Maximum even sum of a pair of given Array

Given an array arr[] of N distinct non-negative integers, the task is to determine if there is an even number that can be represented as the sum of two different elements of arr. If it exists return the maximum such number otherwise return -1.

Examples:

Input: N = 4, arr[] = {2, 3, 4, 5}
Output: 8
Explanation: The values that can be represented as the sum of two distinct elements of arr are 5, 6, 7, 8 and 9. 
We have two even number here, and the maximum is 8.

Input: N = 2, arr[] = {0, 3}
Output: -1
Explanation: The value represented as the sum of two distinct elements of arr is 1. 
We have no even number here, so -1 should be printed.

Naive Approach: The basic approach is: 

Consider each pair sum by nested traversal and consider the maximum of all such pair sums that are even and present in the array.

Follow the steps to solve the problem:

• Initialize the result as 0 at starting.
• Traverse the array using a loop from 0 to N – 2
  • For each element consider pairing with every element in the range i + 1 to N – 1 and check if the pair sum is even or odd.
    • If the sum is even, update the result with the maximum pair sum at each step.
• At last, check if the result is 0 then return -1 otherwise return the result.

Below is the implementation of the above approach.

C++

// C++ program for the above approach
#include <bits/stdc++.h>

using namespace std;

// Function to find maximum even pair sum
int maxEvenSumPair(int arr[], int n)
{
    int result = 0;
    for (int i = 0; i < (n - 1); i++) {
        for (int j = (i + 1); j < n; j++) {

            // Calculate the sum for each
            // pair
            int curSum = arr[i] + arr[j];

            // If sum is even than consider
            // it for result
            if (curSum % 2 == 0)

                // Maintain maximum pair
                // sum in result
                result = max(result, curSum);
        }
    }

    // If result is zero that means
    // no pair with even sum is present
    if (result == 0)
        return -1;

    return result;
}

// Driver code
int main()
{

    int arr[] = { 2, 3, 4, 5 };
    int size = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << maxEvenSumPair(arr, size);

    return 0;
}

Java

// Java program for the above approach
class GFG {

  // Function to find maximum even pair sum
  public static int maxEvenSumPair(int arr[], int n) {
    int result = 0;
    for (int i = 0; i < (n - 1); i++) {
      for (int j = (i + 1); j < n; j++) {

        // Calculate the sum for each
        // pair
        int curSum = arr[i] + arr[j];

        // If sum is even than consider
        // it for result
        if (curSum % 2 == 0)

          // Maintain maximum pair
          // sum in result
          result = max(result, curSum);
      }
    }

    // If result is zero that means
    // no pair with even sum is present
    if (result == 0)
      return -1;

    return result;
  }

  public static int max(int a, int b) {
    if (a > b) return a;
    return b;
  }

  // Driver code
  public static void main(String[] args) {
    int arr[] = { 2, 3, 4, 5 };
    int size = arr.length;

    // Function call
    System.out.println(maxEvenSumPair(arr, size));
  }
}

// This code is contributed by ajaymakvana

Python3

# Python3 program for the above approach

# Function to find maximum even pair sum
def maxEvenSumPair(arr,  n) :

    result = 0;
    for i in range(n - 1) :
        for j in range(i + 1, n) :

            # Calculate the sum for each
            # pair
            curSum = arr[i] + arr[j];

            # If sum is even than consider
            # it for result
            if (curSum % 2 == 0) :

                # Maintain maximum pair
                # sum in result
                result = max(result, curSum);

    # If result is zero that means
    # no pair with even sum is present
    if (result == 0) :
        return -1;

    return result;

# Driver code
if __name__ == "__main__" :

    arr = [ 2, 3, 4, 5 ];
    size = len(arr);

    # Function call
    print(maxEvenSumPair(arr, size));

    # This code is contributed by AnkThon

C#

using System;
class HelloWorld
{

    // Function to find maximum even pair sum
    static int maxEvenSumPair(int[] arr, int n)
    {
        int result = 0;
        for (int i = 0; i < (n - 1); i++) {
            for (int j = (i + 1); j < n; j++) {

                // Calculate the sum for each
                // pair
                int curSum = arr[i] + arr[j];

                // If sum is even than consider
                // it for result
                if (curSum % 2 == 0)

                    // Maintain maximum pair
                    // sum in result
                    result = Math.Max(result, curSum);
            }
        }

        // If result is zero that means
        // no pair with even sum is present
        if (result == 0)
            return -1;

        return result;
    }
    static void Main()
    {
        int[] arr = { 2, 3, 4, 5 };
        int size = 4;

        // Function call
        Console.Write(maxEvenSumPair(arr, size));
    }
}

// This code is contributed by garg28harsh.

Javascript

<script>
       // JavaScript code for the above approach


       // Function to find maximum even pair sum
       function maxEvenSumPair(arr, n) {
           let result = 0;
           for (let i = 0; i < (n - 1); i++) {
               for (let j = (i + 1); j < n; j++) {

                   // Calculate the sum for each
                   // pair
                   let curSum = arr[i] + arr[j];

                   // If sum is even than consider
                   // it for result
                   if (curSum % 2 == 0)

                       // Maintain maximum pair
                       // sum in result
                       result = Math.max(result, curSum);
               }
           }

           // If result is zero that means
           // no pair with even sum is present
           if (result == 0)
               return -1;

           return result;
       }

       // Driver code

       let arr = [2, 3, 4, 5];
       let size = arr.length;

       // Function call
       document.write(maxEvenSumPair(arr, size));


// This code is contributed by Potta Lokesh

   </script>

Output

8

Time Complexity: O(N²)
Auxiliary Space: O(1)

Find maximum sum Pair by finding maximum even and odd numbers:

To solve the problem efficiently follow the below idea:

The key point/ idea is that sum of any two even numbers or any two odd numbers is always even. Find two greatest even numbers and two greatest odd numbers, the final result will be the maximum sum of the two odd numbers and the sum of the two even numbers.

Follow the steps to solve the problem: 

• Initialize first and second maximum odd and even numbers equal to -1.
• Traverse the array to find the first maximum even and first maximum odd number.
• Traverse the array a second time to find the second maximum odd and second maximum even number.
• Calculate the sum of the pair of maximum two odd and the pair of maximum two even numbers and consider the result equal to the maximum among them.
• If the result is zero it means no even sum pair exist so return -1, otherwise return the maximum result.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find maximum even pair sum
int maxEvenSumPair(int arr[], int n)
{
    int firstEvenMax = -1, secondEvenMax = -1,
        firstOddMax = -1, secondOddMax = -1;

    // First traversal for finding
    // the maximum even and odd number
    for (int i = 0; i < n; i++) {
        if (arr[i] & 1)
            firstOddMax = max(firstOddMax, arr[i]);
        else
            firstEvenMax = max(firstEvenMax, arr[i]);
    }

    // Second traversal for finding
    // the second maximum even and
    // odd number
    for (int i = 0; i < n; i++) {
        if (arr[i] & 1) {
            if (arr[i] != firstOddMax)
                secondOddMax = max(secondOddMax, arr[i]);
        }
        else {
            if (arr[i] != firstEvenMax)
                secondEvenMax = max(secondEvenMax, arr[i]);
        }
    }

    int sumOdd = 0, sumEven = 0;

    // If two even numbers exist in array
    if (firstEvenMax != -1 and secondEvenMax != -1)
        sumEven = firstEvenMax + secondEvenMax;

    // If two odd numbers exist in array
    if (firstOddMax != -1 and secondOddMax != -1)
        sumOdd = firstOddMax + secondOddMax;

    int res = max(sumEven, sumOdd);

    // No even sum pair found so return -1
    if (res == 0)
        return -1;

    return res;
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 4, 5 };
    int size = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    cout << maxEvenSumPair(arr, size);

    return 0;
}

Java

// Java code to implement the approach
import java.io.*;

class GFG {

  // Function to find maximum even pair sum
  static int maxEvenSumPair(int arr[], int n)
  {
    int firstEvenMax = -1, secondEvenMax = -1,
    firstOddMax = -1, secondOddMax = -1;

    // First traversal for finding
    // the maximum even and odd number
    for (int i = 0; i < n; i++) {
      if ((arr[i] & 1)==1)
        firstOddMax = Math.max(firstOddMax, arr[i]);
      else
        firstEvenMax = Math.max(firstEvenMax, arr[i]);
    }

    // Second traversal for finding
    // the second maximum even and
    // odd number
    for (int i = 0; i < n; i++) {
      if ((arr[i] & 1)==1) {
        if (arr[i] != firstOddMax)
          secondOddMax = Math.max(secondOddMax, arr[i]);
      }
      else {
        if (arr[i] != firstEvenMax)
          secondEvenMax = Math.max(secondEvenMax, arr[i]);
      }
    }

    int sumOdd = 0, sumEven = 0;

    // If two even numbers exist in array
    if (firstEvenMax != -1 && secondEvenMax != -1)
      sumEven = firstEvenMax + secondEvenMax;

    // If two odd numbers exist in array
    if (firstOddMax != -1 && secondOddMax != -1)
      sumOdd = firstOddMax + secondOddMax;

    int res = Math.max(sumEven, sumOdd);

    // No even sum pair found so return -1
    if (res == 0)
      return -1;

    return res;
  }

  // Driver code
  public static void main (String[] args) {

    int arr[] = { 2, 3, 4, 5 };
    int size = arr.length;

    // Function Call
    System.out.println(maxEvenSumPair(arr, size));

  }
}

// This code is contributed by Pushpesh Raj.

Python3

# Python3 code to implement the approach

# Function to find maximum even pair sum
def maxEvenSumPair(arr, n) :

    firstEvenMax = -1; secondEvenMax = -1;
    firstOddMax = -1; secondOddMax = -1;

    # First traversal for finding
    # the maximum even and odd number
    for i in range(n) :
        if (arr[i] & 1) :
            firstOddMax = max(firstOddMax, arr[i]);
        else :
            firstEvenMax = max(firstEvenMax, arr[i]);

    # Second traversal for finding
    # the second maximum even and
    # odd number
    for i in range(n) :
        if (arr[i] & 1) :
            if (arr[i] != firstOddMax) :
                secondOddMax = max(secondOddMax, arr[i]);

        else :
            if (arr[i] != firstEvenMax) :
                secondEvenMax = max(secondEvenMax, arr[i]);

    sumOdd = 0; sumEven = 0;

    # If two even numbers exist in array
    if (firstEvenMax != -1 and secondEvenMax != -1) :
        sumEven = firstEvenMax + secondEvenMax;

    # If two odd numbers exist in array
    if (firstOddMax != -1 and secondOddMax != -1) :
        sumOdd = firstOddMax + secondOddMax;

    res = max(sumEven, sumOdd);

    # No even sum pair found so return -1
    if (res == 0) :
        return -1;

    return res;

# Driver code
if __name__ == "__main__" :

    arr = [ 2, 3, 4, 5 ];
    size = len(arr);

    # Function Call
    print(maxEvenSumPair(arr, size));

   # This code is contributed by AnkThon

C#

// C# code to implement the approach
using System;
class GFG {

  // Function to find maximum even pair sum
  static int maxEvenSumPair(int[] arr, int n)
  {
    int firstEvenMax = -1, secondEvenMax = -1,
    firstOddMax = -1, secondOddMax = -1;

    // First traversal for finding
    // the maximum even and odd number
    for (int i = 0; i < n; i++) {
      if ((arr[i] & 1) != 0)
        firstOddMax = Math.Max(firstOddMax, arr[i]);
      else
        firstEvenMax
        = Math.Max(firstEvenMax, arr[i]);
    }

    // Second traversal for finding
    // the second maximum even and
    // odd number
    for (int i = 0; i < n; i++) {
      if ((arr[i] & 1) != 0) {
        if (arr[i] != firstOddMax)
          secondOddMax
          = Math.Max(secondOddMax, arr[i]);
      }
      else {
        if (arr[i] != firstEvenMax)
          secondEvenMax
          = Math.Max(secondEvenMax, arr[i]);
      }
    }

    int sumOdd = 0, sumEven = 0;

    // If two even numbers exist in array
    if (firstEvenMax != -1 && secondEvenMax != -1)
      sumEven = firstEvenMax + secondEvenMax;

    // If two odd numbers exist in array
    if (firstOddMax != -1 && secondOddMax != -1)
      sumOdd = firstOddMax + secondOddMax;

    int res = Math.Max(sumEven, sumOdd);

    // No even sum pair found so return -1
    if (res == 0)
      return -1;

    return res;
  }

  static void Main()
  {
    int[] arr = { 2, 3, 4, 5 };
    int size = arr.Length;

    // Function Call
    Console.Write(maxEvenSumPair(arr, size));
  }
}

// This code is contributed by garg28harsh.

Javascript

<script>

// Javascript code to implement the approach

// Function to find maximum even pair sum
function maxEvenSumPair(arr, n)
{
    let firstEvenMax = -1;
    let secondEvenMax = -1;
    let firstOddMax = -1;
    let secondOddMax = -1;

    // First traversal for finding
    // the maximum even and odd number
    for (let i = 0; i < n; i++) {
        if (arr[i] & 1)
            firstOddMax = Math.max(firstOddMax, arr[i]);
        else
            firstEvenMax = Math.max(firstEvenMax, arr[i]);
    }

    // Second traversal for finding
    // the second maximum even and
    // odd number
    for (let i = 0; i < n; i++) {
        if (arr[i] & 1) {
            if (arr[i] != firstOddMax)
                secondOddMax = Math.max(secondOddMax, arr[i]);
        }
        else {
            if (arr[i] != firstEvenMax)
                secondEvenMax = Math.max(secondEvenMax, arr[i]);
        }
    }

    let sumOdd = 0;
    let sumEven = 0;

    // If two even numbers exist in array
    if (firstEvenMax != -1 && secondEvenMax != -1)
        sumEven = firstEvenMax + secondEvenMax;

    // If two odd numbers exist in array
    if (firstOddMax != -1 && secondOddMax != -1)
        sumOdd = firstOddMax + secondOddMax;

    let res = Math.max(sumEven, sumOdd);

    // No even sum pair found so return -1
    if (res == 0)
        return -1;

    return res;
}

// Driver code
let arr = [ 2, 3, 4, 5 ];
let size = arr.length;

// Function Call
console.log(maxEvenSumPair(arr, size));

// This code is contributed by akashish__

</script>

Output

8

Time Complexity: O(N)
Auxiliary Space: O(1)