Find a pair from the given array with maximum nCr value

Difficulty Level : Medium
Last Updated : 02 Apr, 2019

Given an array arr[] of n positive integers. The task is to find elements arr[i] and arr[j] from the array such that ^arr[i]C_arr[j] is maximum possible. In case of more than 1 valid pairs, print any one of them.

Examples:

Input: arr[] = {3, 1, 2}
Output: 3 2
³C₁ = 3
³C₂ = 3
²C₁ = 2
(3, 1) and (3, 2) are the only pairs with maximum ⁿC_r.

Input: arr[] = {9, 2, 4, 11, 6}
Output: 11 6

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

Approach: ⁿC_r is a monotonic increasing function, that is ^{n + 1}C_r > ⁿC_r. We can use this fact to get close to our answer, we will choose the max n among all the given integers. This way we fixed the value of n.
Now, we have to look for r. Since we know that ⁿC_r = ⁿC_{n – r}, it means ⁿC_r will first reach its maxima and then decrease.

If n is be odd, then our maxima will occur at n / 2 and n / 2 + 1.
For n = 11, we will get the maxima at ¹¹C₅ and ¹¹C₆.

When n is even, then our maxima will occur at n / 2.

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 that gives maximum nCr 
void printMaxValPair(vector<long long>& v, int n) 
{ 
    sort(v.begin(), v.end()); 
  
    // This gives the value of N in nCr 
    long long N = v[n - 1]; 
  
    // Case 1 : When N is odd 
    if (N % 2 == 1) { 
        long long first_maxima = N / 2; 
        long long second_maxima = first_maxima + 1; 
        long long ans1 = 3e18, ans2 = 3e18; 
        long long from_left = -1, from_right = -1; 
        long long from = -1; 
        for (long long i = 0; i < n; ++i) { 
            if (v[i] > first_maxima) { 
                from = i; 
                break; 
            } 
            else { 
                long long diff = first_maxima - v[i]; 
                if (diff < ans1) { 
                    ans1 = diff; 
                    from_left = v[i]; 
                } 
            } 
        } 
        from_right = v[from]; 
        long long diff1 = first_maxima - from_left; 
        long long diff2 = from_right - second_maxima; 
  
        if (diff1 < diff2) 
            cout << N << " " << from_left; 
        else
            cout << N << " " << from_right; 
    } 
  
    // Case 2 : When N is even 
    else { 
        long long maxima = N / 2; 
        long long ans1 = 3e18; 
        long long R = -1; 
        for (long long i = 0; i < n - 1; ++i) { 
            long long diff = abs(v[i] - maxima); 
            if (diff < ans1) { 
                ans1 = diff; 
                R = v[i]; 
            } 
        } 
        cout << N << " " << R; 
    } 
} 
  
// Driver code 
int main() 
{ 
    vector<long long> v = { 1, 1, 2, 3, 6, 1 }; 
    int n = v.size(); 
    printMaxValPair(v, n); 
  
    return 0; 
} 

Java

// Java implementation of the approach  
import java.util.*; 
  
class GFG 
{ 
  
// Function to print the pair that gives maximum nCr  
static void printMaxValPair(Vector<Long> v, int n)  
{  
    Collections.sort(v);  
  
    // This gives the value of N in nCr  
    long N = v.get((int)n - 1);  
  
    // Case 1 : When N is odd  
    if (N % 2 == 1)  
    {  
        long first_maxima = N / 2;  
        long second_maxima = first_maxima + 1;  
        long ans1 =(long) 3e18, ans2 = (long)3e18;  
        long from_left = -1, from_right = -1;  
        long from = -1;  
        for (long i = 0; i < n; ++i) 
        {  
            if (v.get((int)i) > first_maxima)  
            {  
                from = i;  
                break;  
            }  
            else 
            {  
                long diff = first_maxima - v.get((int)i);  
                if (diff < ans1) 
                {  
                    ans1 = diff;  
                    from_left = v.get((int)i);  
                }  
            }  
        }  
        from_right = v.get((int)from);  
        long diff1 = first_maxima - from_left;  
        long diff2 = from_right - second_maxima;  
  
        if (diff1 < diff2)  
            System.out.println( N + " " + from_left);  
        else
            System.out.println( N + " " + from_right);  
    }  
  
    // Case 2 : When N is even  
    else 
    {  
        long maxima = N / 2;  
        long ans1 =(int) 3e18;  
        long R = -1;  
        for (long i = 0; i < n - 1; ++i)  
        {  
            long diff = Math.abs(v.get((int)i) - maxima);  
              
            if (diff < ans1)  
            {  
                ans1 = diff;  
                R = v.get((int)i);  
            }  
        }  
        System.out.println( N + " " + R);  
    }  
}  
  
// Driver code  
public static void main(String args[])  
{  
    long arr[] = { 1, 1, 2, 3, 6, 1}; 
    Vector<Long> v = new Vector<Long>( );  
      
    for(int i = 0; i < arr.length; i++) 
    v.add(arr[i]); 
      
    int n = v.size();  
    printMaxValPair(v, n);  
} 
} 
  
// This code is contributed by Arnab Kundu 

Python3

# Python3 implementation of the approach 
  
# Function to print the pair that 
# gives maximum nCr 
def printMaxValPair(v, n): 
  
    v.sort() 
  
    # This gives the value of N in nCr 
    N = v[n - 1] 
  
    # Case 1 : When N is odd 
    if N % 2 == 1:  
        first_maxima = N // 2
        second_maxima = first_maxima + 1
        ans1, ans2 = 3 * (10 ** 18), 3 * (10 ** 18) 
        from_left, from_right = -1, -1
        _from = -1
        for i in range(0, n):  
            if v[i] > first_maxima:  
                _from = i 
                break
              
            else: 
                diff = first_maxima - v[i] 
                if diff < ans1:  
                    ans1 = diff 
                    from_left = v[i] 
                  
        from_right = v[_from] 
        diff1 = first_maxima - from_left 
        diff2 = from_right - second_maxima 
  
        if diff1 < diff2: 
            print(N, from_left) 
        else: 
            print(N, from_right) 
  
    # Case 2 : When N is even 
    else: 
        maxima = N // 2
        ans1 = 3 * (10 ** 18) 
        R = -1
        for i in range(0, n - 1):  
            diff = abs(v[i] - maxima) 
            if diff < ans1:  
                ans1 = diff 
                R = v[i] 
              
        print(N, R) 
  
# Driver code 
if __name__ == "__main__": 
  
    v = [1, 1, 2, 3, 6, 1]  
    n = len(v) 
    printMaxValPair(v, n) 
  
# This code is contributed by  
# Rituraj Jain 

C#

// C# implementation of the approach  
using System; 
using System.Collections.Generic; 
  
class GFG  
{  
  
// Function to print the pair that gives maximum nCr  
static void printMaxValPair(List<long> v, int n)  
{  
    v.Sort();  
  
    // This gives the value of N in nCr  
    long N = v[(int)n - 1];  
  
    // Case 1 : When N is odd  
    if (N % 2 == 1)  
    {  
        long first_maxima = N / 2;  
        long second_maxima = first_maxima + 1;  
        long ans1 = (long) 3e18, ans2 = (long)3e18;  
        long from_left = -1, from_right = -1;  
        long from = -1;  
        for (long i = 0; i < n; ++i)  
        {  
            if (v[(int)i] > first_maxima)  
            {  
                from = i;  
                break;  
            }  
            else
            {  
                long diff = first_maxima - v[(int)i];  
                if (diff < ans1)  
                {  
                    ans1 = diff;  
                    from_left = v[(int)i];  
                }  
            }  
        }  
        from_right = v[(int)from];  
        long diff1 = first_maxima - from_left;  
        long diff2 = from_right - second_maxima;  
  
        if (diff1 < diff2)  
            Console.WriteLine( N + " " + from_left);  
        else
            Console.WriteLine( N + " " + from_right);  
    }  
  
    // Case 2 : When N is even  
    else
    {  
        long maxima = N / 2;  
        long ans1 = (long)3e18;  
        long R = -1;  
        for (long i = 0; i < n - 1; ++i)  
        {  
            long diff = Math.Abs(v[(int)i] - maxima);  
              
            if (diff < ans1)  
            {  
                ans1 = diff;  
                R = v[(int)i];  
            }  
        }  
        Console.WriteLine( N + " " + R);  
    }  
}  
  
// Driver code  
public static void Main(String []args)  
{  
    long []arr = { 1, 1, 2, 3, 6, 1};  
    List<long> v = new List<long>( );  
      
    for(int i = 0; i < arr.Length; i++)  
    v.Add(arr[i]);  
      
    int n = v.Count;  
    printMaxValPair(v, n);  
}  
}  
  
// This code contributed by Rajput-Ji 

PHP

<?php 
// PHP implementation of the approach 
  
// Function to print the pair that 
// gives maximum nCr 
function printMaxValPair($v, $n) 
{ 
    sort($v); 
  
    // This gives the value of N in nCr 
    $N = $v[$n - 1]; 
  
    // Case 1 : When N is odd 
    if ($N % 2 == 1)  
    { 
        $first_maxima = $N / 2; 
        $second_maxima = $first_maxima + 1; 
        $ans1 = 3e18; 
        $ans2 = 3e18; 
        $from_left = -1; 
        $from_right = -1; 
        $from = -1; 
        for ($i = 0; $i < $n; ++$i)  
        { 
            if ($v[$i] > $first_maxima)  
            { 
                $from = $i; 
                break; 
            } 
            else 
            { 
                $diff = $first_maxima - $v[$i]; 
                if ($diff < $ans1)  
                { 
                    $ans1 = $diff; 
                    $from_left = $v[$i]; 
                } 
            } 
        } 
        $from_right = $v[$from]; 
        $diff1 = $first_maxima - $from_left; 
        $diff2 = $from_right - $second_maxima; 
  
        if ($diff1 < $diff2) 
            echo $N . " " . $from_left; 
        else
            echo $N . " " . $from_right; 
    } 
  
    // Case 2 : When N is even 
    else 
    { 
        $maxima = $N / 2; 
        $ans1 = 3e18; 
        $R = -1; 
        for ($i = 0; $i < $n - 1; ++$i)  
        { 
            $diff = abs($v[$i] - $maxima); 
            if ($diff < $ans1)  
            { 
                $ans1 = $diff; 
                $R = $v[$i]; 
            } 
        } 
        echo $N . " " . $R; 
    } 
} 
  
// Driver code 
$v = array( 1, 1, 2, 3, 6, 1 ); 
$n = count($v); 
printMaxValPair($v, $n); 
  
// This code is contributed by mits 
?> 

Output:

6 3

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

C++

Java

Python3

C#

PHP