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

Given an array of non-negative integers arr[]. The task is to find a pair (n, r) such that value of ⁿC_r is maximum possible r < n.

ⁿC_r = n! / (r! * (n – r)!)

Examples:

Input: arr[] = {5, 2, 3, 4, 1}
Output: n = 5 and r = 2
⁵C₃ = 5! / (3! * (5 – 3)!) = 10
Input: arr[] = {0, 2, 3, 4, 1, 6, 8, 9}
Output: n = 9 and r = 4

Naive approach: A simple approach is to consider each (n, r) pair and find the maximum possible value of ⁿC_r.
Efficient approach: It is known from combinatorics:

When n is odd:
ⁿC₀ < ⁿC₁ ….. < ⁿC_(n-1)/2 = ⁿC_(n+1)/2 > ….. > ⁿC_n-1 > ⁿC_n
When n is even:
ⁿC₀ < ⁿC₁ ….. < ⁿC_n/2> ….. > ⁿC_n-1 > ⁿC_n
Also, ⁿC_r = ⁿC_n-r

It can be observed that ⁿC_r will be maximum when n will be maximum and abs(r – middle) will be minimum. The problem now boils down to finding the largest element in arr[] and r such that abs(r – middle) is minimum.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the pair (n, r)
// such that nCr is maximum possible
void findPair(int arr[], int n)
{
    // Array should contain atleast 2 elements
    if (n < 2) {
        cout << "-1";
        return;
    }
 
    // Maximum element from the array
    int maximum = *max_element(arr, arr + n);
 
    // temp stores abs(middle - arr[i])
    int temp = 10000001, r = 0, middle = maximum / 2;
 
    // Finding r with minimum abs(middle - arr[i])
    for (int i = 0; i < n; i++) {
 
        // When n is even then middle is (maximum / 2)
        if (abs(middle - arr[i]) < temp && n % 2 == 0) {
            temp = abs(middle - arr[i]);
            r = arr[i];
        }
 
        // When n is odd then middle elements are
        // (maximum / 2) and ((maximum / 2) + 1)
        else if (min(abs(middle - arr[i]), abs(middle + 1 - arr[i])) < temp
                 && n % 2 == 1) {
            temp = min(abs(middle - arr[i]), abs(middle + 1 - arr[i]));
            r = arr[i];
        }
    }
 
    cout << "n = " << maximum
         << " and r = " << r;
}
 
// 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 above approach
class GFG
{
     
// Function to print the pair (n, r)
// such that nCr is maximum possible
static void findPair(int arr[], int n)
{
    // Array should contain atleast 2 elements
    if (n < 2)
    {
        System.out.print("-1");
        return;
    }
 
    // Maximum element from the array
    int maximum = arr[0];
    for(int i = 1; i < n; i++)
    maximum = Math.max(maximum, arr[i]);
 
    // temp stores abs(middle - arr[i])
    int temp = 10000001, r = 0, middle = maximum / 2;
 
    // Finding r with minimum abs(middle - arr[i])
    for (int i = 0; i < n; i++)
    {
 
        // When n is even then middle is (maximum / 2)
        if (Math.abs(middle - arr[i]) < temp && n % 2 == 0)
        {
            temp = Math.abs(middle - arr[i]);
            r = arr[i];
        }
 
        // When n is odd then middle elements are
        // (maximum / 2) and ((maximum / 2) + 1)
        else if (Math.min(Math.abs(middle - arr[i]),
                          Math.abs(middle + 1 - arr[i])) <
                                     temp && n % 2 == 1)
        {
            temp = Math.min(Math.abs(middle - arr[i]),
                            Math.abs(middle + 1 - arr[i]));
            r = arr[i];
        }
    }
    System.out.print( "n = " + maximum + " and r = " + r);
}
 
// 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 Arnab Kundu

Python3

# Python3 implementation of the approach
 
# Function to print the pair (n, r)
# such that nCr is maximum possible
 
 
def find_pair(arr):
 
    current_min_diff = float('inf')
    n = max(arr)
    middle = n / 2
 
    for elem in arr:
        diff = abs(elem - middle)
        if diff < current_min_diff:
            current_min_diff = diff
            r = elem
 
    print("n =", n, "and r =", r)
    return r
 
 
# Driver code
if __name__ == "__main__":
    arr = [0, 2, 3, 4, 1, 6, 8, 9]
    # arr = [3,2,1.5]
    find_pair(arr)
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
     
class GFG
{
     
// Function to print the pair (n, r)
// such that nCr is maximum possible
static void findPair(int []arr, int n)
{
    // Array should contain atleast 2 elements
    if (n < 2)
    {
        Console.Write("-1");
        return;
    }
 
    // Maximum element from the array
    int maximum = arr[0];
    for(int i = 1; i < n; i++)
    maximum = Math.Max(maximum, arr[i]);
 
    // temp stores abs(middle - arr[i])
    int temp = 10000001, r = 0, middle = maximum / 2;
 
    // Finding r with minimum abs(middle - arr[i])
    for (int i = 0; i < n; i++)
    {
 
        // When n is even then middle is (maximum / 2)
        if (Math.Abs(middle - arr[i]) < temp && n % 2 == 0)
        {
            temp = Math.Abs(middle - arr[i]);
            r = arr[i];
        }
 
        // When n is odd then middle elements are
        // (maximum / 2) and ((maximum / 2) + 1)
        else if (Math.Min(Math.Abs(middle - arr[i]),
                          Math.Abs(middle + 1 - arr[i])) <
                                   temp && n % 2 == 1)
        {
            temp = Math.Min(Math.Abs(middle - arr[i]),
                            Math.Abs(middle + 1 - arr[i]));
            r = arr[i];
        }
    }
    Console.Write( "n = " + maximum +
                   " and r = " + r);
}
 
// 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 29AjayKumar

Javascript

<script>
 
 
// Java  scriptimplementation of above approach
 
     
// Function to print the pair (n, r)
// such that nCr is maximum possible
function findPair(arr,n)
{
    // Array should contain atleast 2 elements
    if (n < 2)
    {
       document.write("-1");
        return;
    }
 
    // Maximum element from the array
    let maximum = arr[0];
    for(let i = 1; i < n; i++)
    maximum = Math.max(maximum, arr[i]);
 
    // temp stores abs(middle - arr[i])
    let temp = 10000001, r = 0, middle = maximum / 2;
 
    // Finding r with minimum abs(middle - arr[i])
    for (let i = 0; i < n; i++)
    {
 
        // When n is even then middle is (maximum / 2)
        if (Math.abs(middle - arr[i]) < temp && n % 2 == 0)
        {
            temp = Math.abs(middle - arr[i]);
            r = arr[i];
        }
 
        // When n is odd then middle elements are
        // (maximum / 2) and ((maximum / 2) + 1)
        else if (Math.min(Math.abs(middle - arr[i]),
                          Math.abs(middle + 1 - arr[i])) <
                                     temp && n % 2 == 1)
        {
            temp = Math.min(Math.abs(middle - arr[i]),
                            Math.abs(middle + 1 - arr[i]));
            r = arr[i];
        }
    }
    document.write( "n = " + maximum + " and r = " + r);
}
 
// 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>