# 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].

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ≤ a ≤ …. ≤ 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 ` `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 = 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 = 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 `

Output:

```10
```

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Improved By : AnkitRai01, 29AjayKumar