Find a pair from the given array with maximum nCr value in C++

Concept

With respect of given an array arr[] of n positive integers, the task is to determineelements arr[i] and arr[j] from the array such that arr[i]Carr[j] is at most possible. With respect of more than 1 valid pairs, print any one of them.

Input 

arr[] = {4, 1, 2}

Output 

4 2
4C1 = 4
4C2 = 4
2C1 = 4
(4, 2) is the only pairs with maximum nCr.

Method

ⁿC_r is treated as a monotonic increasing function, that is ⁿ⁺¹C_r > ⁿC_r. We can apply this fact to get close to our answer; we will select the max n among all the given integers. In this way we fixed the value of n.

Now, we concentrate for r. As we know that ⁿC_r = ⁿC_n-r , it indicates nCr will first reach its maxima and then decrease.

For odd value of n, then our maxima will occur at n / 2 and n / 2 + 1.

With respect of n = 11, we will get the maxima at ¹¹C₅ and ¹¹C₆.

For even value of n, then our maxima will occur at n / 2.

With respect of n = 4, we will get the maxima at ⁴C2

Example

 Live Demo

// This is C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Now Function to print the pair that gives maximum nCr
void printMaxValPair1(vector<long long>& v1, int n1){
   sort(v1.begin(), v1.end());
   // This provides the value of N in nCr
   long long N1 = v1[n1 - 1];
   // Case 1 : When N1 is odd
   if (N1 % 2 == 1) {
      long long first_maxima1 = N1 / 2;
      long long second_maxima1 = first_maxima1 + 1;
      long long ans1 = 3e18, ans2 = 3e18;
      long long from_left1 = -1, from_right1 = -1;
      long long from = -1;
      for (long long i = 0; i < n1; ++i) {
         if (v1[i] > first_maxima1) {
            from = i;
            break;
         }
         else {
            long long diff = first_maxima1 - v1[i];
            if (diff < ans1) {
               ans1 = diff;
               from_left1 = v1[i];
            }
         }
      }
      from_right1 = v1[from];
      long long diff1 = first_maxima1 - from_left1;
      long long diff2 = from_right1 - second_maxima1;
      if (diff1 < diff2)
         cout << N1 << " " << from_left1;
      else
         cout << N1 << " " << from_right1;
   }
   // Case 2 : When N1 is even
   else {
      long long maxima = N1 / 2;
      long long ans1 = 3e18;
      long long R = -1;
      for (long long i = 0; i < n1 - 1; ++i) {
         long long diff = abs(v1[i] - maxima);
         if (diff < ans1) {
            ans1 = diff;
            R = v1[i];
         }
      }
      cout << N1 << " " << R;
   }
}
// Driver code
int main(){
   vector<long long> v1 = { 1, 1, 2, 3, 6, 1 };
   int n1 = v1.size();
   printMaxValPair1(v1, n1);
   return 0;
}

Output

6 3