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++
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
n = 9 and r = 8
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