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

• Last Updated : 13 Mar, 2022

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

nCr = 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 nCr.
Efficient approach: It is known from combinatorics:

When n is odd:
nC0 < nC1 ….. < nC(n-1)/2 = nC(n+1)/2 > ….. > nCn-1 > nCn
When n is even:
nC0 < nC1 ….. < nCn/2 > ….. > nCn-1 > nCn
Also, nCr = nCn-r

It can be observed that nCr 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 ``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);` `    ``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;``    ``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

 ``

Output:

`n = 9 and r = 4`

Time Complexity: O(n)

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up