# Count of non decreasing arrays of length N formed with values in range L to R

Given are integers N, L and R, the task is to count the number of non decreasing arrays of length N formed with values in range [L, R] with repetition allowed.

Examples:

Input: N = 4, L = 4, R = 6
Output: 5
All possible arrays are {4, 4, 4, 6}, {4, 4, 5, 6}, {4, 5, 5, 6}, {4, 5, 6, 6} and {4, 6, 6, 6}.

Input: N = 2, L = 5, R = 2
Output: 0
No such combinations exist as L > R.

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

Approach:

• Since it is known that the minimum number is L and maximum number is R in the array.
• If the remaining (N – 2) indices are filled with L, the minimum possible sum is obtained and if the remaining (N-2) indices are filled with R, the maximum possible sum is obtained.
• It can be concluded that there exists a combination of numbers which results in a sum in between the minimum possible and maximum possible sum.
• Therefore, total different number of sums can be computed by:
[(N – 2) * R – (N – 2) * L] + 1 = (N – 2) * (R – L) + 1

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count ` `// of different arrays ` `int` `countSum(``int` `N, ``int` `L, ``int` `R) ` `{ ` ` `  `    ``// No such combination exists ` `    ``if` `(L > R) { ` `        ``return` `0; ` `    ``} ` ` `  `    ``// Arrays formed with single elements ` `    ``if` `(N == 1) { ` `        ``return` `R - L + 1; ` `    ``} ` ` `  `    ``if` `(N > 1) { ` `        ``return` `(N - 2) * (R - L) + 1; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 4, L = 4, R = 6; ` ` `  `    ``cout << countSum(N, L, R); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// Function to return the count ` `// of different arrays ` `static` `int` `countSum(``int` `N, ``int` `L, ``int` `R) ` `{ ` ` `  `    ``// No such combination exists ` `    ``if` `(L > R) ` `    ``{ ` `        ``return` `0``; ` `    ``} ` ` `  `    ``// Arrays formed with single elements ` `    ``if` `(N == ``1``) ` `    ``{ ` `        ``return` `R - L + ``1``; ` `    ``} ` ` `  `    ``if` `(N > ``1``) ` `    ``{ ` `        ``return` `(N - ``2``) * (R - L) + ``1``; ` `    ``} ` `    ``return` `0``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``4``, L = ``4``, R = ``6``; ` ` `  `    ``System.out.print(countSum(N, L, R)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count ` `# of different arrays ` `def` `countSum(N, L, R): ` ` `  `    ``# No such combination exists ` `    ``if` `(L > R): ` `        ``return` `0``; ` ` `  `    ``# Arrays formed with single elements ` `    ``if` `(N ``=``=` `1``): ` `        ``return` `R ``-` `L ``+` `1``; ` `    ``if` `(N > ``1``): ` `        ``return` `(N ``-` `2``) ``*` `(R ``-` `L) ``+` `1``; ` `     `  `    ``return` `0``; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``N, L, R ``=` `4``, ``4``, ``6``; ` ` `  `    ``print``(countSum(N, L, R)); ` ` `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the count ` `// of different arrays ` `static` `int` `countSum(``int` `N, ``int` `L, ``int` `R) ` `{ ` ` `  `    ``// No such combination exists ` `    ``if` `(L > R) ` `    ``{ ` `        ``return` `0; ` `    ``} ` ` `  `    ``// Arrays formed with single elements ` `    ``if` `(N == 1) ` `    ``{ ` `        ``return` `R - L + 1; ` `    ``} ` ` `  `    ``if` `(N > 1) ` `    ``{ ` `        ``return` `(N - 2) * (R - L) + 1; ` `    ``} ` `    ``return` `0; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `N = 4, L = 4, R = 6; ` ` `  `    ``Console.Write(countSum(N, L, R)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```5
```

Time Complexity: O(1)

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