Check if original Array Sum is Odd or Even using Bitwise AND of Array

Given an integer N denoting the size of an array and the bitwise AND (K) of all elements of the array. The task is to determine whether the total sum of the elements is odd or even or cannot be determined.

Examples:

Input: N = 1, K = 11
Output: Odd
Explanation: As there is only one element in the array, the element itself is 11.

Input: N = 1, K = 2
Output: Even
Explanation: As there is only one element in the array. the element itself is 2.

Input: N = 5, K = 4
Output: Impossible

Approach: The solution to the problem is based on the below observation: 

Observation:

• We know that for bitwise AND to be 1, all the bits must be 1. 
  
• So if the given LSB of given bitwise AND of Array is 1, it must mean that LSB of all the elements of Array must have been 1.
• On the other hand, if the LSB of given bitwise AND of Array is 0, it can mean that either all or some of the elements of Array has 0 in their LSB.

Conclusion: Therefore using above observation, it can be said that: 

• For bitwise AND to be odd, all the elements of the array must be odd, as LSB of odd numbers is always 1.
• But if the bitwise AND is even, then nothing can be said about the parity (odd or even) of all the elements.

Follow the steps mentioned below to solve the problem:

• If the LSB of K is 1, it clearly proves that every number has LSB as 1 i.e., every number is odd because if any single number has LSB 0 then LSB of K will be 0.
  • The sum of N odd number is always odd if N is odd.
  • The sum is even and if N is even.
• If the LSB of K is 0 the sum cannot be determined except for the case when N is 1 i.e. when N is 1 sum depends on the parity of K. In another case, it’s impossible to find as the count of even and odd numbers cannot be determined.

Below is the implementation of the above approach.

C++

// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find whether the sum is even,
// odd or impossible to find.
void findtotalsum(int n, int a)
{
    // Separate case when N is 1
    // as it depend on parity of n
    if (n == 1) {
        if (a % 2 == 0)
            cout << "Even" << endl;
        else
            cout << "Odd" << endl;
    }

    // Check if a is odd
    else if (a % 2 != 0) {

        // Checking whether n is odd or even
        if (n % 2 == 0)
            cout << "Even" << endl;
        else
            cout << "Odd" << endl;
    }
    else {

        // When no condition applies
        // its impossible to find sum
        cout << "Impossible" << endl;
    }
}

// Driver code
int main()
{
    long int N, K;
    N = 5, K = 4;
    findtotalsum(N, K);
    return 0;
}

Java

// Java code to minimize number of rotations
import java.util.*;

class GFG {

  // Function to find whether the sum is even,
  // odd or impossible to find.
  static void findtotalsum(int n, int a)
  {

    // Separate case when N is 1
    // as it depend on parity of n
    if (n == 1) {
      if (a % 2 == 0)
        System.out.println("Even");
      else
        System.out.println("Odd");
    }

    // Check if a is odd
    else if (a % 2 != 0) {

      // Checking whether n is odd or even
      if (n % 2 == 0)
        System.out.println("Even");
      else
        System.out.println("Odd");
    }
    else {

      // When no condition applies
      // its impossible to find sum
      System.out.println("Impossible");
    }
  }

  // Driver code
  public static void main (String[] args) {
    int N = 5, K = 4;
    findtotalsum(N, K);
  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python program for the above approach

# Function to find whether the sum is even,
# odd or impossible to find.
def findtotalsum(n, a) :

    # Separate case when N is 1
    # as it depend on parity of n
    if (n == 1) :
        if (a % 2 == 0) :
            print("Even")
        else :
            print( "Odd" )


    # Check if a is odd
    elif (a % 2 != 0) :

        # Checking whether n is odd or even
        if (n % 2 == 0) :
            print("Even")
        else :
            print( "Odd" )

    else :

        # When no condition applies
        # its impossible to find sum
        print( "Impossible")

# Driver code

N = 5
K = 4
findtotalsum(N, K)

# This code is contributed by sanjoy_62.

C#

// C# code to minimize number of rotations
using System;

class GFG {

  // Function to find whether the sum is even,
  // odd or impossible to find.
  static void findtotalsum(int n, int a)
  {

    // Separate case when N is 1
    // as it depend on parity of n
    if (n == 1) {
      if (a % 2 == 0)
        Console.WriteLine("Even");
      else
        Console.WriteLine("Odd");
    }

    // Check if a is odd
    else if (a % 2 != 0) {

      // Checking whether n is odd or even
      if (n % 2 == 0)
        Console.WriteLine("Even");
      else
        Console.WriteLine("Odd");
    }
    else {

      // When no condition applies
      // its impossible to find sum
      Console.WriteLine("Impossible");
    }
  }

  // Driver code
  public static void Main () {
    int N = 5, K = 4;
    findtotalsum(N, K);
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
    // JavaScript code to implement the above approach

    // Function to find whether the sum is even,
    // odd or impossible to find.
    const findtotalsum = (n, a) => {

        // Separate case when N is 1
        // as it depend on parity of n
        if (n == 1) {
            if (a % 2 == 0)
                document.write("Even<br/>");
            else
                document.write("Odd<br/>");
        }

        // Check if a is odd
        else if (a % 2 != 0) {

            // Checking whether n is odd or even
            if (n % 2 == 0)
                document.write("Even<br/>");
            else
                document.write("Odd<br/>");
        }
        else {

            // When no condition applies
            // its impossible to find sum
            document.write("Impossible<br/>");
        }
    }

    // Driver code
    let N, K;
    N = 5, K = 4;
    findtotalsum(N, K);

// This code is contributed by rakeshsahni

</script>

Output

Impossible

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