Find a pair (n,r) in an integer array such that value of nPr is maximum
Last Updated :
12 Jul, 2025
Given an array of non-negative integers arr[], the task is to find a pair (n, r) such that nPr is the 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 the maximum possible.
Input: arr[] = {0, 2, 3, 4, 1, 6, 8, 9}
Output: n = 9 and r = 8
Approach: "Brute Force Approach with Sorting"
Naive approach: A simple approach is to consider each (n, r) pair and calculate nPr value and find the maximum value among them.
One possible approach to solve this problem is to sort the given array in decreasing order and then calculate the value of nPr for each possible value of n and r. The first value of nPr that we obtain will be the maximum possible value.
Algorithm:
- Sort the given array arr in decreasing order.
- Initialize max_n to the first element of arr, and max_r to the second element of arr.
- For each possible value of n (starting from max_n), and for each possible value of r (starting from max_r), calculate the value of nPr.
- If the value of nPr is greater than the current maximum, update the values of max_n and max_r.
- Return the values of max_n and max_r.
C++
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
pair<int, int> find_max_npr(vector<int>& arr) {
sort(arr.begin(), arr.end(), greater<int>());
int max_n = arr[0], max_r = arr[1];
int max_npr = pow(max_n, max_r);
for (int n = max_n; n > 0; --n) {
for (int r = max_r; r > 0; --r) {
int npr = pow(n, r);
if (npr > max_npr) {
max_n = n;
max_r = r;
max_npr = npr;
}
}
}
return make_pair(max_n, max_r);
}
int main() {
vector<int> arr {5, 2, 3, 4, 1};
auto result = find_max_npr(arr);
cout << result.first << " " << result.second << endl; // Output: 5 4
arr = {0, 2, 3, 4, 1, 6, 8, 9};
result = find_max_npr(arr);
cout << result.first << " " << result.second << endl; // Output: 9 8
return 0;
}
import java.util.Arrays;
public class GFG {
public static int[] findMaxNPR(int[] arr) {
// Sort the array in descending order
Arrays.sort(arr);
int maxN = arr[arr.length - 1];
int maxR = arr[arr.length - 2];
int maxNPR = (int) Math.pow(maxN, maxR);
for (int n = maxN; n > 0; n--) {
for (int r = maxR; r > 0; r--) {
int npr = (int) Math.pow(n, r);
if (npr > maxNPR) {
maxN = n;
maxR = r;
maxNPR = npr;
}
}
}
int[] result = {maxN, maxR};
return result;
}
public static void main(String[] args) {
int[] arr1 = {5, 2, 3, 4, 1};
int[] result1 = findMaxNPR(arr1);
System.out.println("Output 1: (" + result1[0] + ", " + result1[1] + ")");
int[] arr2 = {0, 2, 3, 4, 1, 6, 8, 9};
int[] result2 = findMaxNPR(arr2);
System.out.println("Output 2: (" + result2[0] + ", " + result2[1] + ")");
}
}
def find_max_npr(arr):
arr.sort(reverse=True)
max_n, max_r = arr[0], arr[1]
max_npr = max_n ** max_r
for n in range(max_n, 0, -1):
for r in range(max_r, 0, -1):
npr = n ** r
if npr > max_npr:
max_n, max_r = n, r
max_npr = npr
return max_n, max_r
# Example usage:
arr = [5, 2, 3, 4, 1]
print(find_max_npr(arr)) # Output: (5, 4)
arr = [0, 2, 3, 4, 1, 6, 8, 9]
print(find_max_npr(arr)) # Output: (9, 8)
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
// Function to find the pair (n, r) with the maximum value of n^r
static Tuple<int, int> FindMaxNpr(List<int> arr)
{
// Sort the input array in descending order
arr.Sort((a, b) => b.CompareTo(a));
// Get the maximum values of n and r from the sorted array
int max_n = arr[0];
int max_r = arr[1];
int max_npr = (int)Math.Pow(max_n, max_r);
// Iterate through all possible combinations of n and r
for (int n = max_n; n > 0; n--)
{
for (int r = max_r; r > 0; r--)
{
int npr = (int)Math.Pow(n, r);
// If n^r is greater than the current maximum n^r, update the maximum values
if (npr > max_npr)
{
max_n = n;
max_r = r;
max_npr = npr;
}
}
}
// Return the pair (n, r) with the maximum n^r
return Tuple.Create(max_n, max_r);
}
static void Main()
{
List<int> arr = new List<int> { 5, 2, 3, 4, 1 };
var result = FindMaxNpr(arr);
Console.WriteLine(result.Item1 + " " + result.Item2); // Output: 5 4
arr = new List<int> { 0, 2, 3, 4, 1, 6, 8, 9 };
result = FindMaxNpr(arr);
Console.WriteLine(result.Item1 + " " + result.Item2); // Output: 9 8
}
}
function findMaxNPR(arr) {
arr.sort((a, b) => b - a); // Sort the array in descending order
let max_n = arr[0];
let max_r = arr[1];
let max_npr = Math.pow(max_n, max_r); // Calculate the initial max_npr
for (let n = max_n; n > 0; n--) {
for (let r = max_r; r > 0; r--) {
let npr = Math.pow(n, r); // Calculate n^r
if (npr > max_npr) {
max_n = n;
max_r = r;
max_npr = npr; // Update max_n, max_r, and max_npr if npr is greater
}
}
}
return [max_n, max_r];
}
// Example usage:
let arr = [5, 2, 3, 4, 1];
console.log(findMaxNPR(arr)); // Output: [5, 4]
arr = [0, 2, 3, 4, 1, 6, 8, 9];
console.log(findMaxNPR(arr)); // Output: [9, 8]
Time Complexity: The time complexity of this approach is O(n2), where n is the size of the input array.
Auxiliary Space: The auxiliary space required by this approach is O(1).
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 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 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# 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
<script>
// Java Script implementation of the approach
// Function to print the pair (n, r)
// such that nPr is maximum possible
function findPair(arr,n)
{
// There should be atleast 2 elements
if (n < 2)
{
document.write("-1");
return;
}
let 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];
}
}
document.write("n = " + first +
" and r = " + second);
}
// Driver code
let arr = [ 0, 2, 3, 4, 1, 6, 8, 9 ];
let n = arr.length;
findPair(arr, n);
// This code is contributed by sravan kumar
</script>
Time Complexity: O(n)
Auxiliary Space: O(1)
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