Merge d448a4e7e5 into f3f32ae3ca

updates:
- [github.com/astral-sh/ruff-pre-commit: v0.7.3 → v0.7.4](https://github.com/astral-sh/ruff-pre-commit/compare/v0.7.3...v0.7.4)

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>

.pre-commit-config.yaml

@@ -16,7 +16,7 @@ repos:
       - id: auto-walrus
 
   - repo: https://github.com/astral-sh/ruff-pre-commit
-    rev: v0.7.2
+    rev: v0.7.4
     hooks:
       - id: ruff
       - id: ruff-format
@@ -42,7 +42,7 @@ repos:
         pass_filenames: false
 
   - repo: https://github.com/abravalheri/validate-pyproject
-    rev: v0.22
+    rev: v0.23
     hooks:
       - id: validate-pyproject
 

data_structures/arrays/sudoku_solver.py

@@ -172,7 +172,7 @@ def solved(values):
 
 def from_file(filename, sep="\n"):
     "Parse a file into a list of strings, separated by sep."
-    return open(filename).read().strip().split(sep)  # noqa: SIM115
+    return open(filename).read().strip().split(sep)
 
 
 def random_puzzle(assignments=17):

scripts/validate_filenames.py

@@ -1,4 +1,4 @@
-#!/usr/bin/env python3
+#!python
 import os
 
 try:

strings/min_window_substring.py

@@ -0,0 +1,89 @@
+def min_window(search_str: str, target_letters: str) -> str:
+    """
+    Given a string to search, and another string of target char_dict,
+    return the smallest substring of the search string that contains
+    all target char_dict.
+
+    This is somewhat modified from my solution to the problem
+    "Minimum Window Substring" on leetcode.
+    https://leetcode.com/problems/minimum-window-substring/description/
+
+    >>> min_window("Hello World", "lWl")
+    'llo W'
+    >>> min_window("Hello World", "f")
+    ''
+
+    This solution uses a sliding window, alternating between shifting
+    the end of the window right until all target char_dict are contained
+    in the window, and shifting the start of the window right until the
+    window no longer contains every target character.
+
+    Time complexity: O(target_count + search_len) ->
+        The algorithm checks a dictionary at most twice for each character
+        in search_str.
+
+    Space complexity: O(search_len) ->
+        The primary contributor to additional space is the building of a
+        dictionary using the search string.
+    """
+
+    target_count = len(target_letters)
+    search_len = len(search_str)
+
+    # Return if not possible due to string lengths.
+    if search_len < target_count:
+        return ""
+
+    # Build dictionary with counts for each letter in target_letters
+    char_dict = {}
+    for ch in target_letters:
+        if ch not in char_dict:
+            char_dict[ch] = 1
+        else:
+            char_dict[ch] += 1
+
+    # Initialize window
+    window_start = 0
+    window_end = 0
+
+    exists = False
+    min_window_len = search_len + 1
+
+    # Start sliding window algorithm
+    while window_end < search_len:
+        # Slide window end right until all search characters are contained
+        while target_count > 0 and window_end < search_len:
+            cur = search_str[window_end]
+            if cur in char_dict:
+                char_dict[cur] -= 1
+                if char_dict[cur] >= 0:
+                    target_count -= 1
+            window_end += 1
+        temp = window_end - window_start
+
+        # Check if window is the smallest found so far
+        if target_count == 0 and temp < min_window_len:
+            min_window = [window_start, window_end]
+            exists = True
+            min_window_len = temp
+
+        # Slide window start right until a search character exits the window
+        while target_count == 0 and window_start < window_end:
+            cur = search_str[window_start]
+            window_start += 1
+            if cur in char_dict:
+                char_dict[cur] += 1
+                if char_dict[cur] > 0:
+                    break
+        temp = window_end - window_start + 1
+
+        # Check if window is the smallest found so far
+        if temp < min_window_len and target_count == 0:
+            min_window = [window_start - 1, window_end]
+            min_window_len = temp
+        target_count = 1
+
+    if exists:
+        return search_str[min_window[0] : min_window[1]]
+    else:
+        return ""