# Count all subarrays whose sum can be split as difference of squares of two Integers

Given an array arr[], the task is to count all subarrays whose sum can be split as the difference of the squares of two integers.

Examples:

Input: arr[] = {1, 3, 5}
Output: 6
Explanation:
There are six subarrays which can be formed from the array whose sum can be split as the difference of squares of two integers. They are:
Sum of the subarray {1} = 1 = 12 – 02
Sum of the subarray {3} = 3 = 22 – 12
Sum of the subarray {5} = 5 = 32 – 22
Sum of the subarray {1, 3} = 4 = 22 – 02
Sum of the subarray {3, 5} = 8 = 32 – 12
Sum of the subarray {1, 3, 5} = 9 = 52 – 42

Input: arr[] = {1, 2, 3, 4, 5}
Output: 11

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: In order to solve this problem, an observation needs to be made. Any number can be represented as the difference of two squares except those which can be represented in the form ((4 * N) + 2) where N is an integer. Therefore, the following steps can be followed to compute the answer:

• Iterate over the array to find all the possible subarrays of the given array.
• If a number K can be expressed as ((4 * N) + 2) where N is some integer, then K + 2 is always a multiple of 4.
• Therefore, for every sum of the subarray K, check if (K + 2) is a multiple of 4 or not.
• If it is, then that particular value cannot be expressed as the difference of the squares.

Below is the implementation of the above approach:

## C++

 `// C++ program to count all the non-contiguous ` `// subarrays whose sum can be split ` `// as the difference of the squares ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count all the non-contiguous ` `// subarrays whose sum can be split ` `// as the difference of the squares ` `int` `Solve(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `temp = 0, count = 0; ` ` `  `    ``// Loop to iterate over all the possible ` `    ``// subsequences of the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``temp = 0; ` `        ``for` `(``int` `j = i; j < n; j++) { ` ` `  `            ``// Finding the sum of all the ` `            ``// possible subsequences ` `            ``temp += arr[j]; ` ` `  `            ``// Condition to check whether ` `            ``// the number can be split ` `            ``// as difference of squares ` `            ``if` `((temp + 2) % 4 != 0) ` `                ``count++; ` `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 1, 2, 3, 4, 5, ` `                  ``6, 7, 8, 9, 10 }; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << Solve(arr, N); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count all the  ` `// non-contiguous subarrays whose  ` `// sum can be split as the  ` `// difference of the squares ` `import` `java.io.*; ` `import` `java.lang.*; ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to count all the non-contiguous ` `// subarrays whose sum can be split ` `// as the difference of the squares ` `static` `int` `Solve(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `temp = ``0``, count = ``0``; ` ` `  `    ``// Loop to iterate over all the possible ` `    ``// subsequences of the array ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `       ``temp = ``0``; ` `       ``for``(``int` `j = i; j < n; j++) ` `       ``{ ` `            `  `           ``// Finding the sum of all the ` `           ``// possible subsequences ` `           ``temp += arr[j]; ` `            `  `           ``// Condition to check whether ` `           ``// the number can be split ` `           ``// as difference of squares ` `           ``if` `((temp + ``2``) % ``4` `!= ``0``) ` `               ``count++; ` `       ``} ` `    ``} ` `     `  `    ``return` `count; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String [] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ` `                  ``6``, ``7``, ``8``, ``9``, ``10` `}; ` `    ``int` `N = arr.length; ` ` `  `    ``System.out.println(Solve(arr, N)); ` `} ` `} ` ` `  `// This code is contributed by shivanisinghss2110 `

## Python3

 `# Python3 program to count all the non-contiguous ` `# subarrays whose sum can be split ` `# as the difference of the squares ` ` `  `# Function to count all the non-contiguous ` `# subarrays whose sum can be split ` `# as the difference of the squares ` `def` `Solve(arr, n): ` ` `  `    ``temp ``=` `0``; count ``=` `0``; ` ` `  `    ``# Loop to iterate over all the possible ` `    ``# subsequences of the array ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``temp ``=` `0``; ` `        ``for` `j ``in` `range``(i, n): ` ` `  `            ``# Finding the sum of all the ` `            ``# possible subsequences ` `            ``temp ``=` `temp ``+` `arr[j]; ` ` `  `            ``# Condition to check whether ` `            ``# the number can be split ` `            ``# as difference of squares ` `            ``if` `((temp ``+` `2``) ``%` `4` `!``=` `0``): ` `                ``count ``+``=` `1``; ` `         `  `    ``return` `count; ` ` `  `# Driver code ` `arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5``, ` `        ``6``, ``7``, ``8``, ``9``, ``10` `]; ` `N ``=` `len``(arr); ` `print``(Solve(arr, N)); ` ` `  `# This code is contributed by Code_Mech `

## C#

 `// C# program to count all the  ` `// non-contiguous subarrays whose  ` `// sum can be split as the  ` `// difference of the squares ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to count all the  ` `// non-contiguous subarrays whose  ` `// sum can be split as the  ` `// difference of the squares ` `static` `int` `Solve(``int` `[]arr, ``int` `n) ` `{ ` `    ``int` `temp = 0, count = 0; ` ` `  `    ``// Loop to iterate over all  ` `    ``// the possible subsequences  ` `    ``// of the array ` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{ ` `       ``temp = 0; ` `       ``for``(``int` `j = i; j < n; j++) ` `       ``{ ` `            `  `          ``// Finding the sum of all the ` `          ``// possible subsequences ` `          ``temp += arr[j]; ` `           `  `          ``// Condition to check whether ` `          ``// the number can be split ` `          ``// as difference of squares ` `          ``if` `((temp + 2) % 4 != 0) ` `              ``count++; ` `       ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 1, 2, 3, 4, 5, ` `                  ``6, 7, 8, 9, 10 }; ` `    ``int` `N = arr.Length; ` ` `  `    ``Console.Write(Solve(arr, N)); ` `} ` `} ` ` `  `// This code is contributed by shivanisinghss2110 `

Output:

```40
```

Time Complexity: O(N)2 where N is the size of the array

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