# 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 and in O(n).

Example:
arr[] = {10, 3, 5, 6, 2}
prod[] = {180, 600, 360, 300, 900}

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

Algorithm:
1) Construct a temporary array left[] such that left[i] contains product of all elements on left of arr[i] excluding arr[i].
2) Construct another temporary array right[] such that right[i] contains product of all elements on on right of arr[i] excluding arr[i].
3) To get prod[], multiply left[] and right[].

Implementation:

## C/C++

```#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)
{
/* 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();
}
```

## Java

```class ProductArray
{
/* Function to print product array for a given array
arr[] of size n */
void productArray(int arr[], int n)
{
// 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

```

Time Complexity: O(n)
Space Complexity: O(n)
Auxiliary Space: O(n)

The above method can be optimized to work in space complexity O(1). Thanks to Dileep for suggesting the below solution.

## C/C++

```
void productArray(int arr[], int n)
{
int i, temp = 1;

/* Allocate memory for the product array */
int *prod = (int *)malloc(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++)
printf("%d ", prod[i]);

return;
}
```

## Java

```class ProductArray
{
void productArray(int arr[], int n)
{
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
```

Time Complexity: O(n)
Space Complexity: O(n)
Auxiliary Space: O(1)

### Asked in: Accolite,Amazon,D-E-Shaw,Intuit ,Opera

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

Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.

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