Add 'Minimum Window Substring'

Author: nickolashkraus

README.md

@@ -64,6 +64,8 @@ A collection of LeetCode solutions
 
 [Minimum Size Subarray Sum](./src/minimum_size_subarray_sum.py)
 
+[Minimum Window Substring](./src/minimum_window_substring.py)
+
 [Permutation in String](./src/permutation_in_string.py)
 
 [Product of Array Except Self](./src/product_of_array_except_self.py)

src/minimum_window_substring.py

@@ -0,0 +1,53 @@
+"""
+76. Minimum Window Substring
+
+https://leetcode.com/problems/minimum-window-substring
+
+NOTES
+  * Use a sliding window.
+
+Not sure why this is categorized as "hard". Maybe I've just done so many of
+these problems, the solution is intuitive. Frequency map, sliding window, boom,
+done.
+
+Spoke too soon... Spent over thirty minutes with the tricky logic around
+remaining characters. The key is to only decrement/increment `remaining` if the
+count is greater than 0. This ensures the substring possesses all required
+characters without erroneously counting extra occurrences of characters already
+satisfied in the substring.
+"""
+
+import sys
+from collections import Counter
+
+
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        left, right, start, end, minimum_size = 0, 0, 0, 0, sys.maxsize
+
+        # Create a frequency map. Additionally, keep track of the count of
+        # remaining characters needed for the substring to be valid (i.e.,
+        # every character in `t` is included in the substring).
+        counter: Counter[str] = Counter(t)
+        remaining = len(t)
+
+        while right < len(s):
+            if s[right] in counter:
+                #  Only decrement `remaining` if count is greater than 0.
+                if counter[s[right]] > 0:
+                    remaining -= 1
+                counter[s[right]] -= 1
+            while left <= right and remaining == 0:
+                current_size = right - left + 1
+                if current_size < minimum_size:
+                    minimum_size = current_size
+                    start, end = left, right
+                if s[left] in counter:
+                    counter[s[left]] += 1
+                    #  Only increment `remaining` if count is greater than 0.
+                    if counter[s[left]] > 0:
+                        remaining += 1
+                left += 1
+            right += 1
+
+        return s[start:end + 1] if minimum_size < sys.maxsize else ""

tests/test_minimum_window_substring.py

@@ -0,0 +1,27 @@
+"""
+76. Minimum Window Substring
+
+https://leetcode.com/problems/minimum-window-substring
+"""
+
+from unittest import TestCase
+
+from src.minimum_window_substring import Solution
+
+
+class TestSolution(TestCase):
+    def test_1(self):
+        exp = "BANC"
+        assert Solution().minWindow("ADOBECODEBANC", "ABC") == exp
+
+    def test_2(self):
+        exp = "a"
+        assert Solution().minWindow("a", "a") == exp
+
+    def test_3(self):
+        exp = ""
+        assert Solution().minWindow("a", "aa") == exp
+
+    def test_4(self):
+        exp = "ba"
+        assert Solution().minWindow("bba", "ab") == exp