Add minimum number to an array so that the sum becomes even

Given an array, write a program to add the minimum number(should be greater than 0) to the array so that the sum of array becomes even.

Examples: 

Input : 1 2 3 4 5 6 7 8
Output : 2
Explanation : Sum of array is 36, so we 
add minimum number 2 to make the sum even.

Input : 1 2 3 4 5 6 7 8 9
Output : 1

Method 1 (Computing Sum). We calculate the sum of all elements of the array, then we can check if the sum is even minimum number is 2, else minimum number is 1. This method can cause overflow if sum exceeds allowed limit.

Method 2. Instead of calculating the sum of numbers, we keep the count of odd number of elements in the array. If count of odd numbers present is even we return 2, else we return 1. 

For example - Array contains : 1 2 3 4 5 6 7 
Odd number counts is 4. And we know that the sum of even numbers of odd number is even. And sum of even number is always even (that is why, we don't keep count of even numbers). 

Implementation:

// CPP program to add minimum number
// so that the sum of array becomes even
#include <iostream>
using namespace std;

// Function to find out minimum number
int minNum(int arr[], int n)
{
    // Count odd number of terms in array
    int odd = 0;
    for (int i = 0; i < n; i++) 
        if (arr[i] % 2)
            odd += 1;
    
   return (odd % 2)? 1 : 2;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << minNum(arr, n) << "n";

    return 0;
}

// Java program to add minimum number
// so that the sum of array becomes even

class GFG
{
    // Function to find out minimum number
    static int minNum(int arr[], int n)
    {
        // Count odd number of terms in array
        int odd = 0;
        for (int i = 0; i < n; i++)
            if (arr[i] % 2 != 0)
                odd += 1;

        return ((odd % 2) != 0)? 1 : 2;
    }

    // Driver method to test above function
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        int n = arr.length;

        System.out.println(minNum(arr, n));
    }
}

# Python program to add minimum number
# so that the sum of array becomes even

# Function to find out minimum number
def minNum(arr, n):

    # Count odd number of terms in array
    odd = 0
    for i in range(n):
        if (arr[i] % 2):
            odd += 1
    
    if (odd % 2):
        return(1)
    return (2)


# Driver code
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
n = len(arr)
print(minNum(arr, n))

// C# program to add minimum number
// so that the sum of array becomes even
using System;

class GFG
{
    // Function to find out minimum number
    static int minNum(int []arr, int n)
    {
        // Count odd number of terms in array
        int odd = 0;
        for (int i = 0; i < n; i++)
            if (arr[i] % 2 != 0)
                odd += 1;

        return ((odd % 2) != 0)? 1 : 2;
    }

    // Driver Code
    public static void Main()
    {
        int []arr = {1, 2, 3, 4, 5, 6, 7, 8, 9};
        int n = arr.Length;

        Console.Write(minNum(arr, n));
    }
}

// This code is contributed by Nitin Mittal.

<?php
// PHP program to add minimum number
// so that the sum of array becomes even

// Function to find out minimum number
function minNum( $arr, $n)
{
    
    // Count odd number of
    // terms in array
    $odd = 0;
    for ($i = 0; $i < $n; $i++) 
        if ($arr[$i] % 2)
            $odd += 1;
    
    return ($odd % 2)? 1 : 2;
}

    // Driver code
    $arr = array(1, 2, 3, 4, 5, 
                 6, 7, 8, 9);
    $n = count($arr);
    echo minNum($arr, $n) ;

// This code is contributed by anuj_67.
?>

<script>

// Javascript  program to add minimum number
// so that the sum of array becomes even

    // Function to find out minimum number
    function minNum(arr, n)
    {
        // Count odd number of terms in array
        let odd = 0;
        for (let i = 0; i < n; i++)
            if (arr[i] % 2 != 0)
                odd += 1;
  
        return ((odd % 2) != 0)? 1 : 2;
    }
    
// driver program
    
        let arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ];
        let n = arr.length;
  
        document.write(minNum(arr, n));

// This code is contributed by code_hunt.
</script>

1n

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

Method 3. We can also improve the 2 method, we don't need to keep count of number of odd elements present. We can take a boolean variable (initialized as 0). Whenever we find the odd element in the array we perform the NOT(!) operation on the boolean variable. This logical operator inverts the value of the boolean variable (meaning if it is 0, it converts the variable to 1 and vice-versa). 

For example - Array contains : 1 2 3 4 5 
Explanation : variable initialized as 0. 

Traversing the array 

• 1 is odd, applying NOT operation in variable, now variable becomes 1. 
• 2 is even, no operation. 
• 3 is odd, applying NOT operation in variable, now variable becomes 0. 
• 4 is even, no operation. 
• 5 is odd, applying NOT operation in variable, now variable becomes 1.

If variable value is 1 it means odd number of odd elements are present, minimum number to make sum of elements even is by adding 1. 
Else minimum number is 2.

Implementation:

// CPP program to add minimum number
// so that the sum of array becomes even

#include <iostream>
using namespace std;

// Function to find out minimum number
int minNum(int arr[], int n)
{
    // Count odd number of terms in array
    bool odd = 0;
    for (int i = 0; i < n; i++) 
        if (arr[i] % 2)
            odd = !odd;
    
    if (odd)
        return 1;
    return 2;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << minNum(arr, n) << "n";

    return 0;
}

// Java program to add minimum number
// so that the sum of array becomes even

class GFG
{
    // Function to find out minimum number
    static int minNum(int arr[], int n)
    {
        // Count odd number of terms in array
        Boolean odd = false;
        for (int i = 0; i < n; i++)
            if (arr[i] % 2 != 0)
                odd = !odd;

        if (odd)
            return 1;
        return 2;
    }

    //Driver method to test above function
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        int n = arr.length;

        System.out.println(minNum(arr, n));
    }
}

# Python program to add minimum number
# so that the sum of array becomes even

# Function to find out minimum number
def minNum(arr, n):

    # Count odd number of terms in array
    odd = False
    for i in range(n):
        if (arr[i] % 2):
            odd = not odd
    if (odd):
        return 1
    return 2


# Driver code
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
n = len(arr)
print (minNum(arr, n))

// C# program to add minimum number
// so that the sum of array becomes even
using System;

class GFG
{
    
    // Function to find out minimum number
    static int minNum(int []arr, int n)
    {
        // Count odd number of terms in array
        bool odd = false;
        for (int i = 0; i < n; i++)
            if (arr[i] % 2 != 0)
                odd = !odd;

        if (odd)
            return 1;
        return 2;
    }

    //Driver Code
    public static void Main()
    {
        int []arr = {1, 2, 3, 4, 5, 6, 7, 8, 9};
        int n = arr.Length;

        Console.Write(minNum(arr, n));
    }
}

// This code is contributed by Nitin Mittal.

<?php
// PHP program to add minimum number
// so that the sum of array becomes even

// Function to find out minimum number
function minNum($arr, $n)
{
    
    // Count odd number of 
    // terms in array
    $odd = 0;
    for($i = 0; $i < $n; $i++) 
        if ($arr[$i] % 2)
            $odd = !$odd;
    
    if ($odd)
        return 1;
    return 2;
}

    // Driver code
    $arr = array(1, 2, 3, 4, 5, 6, 7, 8, 9);
    $n = sizeof($arr);
    echo minNum($arr, $n) ,"\n";

// This code is contributed by nitin mittal
?>

<script>
    // Javascript program to add minimum number
    // so that the sum of array becomes even
    
    // Function to find out minimum number
    function minNum(arr, n)
    {
        // Count odd number of terms in array
        let odd = false;
        for (let i = 0; i < n; i++)
            if (arr[i] % 2 != 0)
                odd = !odd;
 
        if (odd)
            return 1;
        return 2;
    }
    
    let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
    let n = arr.length;

    document.write(minNum(arr, n));
 
 // This code is contributed by divyesh072019.
</script>

1n

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

Exercise : 
Find the minimum number required to make the sum of elements odd.