Count non decreasing subarrays of size N from N Natural numbers

Given are N natural numbers, the task is to find the count of the subarrays of size N that can be formed using elements from 1 to N such that each element in the subarray is smaller than or equal to the elements to its right (a[i] ≤ a[i+1]).
Examples:

Input: N = 2 
Output: 3 
Explanation:
Given array of N natural numbers: {1, 2} 
Required subarrays that can be formed: [1, 1], [1, 2], [2, 2].
Input: N = 3 
Output: 10 
Explanation:
Given array of N natural numbers: {1, 2, 3} 
Required subarrays that can be formed: [1, 1, 1], [1, 1, 2], [1, 2, 2], [2, 2, 2], [1, 1, 3], [1, 3, 3], [3, 3, 3], [2, 2, 3], [2, 3, 3], [1, 2, 3]. 

Approach:

• Since each element of the array is between 1 to N and the subarrays can have duplicate elements in non-descending order, i.e., a[0] ≤ a[1] ≤ …. ≤ a[N – 1].
• The number of ways of choosing r objects with replacement from n objects is (using Combination with repetition).
• Here r = N and n = N as we can choose from 1 to N. So the count of all the sorted array of length N with elements from 1 to N will be .
• Now this can be further expanded with the help of Binomial Coefficients. The coefficient obtained from this will be the required subarray’s count.

Below is the implementation of above approach: 

C++

// C++ program to count non decreasing subarrays
// of size N from N Natural numbers

#include <bits/stdc++.h>
using namespace std;

// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    int C[k + 1];
    memset(C, 0, sizeof(C));

    // Since nC0 is 1
    C[0] = 1;

    for (int i = 1; i <= n; i++) {

        // Compute next row of pascal triangle using
        // the previous row
        for (int j = min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}

// Function to find the count of required subarrays
int count_of_subarrays(int N)
{

    // The required count is the binomial coefficient
    // as explained in the approach above
    int count = binomialCoeff(2 * N - 1, N);

    return count;
}

// Driver Function
int main()
{

    int N = 3;

    cout << count_of_subarrays(N) << "\n";
}

Java

// Java program to count non decreasing subarrays
// of size N from N Natural numbers
class GFG
{

// Returns value of Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{
    int []C = new int[k + 1];

    // Since nC0 is 1
    C[0] = 1;

    for (int i = 1; i <= n; i++)
    {

        // Compute next row of pascal triangle using
        // the previous row
        for (int j = Math.min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}

// Function to find the count of required subarrays
static int count_of_subarrays(int N)
{

    // The required count is the binomial coefficient
    // as explained in the approach above
    int count = binomialCoeff(2 * N - 1, N);

    return count;
}

// Driver Function
public static void main(String[] args)
{
    int N = 3;
    System.out.print(count_of_subarrays(N)+ "\n");
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to count non decreasing subarrays
# of size N from N Natural numbers

# Returns value of Binomial Coefficient C(n, k)
def binomialCoeff(n, k) :

    C = [0] * (k + 1);

    # Since nC0 is 1
    C[0] = 1;

    for i in range(1, n + 1) :

        # Compute next row of pascal triangle using
        # the previous row
        for j in range(min(i, k), 0, -1) :
            C[j] = C[j] + C[j - 1];

    return C[k];

# Function to find the count of required subarrays
def count_of_subarrays(N) :

    # The required count is the binomial coefficient
    # as explained in the approach above
    count = binomialCoeff(2 * N - 1, N);

    return count;

# Driver Function
if __name__ == "__main__" :

    N = 3;

    print(count_of_subarrays(N)) ;

# This code is contributed by AnkitRai01

C#

// C# program to count non decreasing subarrays
// of size N from N Natural numbers
using System;

class GFG
{

    // Returns value of Binomial Coefficient C(n, k)
    static int binomialCoeff(int n, int k)
    {
        int []C = new int[k + 1];

        // Since nC0 is 1
        C[0] = 1;

        for (int i = 1; i <= n; i++)
        {

            // Compute next row of pascal triangle using
            // the previous row
            for (int j = Math.Min(i, k); j > 0; j--)
                C[j] = C[j] + C[j - 1];
        }
        return C[k];
    }

    // Function to find the count of required subarrays
    static int count_of_subarrays(int N)
    {

        // The required count is the binomial coefficient
        // as explained in the approach above
        int count = binomialCoeff(2 * N - 1, N);

        return count;
    }

    // Driver Function
    public static void Main()
    {
        int N = 3;
        Console.WriteLine(count_of_subarrays(N));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Javascript program to count non decreasing subarrays
// of size N from N Natural numbers

// Returns value of Binomial Coefficient C(n, k)
function binomialCoeff(n, k)
{
    var C = Array(k+1).fill(0);

    // Since nC0 is 1
    C[0] = 1;

    for (var i = 1; i <= n; i++) {

        // Compute next row of pascal triangle using
        // the previous row
        for (var j = Math.min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}

// Function to find the count of required subarrays
function count_of_subarrays(N)
{

    // The required count is the binomial coefficient
    // as explained in the approach above
    var count = binomialCoeff(2 * N - 1, N);

    return count;
}

// Driver Function
var N = 3;
document.write( count_of_subarrays(N));

// This code is contributed by itsok.
</script>

Output:

10

Time Complexity: O(N²)

Auxiliary Space: O(N)