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A Product Array Puzzle | Set 3
  • Difficulty Level : Medium
  • Last Updated : 16 Feb, 2021

Given an array arr[] consisting of N integers, the task is to construct a Product array of the same size without using division (‘/’) operator such that each array element becomes equal to the product of all the elements of arr[] except arr[i].

Examples:

Input: arr[] = {10, 3, 5, 6, 2}
Output: 180 600 360 300 900
Explanation:
3 * 5 * 6 * 2 is the product of all array elements except 10 is 180
10 * 5 * 6 * 2 is the product of all array elements except 3 is 600.
10 * 3 * 6 * 2 is the product of all array elements except 5 is 360.
10 * 3 * 5 * 2 is the product of all array elements except 6 is 300.
10 * 3 * 6 * 5 is the product of all array elements except 2 is 9.

Input: arr[] = {1, 2, 1, 3, 4}
Output: 24 12 24 8 6

Approach: The idea is to use log() and exp() functions instead of log10() and pow(). Below are some observations regarding the same:



  • Suppose M is the multiplication of all the array elements then the element of output array at ith position will be equal M/arr[i].
  • The divisions of two numbers can be performed by using the property of logarithm and exp functions.
    • log(a) - log(b) = log(a/b)
    • exp(log(a/b)) = a/b
    • exp(log(a) - log(b)) = a/b
  • The logarithmic function is not defined for numbers less than zero so to maintain the such cases separately.

Follow the steps below to solve the problem:

  • Initialize two variables, say product = 1 and Z = 1, to store the product of array and count of zero elements.
  • Traverse the array and multiply the product by arr[i] if arr[i] is not equal to 0. Otherwise, increment count of Z by one.
  • Traverse the array arr[] and perform the following:
    • If Z is 1 and arr[i] is not zero then update arr[i] as arr[i] = 0 and continue.
    • Otherwise, if Z is 1 and arr[i] is 0 then update arr[i] as product and continue.
    • Otherwise, if Z is greater than 1 then assign arr[i] as 0 and continue.
    • Now find the value of abs(product)/abs(arr[i]) using the formula discussed above and store it in a variable say curr.
    • If the value of arr[i] and product is negative or if arr[i] and product is positive then assign arr[i] as curr.
    • Otherwise, assign arr[i] as -1*curr.
  • After completing the above steps, print the array arr[].

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Fucntion to form product array
// with O(n) time and O(1) space
void productExceptSelf(int arr[],
                       int N)
{
    // Stores the product of array
    int product = 1;
 
    // Stores the count of zeros
    int z = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If arr[i] is not zero
        if (arr[i])
            product *= arr[i];
 
        // If arr[i] is zero then
        // increment count of z by 1
        z += (arr[i] == 0);
    }
 
    // Stores the absolute value
    // of the product
    int a = abs(product), b;
    for (int i = 0; i < N; i++) {
 
        // If Z is equal to 1
        if (z == 1) {
 
            // If arr[i] is not zero
            if (arr[i])
                arr[i] = 0;
 
            // Else
            else
                arr[i] = product;
            continue;
        }
 
        // If count of 0s at least 2
        else if (z > 1) {
 
            // Assign arr[i] = 0
            arr[i] = 0;
            continue;
        }
 
        // Store absoulute value of arr[i]
        int b = abs(arr[i]);
 
        // Find the value of a/b
        int curr = round(exp(log(a) - log(b)));
 
        // If arr[i] and product both
        // are less than zero
        if (arr[i] < 0 && product < 0)
            arr[i] = curr;
 
        // If arr[i] and product both
        // are greater than zero
        else if (arr[i] > 0 && product > 0)
            arr[i] = curr;
 
        // Else
        else
            arr[i] = -1 * curr;
    }
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
        cout << arr[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 10, 3, 5, 6, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    productExceptSelf(arr, N);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Fucntion to form product array
// with O(n) time and O(1) space
static void productExceptSelf(int arr[],
                       int N)
{
    // Stores the product of array
    int product = 1;
 
    // Stores the count of zeros
    int z = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If arr[i] is not zero
        if (arr[i] != 0)
            product *= arr[i];
 
        // If arr[i] is zero then
        // increment count of z by 1
        if (arr[i] == 0)
            z += 1;
    }
 
    // Stores the absolute value
    // of the product
    int a = Math.abs(product);
    for (int i = 0; i < N; i++) {
 
        // If Z is equal to 1
        if (z == 1) {
 
            // If arr[i] is not zero
            if (arr[i] != 0)
                arr[i] = 0;
 
            // Else
            else
                arr[i] = product;
            continue;
        }
 
        // If count of 0s at least 2
        else if (z > 1) {
 
            // Assign arr[i] = 0
            arr[i] = 0;
            continue;
        }
 
        // Store absoulute value of arr[i]
        int b = Math.abs(arr[i]);
 
        // Find the value of a/b
        int curr = (int)Math.round(Math.exp(Math.log(a) - Math.log(b)));
 
        // If arr[i] and product both
        // are less than zero
        if (arr[i] < 0 && product < 0)
            arr[i] = curr;
 
        // If arr[i] and product both
        // are greater than zero
        else if (arr[i] > 0 && product > 0)
            arr[i] = curr;
 
        // Else
        else
            arr[i] = -1 * curr;
    }
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
        System.out.print(arr[i] + " ");
    }
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 10, 3, 5, 6, 2 };
    int N = arr.length;
 
    // Function Call
    productExceptSelf(arr, N);
}
}
 
// This code is contributed by splevel62.

Python3




# Python 3 program for the above approach
import math
 
# Fucntion to form product array
# with O(n) time and O(1) space
def productExceptSelf(arr, N) :
     
    # Stores the product of array
    product = 1
 
    # Stores the count of zeros
    z = 0
 
    # Traverse the array
    for i in range(N):
 
        # If arr[i] is not zero
        if (arr[i] != 0) :
            product *= arr[i]
 
        # If arr[i] is zero then
        # increment count of z by 1
        if(arr[i] == 0):
            z += 1
     
    # Stores the absolute value
    # of the product
    a = abs(product)
    for i in range(N):
 
        # If Z is equal to 1
        if (z == 1) :
 
            # If arr[i] is not zero
            if (arr[i] != 0) :
                arr[i] = 0
 
            # Else
            else :
                arr[i] = product
            continue
         
 
        # If count of 0s at least 2
        elif (z > 1) :
 
            # Assign arr[i] = 0
            arr[i] = 0
            continue
         
 
        # Store absoulute value of arr[i]
        b = abs(arr[i])
 
        # Find the value of a/b
        curr = round(math.exp(math.log(a) - math.log(b)))
 
        # If arr[i] and product both
        # are less than zero
        if (arr[i] < 0 and product < 0):
            arr[i] = curr
 
        # If arr[i] and product both
        # are greater than zero
        elif (arr[i] > 0 and product > 0):
            arr[i] = curr
 
        # Else
        else:
            arr[i] = -1 * curr
     
    # Traverse the array arr[]
    for i in range(N):
        print(arr[i], end = " ")
     
# Driver Code
arr = [ 10, 3, 5, 6, 2 ]
N = len(arr)
 
# Function Call
productExceptSelf(arr, N)
 
# This code is contributed by code_hunt.

C#




// C# program for the above approach
using System;
class GFG
{
     
// Fucntion to form product array
// with O(n) time and O(1) space
static void productExceptSelf(int[] arr,
                       int N)
{
    // Stores the product of array
    int product = 1;
 
    // Stores the count of zeros
    int z = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
 
        // If arr[i] is not zero
        if (arr[i] != 0)
            product *= arr[i];
 
        // If arr[i] is zero then
        // increment count of z by 1
        if (arr[i] == 0)
            z += 1;
    }
 
    // Stores the absolute value
    // of the product
    int a = Math.Abs(product);
    for (int i = 0; i < N; i++)
    {
 
        // If Z is equal to 1
        if (z == 1)
        {
 
            // If arr[i] is not zero
            if (arr[i] != 0)
                arr[i] = 0;
 
            // Else
            else
                arr[i] = product;
            continue;
        }
 
        // If count of 0s at least 2
        else if (z > 1)
        {
 
            // Assign arr[i] = 0
            arr[i] = 0;
            continue;
        }
 
        // Store absoulute value of arr[i]
        int b = Math.Abs(arr[i]);
 
        // Find the value of a/b
        int curr = (int)Math.Round(Math.Exp(Math.Log(a) - Math.Log(b)));
 
        // If arr[i] and product both
        // are less than zero
        if (arr[i] < 0 && product < 0)
            arr[i] = curr;
 
        // If arr[i] and product both
        // are greater than zero
        else if (arr[i] > 0 && product > 0)
            arr[i] = curr;
 
        // Else
        else
            arr[i] = -1 * curr;
    }
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++)
    {
        Console.Write(arr[i] + " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] arr = { 10, 3, 5, 6, 2 };
    int N = arr.Length;
 
    // Function Call
    productExceptSelf(arr, N);
}
}
 
// This code is contributed by sanjoy_62.
Output: 
180 600 360 300 900

 

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


 

Alternate Approaches: Please refer to the previous posts of this article for alternate approaches:

 

 

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