i < n; i++)
            prod *= arr[i];

        // we know x/y mathematically is same
        // as x*(y to power -1)
        for (int i = 0; i < n; i++)
            System.out.print(
                (int)prod * Math.pow(arr[i], -1) + " ");
    }

    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 10, 3, 5, 6, 2 };
        int n = arr.length;
        solve(arr, n);
    }
}
// This code is contributed by Sitesh Roy

Python3

# Python program for product array puzzle
# with O(n) time and O(1) space.
def solve(arr, n):

    # Initialize a variable to store the
    # total product of the array elements
    prod = 1
    for i in arr:
        prod *= i

    # we know x / y mathematically is same
    # as x*(y to power -1)
    for i in arr:
        print(int(prod*(i**-1)), end =" ")

# Driver Code
arr = [10, 3, 5, 6, 2]
n = len(arr)
solve(arr, n)


# This code is contributed by Sitesh Roy

Javascript

<script>

// Javascript program for product array puzzle
// with O(n) time and O(1) space.

    function solve(arr, n)
    {
        // Initialize a variable to store the
        // total product of the array elements
        let prod = 1;
        for (let i = 0; i < n; i++)
            prod *= arr[i];

        // we know x/y mathematically is same
        // as x*(y to power -1)
        for (let i = 0; i < n; i++)
            document.write(
                prod * Math.pow(arr[i], -1) + " ");
    }

// Driver program
        let arr = [10, 3, 5, 6, 2 ];
        let n = arr.length;
        solve(arr, n);

// This code is contributed by code_hunt.
</script>

Output

180 600 360 300 900

Complexity Analysis:

• Time complexity: O(n), Only two traversals of the array is required.
• Space complexity: O(1), No extra space is required.

Note: This approach assumes that the array elements are not 0.
This approach is given by Sitesh Roy.