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. 

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 <bits/stdc++.h>
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

Javascript

<script>

// Javascript implementation of the approach

// Function to return the count
// of different arrays
function 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;
    }
}

// Driver code
var N = 4, L = 4, R = 6;
document.write( countSum(N, L, R));

// This code is contributed by noob2000.
</script>

Output:

5

Time Complexity: O(1)
Auxiliary Space: O(1), no extra space is required, so it is a constant.