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

• Difficulty Level : Hard
• Last Updated : 19 May, 2021

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}```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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])).

## 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;
}
```

## 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
```

## Python

```# 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
```

## 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
```

## 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
?>

```

## 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>
```
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.

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))

## 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 Sitesh Roy```

## 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
```

## 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```

## 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
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