Count of decreasing pairs formed from numbers 1 to N

Given an integer N, the task is to count of decreasing pairs from numbers 1 to N.

A pair (x, y) is said to be decreasing if x > y

Examples:

Input: N = 8
Output: 3
Explanation:
Decreasing pairs are: (7, 1), (6, 2), (5, 3).

Input: N = 9
Output: 4
Explanation:
Decreasing pairs are: (8, 1), (7, 2), (6, 3), (5, 4).



Approach: Consider the below cases:

If N = 1 => Count = 0
If N = 2 => Count = 1 {(2, 1)}
If N = 3 => Count = 1 {(3, 1) or (3, 2)}
If N = 4 => Count = 2 {(4, 3), (2, 1)}
If N = 5 => Count = 2 {(5, 4), (3, 2)}
If N = 6 => Count = 3 {(6, 5), (4, 3), (2, 1)}
.
.
and so on

It can be clearly observed that
\text{Count of Decreasing pairs = }\begin{cases} \frac{N}{2} & \text{, if N is odd}\\ \frac{N}{2}-1 & \text{, if N is even} \end{cases}

Below is the implementation of the above approach:

C++

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// C++ program to count decreasing
// pairs formed from numbers 1 to N
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to count the
// possible number of pairs
int divParts(int N)
{
    if (N % 2 == 0)
  
        // if the number is even
        // then the answer in (N/2)-1
        cout << (N / 2) - 1 << endl;
  
    else
        // if the number is odd
        // then the answer in N/2
        cout << N / 2 << endl;
}
  
// Driver code
int main()
{
    int N = 8;
  
    divParts(N);
  
    return 0;
}

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Java

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// Java program to count decreasing
// pairs formed from numbers 1 to N
import java.util.*;
class GFG{
      
// Function to count the
// possible number of pairs
static void divParts(int N)
{
    if (N % 2 == 0)
  
        // if the number is even
        // then the answer in (N/2)-1
        System.out.println((N / 2) - 1);
  
    else
        // if the number is odd
        // then the answer in N/2
        System.out.println((N / 2));
}
  
// Driver code
public static void main(String[] args)
{
    int N = 8;
  
    divParts(N);
}
}
  
// This code is contributed by offbeat

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Python3

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# Python3 program to count decreasing
# pairs formed from numbers 1 to N
  
# Function to count the
# possible number of pairs
def divParts(N):
  
    if (N % 2 == 0):
  
        # if the number is even
        # then the answer in (N/2)-1
        print((N / 2) - 1);
  
    else:
          
        # if the number is odd
        # then the answer in N/2
        print(N / 2);
  
# Driver code
N = 8;
divParts(N);
  
# This code is contributed by Code_Mech

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C#

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// C# program to count decreasing
// pairs formed from numbers 1 to N
using System;
class GFG{
      
// Function to count the
// possible number of pairs
static void divParts(int N)
{
    if (N % 2 == 0)
  
        // if the number is even
        // then the answer in (N/2)-1
        Console.WriteLine((N / 2) - 1);
  
    else
        // if the number is odd
        // then the answer in N/2
        Console.WriteLine((N / 2));
}
  
// Driver code
public static void Main()
{
    int N = 8;
  
    divParts(N);
}
}
  
// This code is contributed by Code_Mech

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Output:

3

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Improved By : offbeat, Code_Mech