Java Program for Number of pairs with maximum sum

Last Updated : 19 Oct, 2023

Write a java program for a given 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<j is 4, which is given by 3 pairs, so the answer is 3 the pairs are (2, 2), (2, 2) and (2, 2)

Input: arr[] = {1, 4, 3, 3, 5, 1}
Output: 1
Explanation: The pair 4, 5 yields the maximum sum i.e, 9 which is given by 1 pair only

Java program for Number of pairs with maximum sum using Naive Approach:

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<j. Find the pair with the maximum possible sum, again traverse for all pairs and keep the count of the number of pairs which gives the pair sum equal to maximum .

Below is the Implementation of the above Approach:

Java

// Java program to count pairs
// with maximum sum.
class GFG {

// function to find the number of
// maximum pair sums
static int sum(int a[], int n)
{
    // traverse through all the pairs
    int maxSum = Integer.MIN_VALUE;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            maxSum = Math.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
public static void main(String[] args)
{
    int array[] = { 1, 1, 1, 2, 2, 2 };
    int n = array.length;
    System.out.println(sum(array, n));
}
}

// This code is contributed by Prerna Saini

Output

Time complexity: O(n2)
Auxiliary Space: O(1)

Efficient Method:

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.

Below is the Implementation of the above Approach:

Java

// Java program to count pairs 
// with maximum sum.
import java.io.*;
class GFG {
    
// function to find the number 
// of maximum pair sums
static int sum(int a[], int n)
{
    // Find maximum and second maximum 
    // elements. Also find their counts.
    int maxVal = a[0], maxCount = 1;
    int secondMax = Integer.MIN_VALUE,
        secondMaxCount = 0;
    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 
public static void main(String[] args)
{
    int array[] = { 1, 1, 1, 2, 2, 2, 3 };
    int n = array.length;
    System.out.println(sum(array, n));
}
}

// This code is contributed by Prerna Saini