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

C++Server Side ProgrammingProgramming

## 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

nCr is treated as a monotonic increasing function, that is n+1Cr > nCr. 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 nCr = nCn-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
Published on 23-Jul-2020 07:15:58