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# Number of times an array can be partitioned repetitively into two subarrays with equal sum

• Last Updated : 04 Sep, 2021

Given an array arr[] of size N, the task is to find the number of times the array can be partitioned repetitively into two subarrays such that the sum of the elements of both the subarrays is the same.

Examples:

Input: arr[] = { 2, 2, 2, 2 }
Output:
Explanation:
1. Make the first partition after index 1. Remaining arrays are {2, 2} on the right side and left side both.
2. Consider the left subarray {2, 2}. Make a partition after index 0 of this left subarray.
Now two similar subarrays with one element each i.e. {2} are formed which cannot be sub-divided.
3. Consider the right subarray {2, 2}. Make a partition after index 0 of this left subarray.
Now two similar subarrays with one element each i.e. {2} are formed which cannot be sub-divided.
Hence the output is 3 as the array was partitioned 3 times.

Input: arr[] = {12, 3, 3, 0, 3, 3}
Output:
Explanation:
1. The first partition is after index 0. Remaining array is arr[] = {3, 3, 0, 3, 3}.
2. The second partition is after index 1. The remaining array is {3, 3}, and {0, 3, 3}.
3. The third partition is after index 0 in array {3, 3}.
4. The fourth partition is after 1 in the array {0, 3, 3}
The remaining array is {0, 3}, and {3} which cannot be sub-divided.
Hence the output is 4.

Approach: The idea is to use Recursion. Below are the steps:

1. Find the prefix-sum of the given array arr[] and store it in an array pref[].
2. Iterate from the start position to the end position.
3. For each possible partition index(say K), if prefix_sum[K] – prefix_sum[start-1] = prefix_sum[end] – prefix_sum[k] then the partition is valid.
4. If a partition is valid in the above step then proceed with the left and right sub-arrays separately and determine whether these two subarrays form a valid partition or not.
5. Repeat the step 3 and 4 for both the left and right partition until any further partition isn’t possible.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Recursion Function to calculate the``// possible splitting``int` `splitArray(``int` `start, ``int` `end,``               ``int``* arr,``               ``int``* prefix_sum)``{``    ``// If there are less than``    ``// two elements, we cannot``    ``// partition the sub-array.``    ``if` `(start >= end)``        ``return` `0;` `    ``// Iterate from the start``    ``// to end-1.``    ``for` `(``int` `k = start; k < end; ++k) {` `        ``if` `((prefix_sum[k] - prefix_sum[start - 1])``            ``== (prefix_sum[end] - prefix_sum[k])) {` `            ``// Recursive call to the left``            ``// and the right sub-array.``            ``return` `1 + splitArray(start,``                                  ``k,``                                  ``arr,``                                  ``prefix_sum)``                   ``+ splitArray(k + 1,``                                ``end,``                                ``arr,``                                ``prefix_sum);``        ``}``    ``}` `    ``// If there is no such partition,``    ``// then return 0``    ``return` `0;``}` `// Function to find the total splitting``void` `solve(``int` `arr[], ``int` `n)``{` `    ``// Prefix array to store``    ``// the prefix-sum using``    ``// 1 based indexing``    ``int` `prefix_sum[n + 1];` `    ``prefix_sum = 0;` `    ``// Store the prefix-sum``    ``for` `(``int` `i = 1; i <= n; ++i) {``        ``prefix_sum[i] = prefix_sum[i - 1]``                        ``+ arr[i - 1];``    ``}` `    ``// Function Call to count the``    ``// number of splitting``    ``cout << splitArray(1, n,``                       ``arr,``                       ``prefix_sum);``}` `// Driver Code``int` `main()``{``    ``// Given array``    ``int` `arr[] = { 12, 3, 3, 0, 3, 3 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function call``    ``solve(arr, N);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{` `// Recursion Function to calculate the``// possible splitting``static` `int` `splitArray(``int` `start, ``int` `end,``                      ``int``[] arr,``                      ``int``[] prefix_sum)``{``    ``// If there are less than``    ``// two elements, we cannot``    ``// partition the sub-array.``    ``if` `(start >= end)``        ``return` `0``;` `    ``// Iterate from the start``    ``// to end-1.``    ``for` `(``int` `k = start; k < end; ++k)``    ``{``        ``if` `((prefix_sum[k] - prefix_sum[start - ``1``]) ==``            ``(prefix_sum[end] - prefix_sum[k]))``        ``{` `            ``// Recursive call to the left``            ``// and the right sub-array.``            ``return` `1` `+ splitArray(start, k, arr, prefix_sum) +``                       ``splitArray(k + ``1``, end, arr, prefix_sum);``        ``}``    ``}` `    ``// If there is no such partition,``    ``// then return 0``    ``return` `0``;``}` `// Function to find the total splitting``static` `void` `solve(``int` `arr[], ``int` `n)``{` `    ``// Prefix array to store``    ``// the prefix-sum using``    ``// 1 based indexing``    ``int` `[]prefix_sum = ``new` `int``[n + ``1``];` `    ``prefix_sum[``0``] = ``0``;` `    ``// Store the prefix-sum``    ``for` `(``int` `i = ``1``; i <= n; ++i)``    ``{``        ``prefix_sum[i] = prefix_sum[i - ``1``] +``                               ``arr[i - ``1``];``    ``}` `    ``// Function Call to count the``    ``// number of splitting``    ``System.out.print(splitArray(``1``, n, arr,``                                ``prefix_sum));``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``// Given array``    ``int` `arr[] = { ``12``, ``3``, ``3``, ``0``, ``3``, ``3` `};``    ``int` `N = arr.length;` `    ``// Function call``    ``solve(arr, N);``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program for the above approach` `# Recursion Function to calculate the``# possible splitting``def` `splitArray(start, end, arr, prefix_sum):``    ` `    ``# If there are less than``    ``# two elements, we cannot``    ``# partition the sub-array.``    ``if` `(start >``=` `end):``        ``return` `0` `    ``# Iterate from the start``    ``# to end-1.``    ``for` `k ``in` `range``(start, end):``        ``if` `((prefix_sum[k] ``-` `prefix_sum[start ``-` `1``]) ``=``=``            ``(prefix_sum[end] ``-` `prefix_sum[k])) :` `            ``# Recursive call to the left``            ``# and the right sub-array.``            ``return` `(``1` `+` `splitArray(start, k, arr,``                                   ``prefix_sum) ``+``                        ``splitArray(k ``+` `1``, end, arr,``                                   ``prefix_sum))``        ` `    ``# If there is no such partition,``    ``# then return 0``    ``return` `0` `# Function to find the total splitting``def` `solve(arr, n):` `    ``# Prefix array to store``    ``# the prefix-sum using``    ``# 1 based indexing``    ``prefix_sum ``=` `[``0``] ``*` `(n ``+` `1``)` `    ``prefix_sum[``0``] ``=` `0` `    ``# Store the prefix-sum``    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``prefix_sum[i] ``=` `(prefix_sum[i ``-` `1``] ``+``                                ``arr[i ``-` `1``])``    ` `    ``# Function Call to count the``    ``# number of splitting``    ``print``(splitArray(``1``, n, arr, prefix_sum))` `# Driver Code` `# Given array``arr ``=` `[ ``12``, ``3``, ``3``, ``0``, ``3``, ``3` `]``N ``=` `len``(arr)` `# Function call``solve(arr, N)` `# This code is contributed by sanjoy_62`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Recursion Function to calculate the``// possible splitting``static` `int` `splitArray(``int` `start, ``int` `end,``                      ``int``[] arr,``                      ``int``[] prefix_sum)``{``    ` `    ``// If there are less than``    ``// two elements, we cannot``    ``// partition the sub-array.``    ``if` `(start >= end)``        ``return` `0;` `    ``// Iterate from the start``    ``// to end-1.``    ``for``(``int` `k = start; k < end; ++k)``    ``{``       ``if` `((prefix_sum[k] -``            ``prefix_sum[start - 1]) ==``           ``(prefix_sum[end] -``            ``prefix_sum[k]))``       ``{``           ` `           ``// Recursive call to the left``           ``// and the right sub-array.``           ``return` `1 + splitArray(start, k, arr,``                                 ``prefix_sum) +``                      ``splitArray(k + 1, end, arr,``                                 ``prefix_sum);``       ``}``    ``}``    ` `    ``// If there is no such partition,``    ``// then return 0``    ``return` `0;``}` `// Function to find the total splitting``static` `void` `solve(``int` `[]arr, ``int` `n)``{``    ` `    ``// Prefix array to store``    ``// the prefix-sum using``    ``// 1 based indexing``    ``int` `[]prefix_sum = ``new` `int``[n + 1];` `    ``prefix_sum = 0;` `    ``// Store the prefix-sum``    ``for``(``int` `i = 1; i <= n; ++i)``    ``{``       ``prefix_sum[i] = prefix_sum[i - 1] +``                              ``arr[i - 1];``    ``}` `    ``// Function Call to count the``    ``// number of splitting``    ``Console.Write(splitArray(1, n, arr,``                             ``prefix_sum));``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ` `    ``// Given array``    ``int` `[]arr = { 12, 3, 3, 0, 3, 3 };``    ``int` `N = arr.Length;` `    ``// Function call``    ``solve(arr, N);``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output:
`4`

Time Complexity: O(N2)
Auxiliary Space: O(N)

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