Counting Swaps for Equal Sum Subarrays

Given an integer N, consider that there is an array of first N natural numbers, the task is to output the number of swaps such that there should be two subarrays of equal sum.

Note: Each number must be a part of only one subarray. Formally, no number should be there such that it is not a part of any subarray.

Examples:

Input: N = 4
Output: 2
Explanation: Consider array of first N naturals is A[] = {1, 2, 3, 4}. Then, there are two possible swaps, which divides first N natural numbers into two equal sum subarrays.

• First swap: Swap A[0] and A[2], then A[] = {3, 2, 1, 4}. Then A[] can be divided into two subarrays as {3, 2} and {1, 4}. Each subarray has sum equal to 5.
• First swap: Swap A[1] and A[4], then A[] = {1, 4, 3, 2}. Then A[] can be divided into two subarrays as {1, 4} and {3, 2}. Each subarray has sum equal to 5.
  There are two possible swaps. Therefore, output is 2.

Input: N = 2
Output: 0
Explanation: It can be verified that there are no valid swaps according to the given problem statement.

Approach: Implement the idea below to solve the problem

The problem is based on the Mathematics and can be solved using the mathematical observations.

Steps were taken to solve the problem:

• If (N%4 == 0 || N%4 == 1), then output 0.
• Else declare a variable let say Cnt and initialize it equal to N*(N+1)/2
  • Divide Cnt by 4
  • Declare a variable let say X and initialize it equal to Sqrt(2*cnt) + 1
  • While((X*(X+1))/2 > cnt)
    • X–;
  • Cnt -= (X*(X+1))/2
  • Declare a variable let say Ans and initialize it equal to N-X
  • If(cnt == 0)
    • Ans += X*(X-1)/2 + (N-X)*(N-X-1)/2
  • Output the value stored in Ans. n

Below is the implementation of the above approach:

C++

// C++ code to imlement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the number of swaps
void Swap_counts(int N)
{
    // At these conditions, no swaps will be possible
    if (N % 4 == 1 || N % 4 == 2)
        cout << 0 << endl;
    else {
        int cnt = N * (N + 1) / 4;

        int X = (int)sqrt(2 * cnt) + 1;

        while ((X * (X + 1)) / 2 > cnt)
            X--;

        cnt -= (X * (X + 1)) / 2;

        int ans = N - X;

        if (cnt == 0)
            ans += (X * (X - 1)) / 2
                   + ((N - X) * (N - X - 1)) / 2;

        cout << ans << endl;
    }
}

int main()
{
    // Input
    int N = 4;

    // Function call
    Swap_counts(N);

    return 0;
}

Java

// Java code to imlement the approach

import java.util.*;

public class Main {

    // Driver Function
    public static void main(String[] args)
    {

        // Inputs
        int N = 4;

        // Function calls
        Swap_counts(N);
    }
    public static void Swap_counts(int N)
    {

        // At this conditions no
        // swaps will be possible
        if (N % 4 == 1 || N % 4 == 2)
            System.out.println(0);
        else {
            int cnt = N * (N + 1);
            cnt /= 4;

            int X = (int)Math.sqrt(2 * cnt) + 1;

            while ((X * (X + 1)) / 2 > cnt)
                X--;

            cnt -= (X * (X + 1)) / 2;

            int ans = N - X;

            if (cnt == 0)
                ans += X * (X - 1) / 2
                       + (N - X) * (N - X - 1) / 2;

            System.out.println(ans);
        }
    }
}

Python3

import math

def swap_counts(N):
    # At these conditions, no swaps will be possible
    if (N % 4 == 1 or N % 4 == 2):
        print(0)
    else:
        cnt = N * (N + 1) // 4

        X = int(math.sqrt(2 * cnt)) + 1

        while ((X * (X + 1)) // 2 > cnt):
            X -= 1

        cnt -= (X * (X + 1)) // 2

        ans = N - X

        if (cnt == 0):
            ans += (X * (X - 1)) // 2 + ((N - X) * (N - X - 1)) // 2

        print(ans)

# Driver Function
if __name__ == "__main__":
    # Inputs
    N = 4

    # Function calls
    swap_counts(N)
#This code is contributed by Rohit Singh

C#

// C# code to implement the approach

using System;

public class GFG {
    // Driver Function
    public static void Main()
    {
        // Inputs
        int N = 4;

        // Function calls
        Swap_counts(N);
    }
    public static void Swap_counts(int N)
    {

        // At this conditions no
        // swaps will be possible
        if (N % 4 == 1 || N % 4 == 2)
            Console.WriteLine(0);
        else {
            int cnt = N * (N + 1);
            cnt /= 4;

            int X = (int)Math.Sqrt(2 * cnt) + 1;

            while ((X * (X + 1)) / 2 > cnt)
                X--;

            cnt -= (X * (X + 1)) / 2;

            int ans = N - X;

            if (cnt == 0)
                ans += X * (X - 1) / 2
                       + (N - X) * (N - X - 1) / 2;

            Console.WriteLine(ans);
        }
    }
}

// This code is contributed by ragul21

Javascript

// Function to calculate the number of swaps
function swapCounts(N) {
    // At these conditions, no swaps will be possible
    if (N % 4 === 1 || N % 4 === 2) {
        console.log(0);
    } else {
        let cnt = (N * (N + 1)) / 4;
        let X = Math.floor(Math.sqrt(2 * cnt)) + 1;

        while ((X * (X + 1)) / 2 > cnt) {
            X--;
        }

        cnt -= (X * (X + 1)) / 2;
        let ans = N - X;

        if (cnt === 0) {
            ans += (X * (X - 1)) / 2 + (N - X) * (N - X - 1) / 2;
        }

        console.log(ans);
    }
}

// Driver function
function main() {
    // Inputs
    let N = 4;

    // Function call
    swapCounts(N);
}

// Run the program
main();

Output

2

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