coderjack
diff --git a/‎strings/can_string_be_rearranged_as_palindrome.py‎
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+# Created by susmith98
+
+from collections import Counter
+from timeit import timeit
+
+# Problem Description:
+# Check if characters of the given string can be rearranged to form a palindrome.
+# Counter is faster for long strings and non-Counter is faster for short strings.
+
+
+def can_string_be_rearranged_as_palindrome_counter(input_str: str = "",) -> bool:
+    """
+    A Palindrome is a String that reads the same forward as it does backwards.
+    Examples of Palindromes mom, dad, malayalam
+    >>> can_string_be_rearranged_as_palindrome_counter("Momo")
+    True
+    >>> can_string_be_rearranged_as_palindrome_counter("Mother")
+    False
+    >>> can_string_be_rearranged_as_palindrome_counter("Father")
+    False
+    >>> can_string_be_rearranged_as_palindrome_counter("A man a plan a canal Panama")
+    True
+    """
+    return sum(c % 2 for c in Counter(input_str.replace(" ", "").lower()).values()) < 2
+
+
+def can_string_be_rearranged_as_palindrome(input_str: str = "") -> bool:
+    """
+    A Palindrome is a String that reads the same forward as it does backwards.
+    Examples of Palindromes mom, dad, malayalam
+    >>> can_string_be_rearranged_as_palindrome("Momo")
+    True
+    >>> can_string_be_rearranged_as_palindrome("Mother")
+    False
+    >>> can_string_be_rearranged_as_palindrome("Father")
+    False
+    >>> can_string_be_rearranged_as_palindrome_counter("A man a plan a canal Panama")
+    True
+    """
+    if len(input_str) == 0:
+        return True
+    lower_case_input_str = input_str.replace(" ", "").lower()
+    # character_freq_dict: Stores the frequency of every character in the input string
+    character_freq_dict = {}
+
+    for character in lower_case_input_str:
+        character_freq_dict[character] = character_freq_dict.get(character, 0) + 1
+    """
+    Above line of code is equivalent to:
+    1) Getting the frequency of current character till previous index
+    >>> character_freq =  character_freq_dict.get(character, 0)
+    2) Incrementing the frequency of current character by 1
+    >>> character_freq = character_freq + 1
+    3) Updating the frequency of current character
+    >>> character_freq_dict[character] = character_freq
+    """
+    """
+    OBSERVATIONS:
+    Even length palindrome
+    -> Every character appears even no.of times.
+    Odd length palindrome
+    -> Every character appears even no.of times except for one character.
+    LOGIC:
+    Step 1: We'll count number of characters that appear odd number of times i.e oddChar
+    Step 2:If we find more than 1 character that appears odd number of times,
+    It is not possible to rearrange as a palindrome
+    """
+    oddChar = 0
+
+    for character_count in character_freq_dict.values():
+        if character_count % 2:
+            oddChar += 1
+    if oddChar > 1:
+        return False
+    return True
+
+
+def benchmark(input_str: str = "") -> None:
+    """
+    Benchmark code for comparing above 2 functions
+    """
+    print("\nFor string = ", input_str, ":")
+    print(
+        "> can_string_be_rearranged_as_palindrome_counter()",
+        "\tans =",
+        can_string_be_rearranged_as_palindrome_counter(input_str),
+        "\ttime =",
+        timeit(
+            "z.can_string_be_rearranged_as_palindrome_counter(z.check_str)",
+            setup="import __main__ as z",
+        ),
+        "seconds",
+    )
+    print(
+        "> can_string_be_rearranged_as_palindrome()",
+        "\tans =",
+        can_string_be_rearranged_as_palindrome(input_str),
+        "\ttime =",
+        timeit(
+            "z.can_string_be_rearranged_as_palindrome(z.check_str)",
+            setup="import __main__ as z",
+        ),
+        "seconds",
+    )
+
+
+if __name__ == "__main__":
+    check_str = input(
+        "Enter string to determine if it can be rearranged as a palindrome or not: "
+    ).strip()
+    benchmark(check_str)
+    status = can_string_be_rearranged_as_palindrome_counter(check_str)
+    print(f"{check_str} can {'' if status else 'not '}be rearranged as a palindrome")