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

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++

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// C++ program to count all the non-contiguous
// subarrays whose sum can be split
// as the difference of the squares
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

40

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

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