# 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 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); ` ` `  `    ``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 `

Output:

```n = 9 and r = 4
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