# 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[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

 ``

Output:

`n = 9 and r = 4`

Time Complexity: O(n)

Auxiliary Space: O(1)

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