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])).

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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])))),
      
    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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>
chevron_right

Output:
The product array is: 
180 600 360 300 900

Complexity Analysis:

This approach is contributed by Abhishek Rajput.

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))

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 << prod * (int)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 Sitesh Roy
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

Output:
180 600 360 300 900

Complexity Analysis:

This approach is given by Sitesh Roy.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.





Article Tags :