check whether integer is even or odd using bit manupulation (TheAlgorithms#7099)

* even_or_not file added

* Updated DIRECTORY.md

* modified DIRECTORY.md

* Update bit_manipulation/even_or_not.py

* updating DIRECTORY.md

* Rename even_or_not.py to is_even.py

* updating DIRECTORY.md

Co-authored-by: luciferx48 <laukik.22010776@gmail.com>
Co-authored-by: Christian Clauss <cclauss@me.com>
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>

Author: 4 people

DIRECTORY.md

@@ -45,6 +45,7 @@
   * [Count 1S Brian Kernighan Method](bit_manipulation/count_1s_brian_kernighan_method.py)
   * [Count Number Of One Bits](bit_manipulation/count_number_of_one_bits.py)
   * [Gray Code Sequence](bit_manipulation/gray_code_sequence.py)
+  * [Is Even](bit_manipulation/is_even.py)
   * [Reverse Bits](bit_manipulation/reverse_bits.py)
   * [Single Bit Manipulation Operations](bit_manipulation/single_bit_manipulation_operations.py)
 

bit_manipulation/is_even.py

@@ -0,0 +1,37 @@
+def is_even(number: int) -> bool:
+    """
+    return true if the input integer is even
+    Explanation: Lets take a look at the following deicmal to binary conversions
+    2 => 10
+    14 => 1110
+    100 => 1100100
+    3 => 11
+    13 => 1101
+    101 => 1100101
+    from the above examples we can observe that
+    for all the odd integers there is always 1 set bit at the end
+    also, 1 in binary can be represented as 001, 00001, or 0000001
+    so for any odd integer n => n&1 is always equlas 1 else the integer is even
+
+    >>> is_even(1)
+    False
+    >>> is_even(4)
+    True
+    >>> is_even(9)
+    False
+    >>> is_even(15)
+    False
+    >>> is_even(40)
+    True
+    >>> is_even(100)
+    True
+    >>> is_even(101)
+    False
+    """
+    return number & 1 == 0
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()