Count numbers present in partitions of N

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Given an integer N, the task is to count the numbers in ordered integer partitions of N.
Examples:

Input: N = 3
Output: 8
Integer partitions of N(=3) are {{1 + 1 + 1}, {1 + 2}, {2 + 1}, {3}}.
Numbers in integer partition of N are:{1, 1, 1, 1, 2, 2, 1, 3}
Therefore, the count of numbers in integer partitions of N(=3) is 8.

Input: N = 4
Output: 20

Approach: The problem can be solved based on the following observations:

Count of ways to partition N into exactly k partitions =

$\binom{n-1}{k-1}$

Therefore, the count of numbers in ordered integer partitions of N is

$\begin{align*} = \sum_{k=1}^nk\binom{n-1}{k-1}&=\sum_{k=0}^{n-1}(k+1)\binom{n-1}k\\ &=(n-1)\sum_{k=0}^{n-1}\binom{n-2}{k-1}+\sum_{k=0}^{n-1}\binom{n-1}k\\ &=(n-1)2^{n-2}+2^{n-1}\\ &=(n+1)2^{n-2}\, . \end{align*}$

Below is the implementation of the above approach:

C++

// C++ program to implement 
// the above approach  
#include <bits/stdc++.h> 
using namespace std;
 
// Function to count of numbers in
// ordered partitions of N
int CtOfNums(int N)
{
  
    // Stores count the numbers in
    // ordered integer partitions
    int res = (N + 1) * (1 << (N - 2));
  
    return round(res);
}
  
// Driver Code
int main()
{
    int N = 3;
     
    cout << CtOfNums(N); 
}
 
// This code is contributed by code_hunt

Java

// Java program to implement 
// the above approach  
import java.io.*; 
 
class GFG{
  
// Function to count of numbers in
// ordered partitions of N
static int CtOfNums(int N)
{
  
    // Stores count the numbers in
    // ordered integer partitions
    int res = (N + 1) * (1 << (N - 2));
  
    return Math.round(res);
}
  
// Driver Code
public static void main (String[] args) 
{
    int N = 3;
  
    System.out.print(CtOfNums(N));
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program to implement
# the above approach 
 
# Function to count of numbers in
# ordered partitions of N
def CtOfNums(N): 
 
    # Stores count the numbers in
    # ordered integer partitions
    res = (N + 1) * (1<<(N - 2))
 
    return round(res) 
 
# Driver code 
if __name__ == '__main__':
    N = 3
    print(CtOfNums(N)) 

C#

// C# program to implement 
// the above approach  
using System;
 
class GFG{
  
// Function to count of numbers in
// ordered partitions of N
static int CtOfNums(int N)
{
  
    // Stores count the numbers in
    // ordered integer partitions
    double res = (N + 1) * (1 << (N - 2));
  
    return (int)Math.Round(res);
}
  
// Driver Code
public static void Main () 
{
    int N = 3;
  
    Console.Write(CtOfNums(N));
}
}
 
// This code is contributed by code_hunt

Javascript

<script>
 
// Javascript program to implement 
// the above approach  
 
// Function to count of numbers in
// ordered partitions of N
function CtOfNums(N)
{
  
    // Stores count the numbers in
    // ordered integer partitions
    var res = (N + 1) * (1 << (N - 2));
  
    return Math.round(res);
}
  
// Driver Code
var N = 3;
document.write(CtOfNums(N)); 
 
</script>

Output:

Time Complexity: O(log₂N)
Auxiliary Space: O(1)

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