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

  • Difficulty Level : Hard
  • Last Updated : 25 May, 2021

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:
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:
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)
 


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