C++ Program for Number of pairs with maximum sum

Given an array arr[], count number of pairs arr[i], arr[j] such that arr[i] + arr[j] is maximum and i < j.

Example :
Input  : arr[] = {1, 1, 1, 2, 2, 2}
Output : 3
Explanation: The maximum possible pair 
sum where i



Method 1 (Naive) 
Traverse a loop i from 0 to n, i.e length of the array and another loop j from i+1 to n to find all possible pairs with i


            C++

    
// CPP program to count pairs with maximum sum.
#include <bits/stdc++.h>
using namespace std;

// function to find the number of maximum pair sums
int sum(int a[], int n)
{
    // traverse through all the pairs
    int maxSum = INT_MIN;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            maxSum = max(maxSum, a[i] + a[j]);

    // traverse through all pairs and keep a count
    // of the number of maximum pairs
    int c = 0;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (a[i] + a[j] == maxSum)
                c++;
    return c;
}

// driver program to test the above function
int main()
{
    int array[] = { 1, 1, 1, 2, 2, 2 };
    int n = sizeof(array) / sizeof(array[0]);
    cout << sum(array, n);
    return 0;
}


Output : 

3


Time complexity:O(n^2)



Method 2 (Efficient) 
If we take a closer look, we can notice following facts. 

Maximum element is always part of solution
If maximum element appears more than once, then result is maxCount * (maxCount - 1)/2. We basically need to choose 2 elements from maxCount (maxCountC2).
If maximum element appears once, then result is equal to count of second maximum element. We can form a pair with every second max and max

            C++

    
// CPP program to count pairs with maximum sum.
#include <bits/stdc++.h>
using namespace std;

// function to find the number of maximum pair sums
int sum(int a[], int n)
{
    // Find maximum and second maximum elements.
    // Also find their counts.
    int maxVal = a[0], maxCount = 1;
    int secondMax = INT_MIN, secondMaxCount;
    for (int i = 1; i < n; i++) {
        if (a[i] == maxVal)
            maxCount++;
        else if (a[i] > maxVal) {
            secondMax = maxVal;
            secondMaxCount = maxCount;
            maxVal = a[i];
            maxCount = 1;
        }
        else if (a[i] == secondMax) {
            secondMax = a[i];
            secondMaxCount++;
        }
        else if (a[i] > secondMax) {
            secondMax = a[i];
            secondMaxCount = 1;
        }
    }

    // If maximum element appears more than once.
    if (maxCount > 1)
        return maxCount * (maxCount - 1) / 2;

    // If maximum element appears only once.
    return secondMaxCount;
}

// driver program to test the above function
int main()
{
    int array[] = { 1, 1, 1, 2, 2, 2, 3 };
    int n = sizeof(array) / sizeof(array[0]);
    cout << sum(array, n);
    return 0;
}


Output : 
 

3


Time complexity:O(n)
 
Please refer complete article on Number of pairs with maximum sum for more details!

                            
                            
                        


                                                                              

                                        

                                                                                
                                                                

                            

    

        
        
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