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.