A Product Array Puzzle | Set 3

• Difficulty Level : Medium
• Last Updated : 26 Sep, 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.
• • • • 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 using namespace std; // Function to form product array// with O(n) time and O(1) spacevoid 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 absolute 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 Codeint main(){    int arr[] = { 10, 3, 5, 6, 2 };    int N = sizeof(arr) / sizeof(arr);     // Function Call    productExceptSelf(arr, N);     return 0;}

Java

 // Java program to implement// the above approachimport java.util.*; class GFG{ // Function to form product array// with O(n) time and O(1) spacestatic 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 absolute 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 Codepublic 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 approachimport math # Function to form product array# with O(n) time and O(1) spacedef 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 absolute 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 Codearr = [ 10, 3, 5, 6, 2 ]N = len(arr) # Function CallproductExceptSelf(arr, N) # This code is contributed by code_hunt.

C#

 // C# program for the above approachusing System;class GFG{     // Function to form product array// with O(n) time and O(1) spacestatic 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 absolute 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 Codepublic 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.

Javascript


Output:
180 600 360 300 900

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