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
Below is the implementation of the above approach:
C++
// 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 pairsint 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 codeint main(){ int N = 8; divParts(N); return 0;} |
Java
// Java program to count decreasing// pairs formed from numbers 1 to Nimport java.util.*;class GFG{ // Function to count the// possible number of pairsstatic 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 codepublic static void main(String[] args){ int N = 8; divParts(N);}} // This code is contributed by offbeat |
Python3
# Python3 program to count decreasing# pairs formed from numbers 1 to N # Function to count the# possible number of pairsdef 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 codeN = 8;divParts(N); # This code is contributed by Code_Mech |
C#
// C# program to count decreasing// pairs formed from numbers 1 to Nusing System;class GFG{ // Function to count the// possible number of pairsstatic 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 codepublic static void Main(){ int N = 8; divParts(N);}} // This code is contributed by Code_Mech |
Output:
3
