Find a pair (n,r) in an integer array such that value of nPr is maximum

Given an array of non-negative integers arr[], the task is to find a pair (n, r) such that nPr is maximum possible and r ≤ n.

nPr = n! / (n – r)!

Examples:



Input: arr[] = {5, 2, 3, 4, 1}
Output: n = 5 and r = 4
5P4 = 5! / (5 – 4)! = 120 which is maximum possible.

Input: arr[] = {0, 2, 3, 4, 1, 6, 8, 9}
Output: n = 9 and r = 8

Naive approach: A simple approach is to consider each (n, r) pair and calculate nPr value and find the maximum value among them.

Efficient approach: Since nPr = n! / (n – r)! = n * (n – 1) * (n – 2) * … * (n – r + 1).
With little mathematics, it can be shown that nPr will be maximum when n is maximum and n – r is minimum. The problem now boils down to finding the largest 2 elements from the given array.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function to print the pair (n, r)
// such that nPr is maximum possible
void findPair(int arr[], int n)
{
  
    // There should be atleast 2 elements
    if (n < 2) {
        cout << "-1";
        return;
    }
  
    int i, first, second;
    first = second = -1;
  
    // Findex the largest 2 elements
    for (i = 0; i < n; i++) {
        if (arr[i] > first) {
            second = first;
            first = arr[i];
        }
        else if (arr[i] > second) {
            second = arr[i];
        }
    }
  
    cout << "n = " << first
         << " and r = " << second;
}
  
// Driver code
int main()
{
    int arr[] = { 0, 2, 3, 4, 1, 6, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    findPair(arr, n);
  
    return 0;
}

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Java














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// Java implementation of the approach  


class GFG  


{  


      


    // Function to print the pair (n, r)  


    // such that nPr is maximum possible  


    static void findPair(int arr[], int n)  


    {  


      


        // There should be atleast 2 elements  


        if (n < 2)  


        {  


            System.out.print("-1");  


            return;  


        }  


      


        int i, first, second;  


        first = second = -1;  


      


        // Findex the largest 2 elements  


        for (i = 0; i < n; i++) 


        {  


            if (arr[i] > first)  


            {  


                second = first;  


                first = arr[i];  


            }  


            else if (arr[i] > second)  


            {  


                second = arr[i];  


            }  


        }  


      


        System.out.println("n = " + first +  


                           " and r = " + second);  


    }  


      


    // Driver code  


    public static void main(String args[])  


    {  


        int arr[] = { 0, 2, 3, 4, 1, 6, 8, 9 };  


        int n = arr.length;  


      


        findPair(arr, n);  


    }  


} 


  


// This code is contributed by AnkitRai01  



















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Python3














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# Python3 implementation of the approach 


  


# Function to print the pair (n, r) 


# such that nPr is maximum possible 


def findPair(arr, n): 


      


    # There should be atleast 2 elements 


    if (n < 2): 


        print("-1") 


        return


  


    i = 0


    first = -1


    second = -1


  


    # Findex the largest 2 elements 


    for i in range(n): 


        if (arr[i] > first): 


            second = first 


            first = arr[i] 


        elif (arr[i] > second): 


            second = arr[i] 


  


    print("n =", first, "and r =", second) 


  


# Driver code 


arr = [0, 2, 3, 4, 1, 6, 8, 9] 


n = len(arr) 


  


findPair(arr, n) 


  


# This code is contributed by mohit kumar 



















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C#














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// C# implementation of the approach  


using System; 


class GFG  


{  


      


    // Function to print the pair (n, r)  


    // such that nPr is maximum possible  


    static void findPair(int[] arr, int n)  


    {  


      


        // There should be atleast 2 elements  


        if (n < 2)  


        {  


            Console.Write("-1");  


            return;  


        }  


      


        int i, first, second;  


        first = second = -1;  


      


        // Findex the largest 2 elements  


        for (i = 0; i < n; i++) 


        {  


            if (arr[i] > first)  


            {  


                second = first;  


                first = arr[i];  


            }  


            else if (arr[i] > second)  


            {  


                second = arr[i];  


            }  


        }  


      


        Console.WriteLine("n = " + first +  


                          " and r = " + second);  


    }  


      


    // Driver code  


    public static void Main()  


    {  


        int[] arr = { 0, 2, 3, 4, 1, 6, 8, 9 };  


        int n = arr.Length;  


      


        findPair(arr, n);  


    }  


} 


  


// This code is contributed by CodeMech 



















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Output:


n = 9 and r = 8





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