feat(strings): check permutations with map/dict

Author: BrianLusina

pystrings/permutation/__init__.py

@@ -1,4 +1,7 @@
-def check_permutation(input_str_1: str, input_str_2: str) -> bool:
+from typing import Dict
+
+
+def check_permutation_with_sorting(input_str_1: str, input_str_2: str) -> bool:
     """
     Check if two strings are permutations of each other.
 
@@ -32,3 +35,52 @@ def check_permutation(input_str_1: str, input_str_2: str) -> bool:
             return False
 
     return True
+
+def check_permutation_with_map(input_str_1: str, input_str_2: str) -> bool:
+    """
+    Check if two strings are permutations of each other.
+
+    This function determines whether two input strings are permutations of one
+    another by comparing the frequency of characters in each string. It considers
+    case sensitivity and whitespace as significant when evaluating the two strings.
+
+    Complexity:
+    Time: O(n) as iteration is handled on each string where n is the number of characters in each string
+    Space: O(n) as a dictionary is used to store characters for lookups
+
+    Args:
+        input_str_1 (str): The first input string to compare.
+        input_str_2 (str): The second input string to compare.
+
+    Returns:
+    bool
+        True if the two strings are permutations of each other, False otherwise.
+    """
+    # strings of different lengths can not be permutations of each other
+    if len(input_str_1) != len(input_str_2):
+        return False
+
+    input_str_1_lower = input_str_1.lower()
+    input_str_2_lower = input_str_2.lower()
+
+    char_count: Dict[str, int] = dict()
+
+    # Now, if input_str_1 and input_str_2 are permutations of each other, the two for loops will counterbalance each
+    # other and the values of all the keys in char_count should be 0.
+    # We find this out at the end where we evaluate count == 0 for all the keys in char_count. Using the all function,
+    # we combine the evaluation results from all the iterations and return it from the function. The all() function
+    # returns True if all items in an iterable are True. Otherwise, it returns False.
+
+    for char in input_str_1_lower:
+        if char in char_count:
+            char_count[char] += 1
+        else:
+            char_count[char] = 1
+
+    for char in input_str_2_lower:
+        if char in char_count:
+            char_count[char] -= 1
+        else:
+            return False
+
+    return all(count == 0 for count in char_count.values())

pystrings/permutation/test_check_permutation.py

@@ -1,20 +1,36 @@
 import unittest
-from . import check_permutation
+from . import check_permutation_with_sorting, check_permutation_with_map
 
 
-class CheckPermutationTestCases(unittest.TestCase):
+class CheckPermutationWithSortingTestCases(unittest.TestCase):
     def test_1(self):
         """should return True for 'google' and 'ooggle'"""
         word_1 = "google"
         word_2 = "ooggle"
-        actual = check_permutation(word_1, word_2)
+        actual = check_permutation_with_sorting(word_1, word_2)
         self.assertTrue(actual)
 
     def test_2(self):
         """should return False for 'not' and 'top'"""
         word_1 = "not"
         word_2 = "top"
-        actual = check_permutation(word_1, word_2)
+        actual = check_permutation_with_sorting(word_1, word_2)
+        self.assertFalse(actual)
+
+
+class CheckPermutationWithMapTestCases(unittest.TestCase):
+    def test_1(self):
+        """should return True for 'google' and 'ooggle'"""
+        word_1 = "google"
+        word_2 = "ooggle"
+        actual = check_permutation_with_map(word_1, word_2)
+        self.assertTrue(actual)
+
+    def test_2(self):
+        """should return False for 'not' and 'top'"""
+        word_1 = "not"
+        word_2 = "top"
+        actual = check_permutation_with_map(word_1, word_2)
         self.assertFalse(actual)
 
 if __name__ == '__main__':