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

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

Brute Force Approach:

To solve this problem is to iterate through the array and keep track of the product of the elements from the beginning of the array to the current index and from the current index to the end of the array. For each index i, if the product of the two subarrays is equal, return i. If no such index is found, return -1.

Implementation of the above approach:

C++

 `#include ` `using` `namespace` `std;`   `// Function to find the smallest index that splits the array` `// into two subarrays with equal product` `void` `prodEquilibrium(``int` `arr[], ``int` `N)` `{` `    ``int` `leftProduct = 1;` `    ``int` `rightProduct = 1;` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// calculate product of elements from 0 to i` `        ``leftProduct *= arr[i];`   `        ``// calculate product of elements from i+1 to N-1` `        ``for` `(``int` `j = i + 1; j < N; j++) {` `            ``rightProduct *= arr[j];` `        ``}`   `        ``// if the product of the two subarrays is equal,` `        ``// return i` `        ``if` `(leftProduct == rightProduct) {` `            ``cout << i + 1 << endl;` `            ``return``;` `        ``}`   `        ``// reset rightProduct for the next iteration` `        ``rightProduct = 1;` `    ``}`   `    ``// if no such index is found, return -1` `    ``cout << -1 << endl;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 3, 3, 2, 1 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``prodEquilibrium(arr, N);`   `    ``return` `0;` `}`

Java

 `import` `java.util.*;`   `public` `class` `Main {`   `    ``// Function to find the smallest index that splits the` `    ``// array into two subarrays with equal product` `    ``static` `void` `prodEquilibrium(``int``[] arr, ``int` `N)` `    ``{` `        ``int` `leftProduct = ``1``;` `        ``int` `rightProduct = ``1``;` `        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``// calculate product of elements from 0 to i` `            ``leftProduct *= arr[i];`   `            ``// calculate product of elements from i+1 to N-1` `            ``for` `(``int` `j = i + ``1``; j < N; j++) {` `                ``rightProduct *= arr[j];` `            ``}`   `            ``// if the product of the two subarrays is equal,` `            ``// return i` `            ``if` `(leftProduct == rightProduct) {` `                ``System.out.println(i + ``1``);` `                ``return``;` `            ``}`   `            ``// reset rightProduct for the next iteration` `            ``rightProduct = ``1``;` `        ``}`   `        ``// if no such index is found, return -1` `        ``System.out.println(-``1``);` `    ``}`   `    ``// 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 Prajwal Kandekar`

Python3

 `def` `prodEquilibrium(arr, N):` `    ``leftProduct ``=` `1` `    ``rightProduct ``=` `1` `    ``for` `i ``in` `range``(N):`   `        ``# calculate product of elements from 0 to i` `        ``leftProduct ``*``=` `arr[i]`   `        ``# calculate product of elements from i+1 to N-1` `        ``for` `j ``in` `range``(i``+``1``, N):` `            ``rightProduct ``*``=` `arr[j]`   `        ``# if the product of the two subarrays is equal, return i` `        ``if` `leftProduct ``=``=` `rightProduct:` `            ``print``(i``+``1``)` `            ``return`   `        ``# reset rightProduct for the next iteration` `        ``rightProduct ``=` `1`   `    ``# if no such index is found, return -1` `    ``print``(``-``1``)`     `# Driver Code` `arr ``=` `[``1``, ``2``, ``3``, ``3``, ``2``, ``1``]` `N ``=` `len``(arr)` `prodEquilibrium(arr, N)`   `# This code is contributed by Prajwal Kandekar`

C#

 `using` `System;`   `public` `class` `Program` `{` `  ``// Function to find the smallest index that splits the array` `  ``// into two subarrays with equal product` `  ``public` `static` `void` `ProdEquilibrium(``int``[] arr, ``int` `N)` `  ``{` `    ``int` `leftProduct = 1;` `    ``int` `rightProduct = 1;` `    ``for` `(``int` `i = 0; i < N; i++)` `    ``{` `      ``// calculate product of elements from 0 to i` `      ``leftProduct *= arr[i];`   `      ``// calculate product of elements from i+1 to N-1` `      ``for` `(``int` `j = i + 1; j < N; j++)` `      ``{` `        ``rightProduct *= arr[j];` `      ``}`   `      ``// if the product of the two subarrays is equal,` `      ``// return i` `      ``if` `(leftProduct == rightProduct)` `      ``{` `        ``Console.WriteLine(i + 1);` `        ``return``;` `      ``}`   `      ``// reset rightProduct for the next iteration` `      ``rightProduct = 1;` `    ``}`   `    ``// if no such index is found, return -1` `    ``Console.WriteLine(``"-1"``);` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main()` `  ``{` `    ``int``[] arr = { 1, 2, 3, 3, 2, 1 };` `    ``int` `N = arr.Length;` `    ``ProdEquilibrium(arr, N);` `  ``}` `}`

Javascript

 `// Javascript code for the above approach`   `// Function to find the smallest index that splits the array into` `// two subarrays with equal product` `function` `prodEquilibrium(arr, N) {` `    ``let leftProduct = 1;` `    ``let rightProduct = 1;` `    ``for` `(let i = 0; i < N; i++) {` `        `  `        ``// calculate product of elements from 0 to i` `        ``leftProduct *= arr[i];` `    `  `        ``// calculate product of elements from i+1 to N-1` `        ``for` `(let j = i + 1; j < N; j++) {` `            ``rightProduct *= arr[j];` `        ``}` `    `  `        ``// if the product of the two subarrays is equal, return i` `        ``if` `(leftProduct === rightProduct) {` `            ``console.log(i + 1);` `            ``return``;` `        ``}` `    `  `        ``// reset rightProduct for the next iteration` `        ``rightProduct = 1;` `    ``}`   `    ``// if no such index is found, return -1` `    ``console.log(-1);` `}`   `// Driver Code` `let arr = [1, 2, 3, 3, 2, 1];` `let N = arr.length;` `prodEquilibrium(arr, N);`

Output

`3`

Time Complexity: O(N^2)

Space Complexity: O(1)

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 product` `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` `            ``cout << i + 1 << endl;` `            ``return``;` `        ``}` `    ``}`   `    ``// If no partition exists, then` `    ``// print -1.` `    ``cout << -1 << endl;` `}`   `// Driver Code` `int` `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 approach` `import` `java.util.*;`   `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` `            ``System.out.print(i + ``1` `+ ``"\n"``);` `            ``return``;` `        ``}` `    ``}`   `    ``// If no partition exists, then` `    ``// print -1.` `    ``System.out.print(-``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 29AjayKumar`

Python3

 `# Python 3 program for the above approach`   `# Function to find the smallest` `# index that splits the array into` `# two subarrays with equal product` `def` `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 Code` `if` `__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 approach` `using` `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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