solve Find the Difference

Author: saubhik

problems/find_the_difference.py

@@ -0,0 +1,44 @@
+class Solution:
+    # Time Complexity: O(n), where n is the length of t or s.
+    # Space Complexity: O(1)
+    def findTheDifference(self, s: str, t: str) -> str:
+        counts = [0] * 26
+        for ch in s:
+            counts[ord(ch) - ord("a")] += 1
+        for ch in t:
+            if counts[ord(ch) - ord("a")] == 0:
+                return ch
+            counts[ord(ch) - ord("a")] -= 1
+        return ""
+
+
+class OfficialSolution:
+    """
+    First approach is sorting and comparing character by character.
+    Time Complexity: O(nlgn).
+    Space Complexity: O(1).
+
+    Second approach is using hash map of 26 keys, which is same as my solution above.
+    Time Complexity: O(n).
+    Space Complexity: O(1).
+
+    Third approach is bit manipulation. Please check the doc string of the function
+    below.
+    """
+
+    def findTheDifferenceApproach3(self, s: str, t: str) -> str:
+        """
+        To use bitwise XOR operation on all elements. XOR would help to eliminate the
+        alike and only leave the odd duckling.
+        Thus, the left over bits after XORing all the characters from string s and
+        string t would be from the extra character of string t.
+
+        Time Complexity: O(n), where N is the length of s or t.
+        Space Complexity: O(1).
+        """
+        xored = 0
+        for ch in s:
+            xored ^= ord(ch)
+        for ch in t:
+            xored ^= ord(ch)
+        return chr(xored)

tests/test_find_the_difference .py

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+import unittest
+
+from find_the_difference import Solution, OfficialSolution
+
+
+class TestFindTheDifference(unittest.TestCase):
+    def test_example_1(self):
+        assert Solution().findTheDifference(s="abcd", t="abcde") == "e"
+        assert OfficialSolution().findTheDifferenceApproach3(s="abcd", t="abcde") == "e"