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# Smallest index that splits an array into two subarrays with equal product

• Difficulty Level : Medium
• Last Updated : 15 Jun, 2021

Given an array(1-based indexing) arr[] consisting of N non zero integers, the task is to find the leftmost index i such that the product of all the elements of the subarrays arr[1, i] and arr[i + 1, N] is the same.

Examples:

Input: arr[] = {1, 2, 3, 3, 2, 1}
Output: 3
Explanation: Index 3 generates subarray {arr[1], arr[3]} with product 6 (= 1 * 2 * 3) and {arr[4], arr[6]} with product 6 ( = 3 * 2 * 1).

Input: arr = {3, 2, 6}
Output: 2

Approach: Follow the steps below to solve the problem:

• Initialize a variable, say product, > that stores the product of all the array elements.
• Traverse the given array and find the product of all the array elements store it in the product.
• Initialize two variables left and right to 1 that stores the product of the left and the right subarray
• Traverse the given array and perform the following steps:
• Multiply the value of left by arr[i].
• Divide the value of right by arr[i].
• If the value of left is equal to right, then print the value of the current index i as the resultant index and break out of the loop.
• After completing the above steps, if any such index doesn’t exist, then print “-1” as the result.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std; // Function to find the smallest// index that splits the array into// two subarrays with equal productvoid prodEquilibrium(int arr[], int N){    // Stores the product of the array    int product = 1;     // Traverse the given array    for (int i = 0; i < N; i++) {        product *= arr[i];    }     // Stores the product of left    // and the right subarrays    int left = 1;    int right = product;     // Traverse the given array    for (int i = 0; i < N; i++) {         // Update the products        left = left * arr[i];        right = right / arr[i];         // Check if product is equal        if (left == right) {             // Print resultant index            cout << i + 1 << endl;            return;        }    }     // If no partition exists, then    // print -1.    cout << -1 << endl;} // Driver Codeint main(){    int arr[] = { 1, 2, 3, 3, 2, 1 };    int N = sizeof(arr) / sizeof(arr[0]);    prodEquilibrium(arr, N);     return 0;}

## Java

 // Java program for the above approachimport java.util.*; class GFG{ // Function to find the smallest// index that splits the array into// two subarrays with equal productstatic void prodEquilibrium(int arr[], int N){         // Stores the product of the array    int product = 1;     // Traverse the given array    for(int i = 0; i < N; i++)    {        product *= arr[i];    }     // Stores the product of left    // and the right subarrays    int left = 1;    int right = product;     // Traverse the given array    for(int i = 0; i < N; i++)    {                 // Update the products        left = left * arr[i];        right = right / arr[i];         // Check if product is equal        if (left == right)        {                         // Print resultant index            System.out.print(i + 1 + "\n");            return;        }    }     // If no partition exists, then    // print -1.    System.out.print(-1 + "\n");} // Driver Codepublic static void main(String[] args){    int arr[] = { 1, 2, 3, 3, 2, 1 };    int N = arr.length;         prodEquilibrium(arr, N);}} // This code is contributed by 29AjayKumar

## Python3

 # Python 3 program for the above approach # Function to find the smallest# index that splits the array into# two subarrays with equal productdef prodEquilibrium(arr, N):       # Stores the product of the array    product = 1     # Traverse the given array    for i in range(N):        product *= arr[i]     # Stores the product of left    # and the right subarrays    left = 1    right = product     # Traverse the given array    for i in range(N):        # Update the products        left = left * arr[i]        right = right // arr[i]         # Check if product is equal        if (left == right):            # Print resultant index            print(i + 1)            return     # If no partition exists, then    # print -1.    print(-1) # Driver Codeif __name__ == '__main__':    arr = [1, 2, 3, 3, 2, 1]    N = len(arr)    prodEquilibrium(arr, N)         # This code is contributed by ipg2016107.

## C#

 // C# program for the above approachusing System;class GFG {     // Function to find the smallest    // index that splits the array into    // two subarrays with equal product    static void prodEquilibrium(int[] arr, int N)    {         // Stores the product of the array        int product = 1;         // Traverse the given array        for (int i = 0; i < N; i++) {            product *= arr[i];        }         // Stores the product of left        // and the right subarrays        int left = 1;        int right = product;         // Traverse the given array        for (int i = 0; i < N; i++) {             // Update the products            left = left * arr[i];            right = right / arr[i];             // Check if product is equal            if (left == right) {                 // Print resultant index                Console.WriteLine(i + 1 + "\n");                return;            }        }         // If no partition exists, then        // print -1.        Console.WriteLine(-1 + "\n");    }     // Driver Code    public static void Main(string[] args)    {        int[] arr = { 1, 2, 3, 3, 2, 1 };        int N = arr.Length;         prodEquilibrium(arr, N);    }}  // This code is contributed by ukasp.

## Javascript


Output:
3

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

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