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Find a pair (n,r) in an integer array such that value of nCr is maximum
• Last Updated : 18 May, 2020

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: A simple approach is to consider each (n, r) pair and find the aximum 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`
Output:
```n = 9 and r = 4
```

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