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Number of pairs with maximum sum
  • Difficulty Level : Medium
  • Last Updated : 23 Apr, 2021

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<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

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<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 

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;
}

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

Python3




# Python program to count pairs with
# maximum sum
 
def _sum( a, n):
 
    # traverse through all the pairs
    maxSum = -9999999
    for i in range(n):
        for j in range(n):
            maxSum = max(maxSum, a[i] + a[j])
     
    # traverse through all pairs and
    # keep a count of the number
    # of maximum pairs
    c = 0
    for i in range(n):
        for j in range(i+1, n):
            if a[i] + a[j] == maxSum:
                c+=1
    return c
 
# driver code
array = [ 1, 1, 1, 2, 2, 2 ]
n = len(array)
print(_sum(array, n))
 
# This code is contributed by "Abhishek Sharma 44"

C#




// C# program to count pairs
// with maximum sum.
using System;
 
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 = int.MinValue;
         
        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()
    {
        int []array = { 1, 1, 1, 2, 2, 2 };
        int n = array.Length;
        Console.WriteLine(sum(array, n));
    }
}
 
// This code is contributed by anuj_67.

PHP




<?php
// PHP program to count pairs
// with maximum sum.
 
// function to find the number
// of maximum pair sum
function sum( $a, $n)
{
     
    // traverse through all
    // the pairs
    $maxSum = PHP_INT_MIN;
    for($i = 0; $i < $n; $i++)
        for($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
    $c = 0;
    for($i = 0; $i < $n; $i++)
        for($j = $i + 1; $j < $n; $j++)
            if ($a[$i] + $a[$j] == $maxSum)
                $c++;
    return $c;
}
 
    // Driver Code
    $array = array(1, 1, 1, 2, 2, 2);
    $n = count($array);
    echo sum($array, $n);
     
// This code is contributed by anuj_67.
?>

Javascript




<script>
 
// JavaScript program to count pairs
// with maximum sum.
 
// Function to find the number of
// maximum pair sums
function sum(a, n)
{
     
    // traverse through all the pairs
    let maxSum = Number.MIN_VALUE;
    for(let i = 0; i < n; i++)
        for(let 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
    let c = 0;
    for(let i = 0; i < n; i++)
        for(let j = i + 1; j < n; j++)
            if (a[i] + a[j] == maxSum)
                c++;
    return c;
}
 
// Driver Code
let array = [ 1, 1, 1, 2, 2, 2 ];
let n = array.length;
 
document.write(sum(array, n));
 
// This code is contributed by code_hunt  
 
</script>

Output : 

3

Time complexity:O(n^2)

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

  1. Maximum element is always part of solution
  2. If maximum element appears more than once, then result is maxCount * (maxCount – 1)/2. We basically need to choose 2 elements from maxCount (maxCountC2).
  3. 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;
}

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

Python3




# Python 3 program to count
# pairs with maximum sum.
import sys
 
# Function to find the number
# of maximum pair sums
def sum(a, n):
 
    # Find maximum and second maximum elements.
    # Also find their counts.
    maxVal = a[0]; maxCount = 1
    secondMax = sys.maxsize
     
    for i in range(1, n) :
         
        if (a[i] == maxVal) :
            maxCount += 1
         
        elif (a[i] > maxVal) :
            secondMax = maxVal
            secondMaxCount = maxCount
            maxVal = a[i]
            maxCount = 1
         
        elif (a[i] == secondMax) :
            secondMax = a[i]
            secondMaxCount += 1
         
        elif (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 Code
array = [1, 1, 1, 2, 2, 2, 3]
n = len(array)
print(sum(array, n))
 
 
# This code is contributed by Smitha Dinesh Semwal

C#




// C# program to count pairs with maximum
// sum.
using System;
 
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 = int.MinValue;
        int 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()
    {
        int []array = { 1, 1, 1, 2,
                               2, 2, 3 };
        int n = array.Length;
         
        Console.WriteLine(sum(array, n));
    }
}
 
// This code is contributed by anuj_67.

PHP




<?php
// PHP program to count
// pairs with maximum sum.
 
// function to find the number
// of maximum pair sums
function sum( $a, $n)
{
    // Find maximum and second
    // maximum elements. Also
    // find their counts.
    $maxVal = $a[0]; $maxCount = 1;
    $secondMax = PHP_INT_MIN;
    $secondMaxCount;
    for ( $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 Code
$array = array(1, 1, 1, 2,
                2, 2, 3 );
$n = count($array);
echo sum($array, $n);
 
// This code is contributed by anuj_67.
?>

Output : 
 

3

Time complexity:O(n)
 




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