minimum-window-substring py3 brute force

0076_minimum-window-substring/README.md

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+Given two strings `s` and `t` of lengths `m` and `n` respectively, return _the **minimum window substring** of_ `s` _such that every character in_ `t` _(**including duplicates**) is included in the window. If there is no such substring__, return the empty string_ `""`_._
+
+The testcases will be generated such that the answer is **unique**.
+
+A **substring** is a contiguous sequence of characters within the string.
+
+**Example 1:**
+
+    Input: s = "ADOBECODEBANC", t = "ABC"
+    Output: "BANC"
+    Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.
+    
+
+**Example 2:**
+
+    Input: s = "a", t = "a"
+    Output: "a"
+    Explanation: The entire string s is the minimum window.
+    
+
+**Example 3:**
+
+    Input: s = "a", t = "aa"
+    Output: ""
+    Explanation: Both 'a's from t must be included in the window.
+    Since the largest window of s only has one 'a', return empty string.
+    
+
+**Constraints:**
+
+*   `m == s.length`
+*   `n == t.length`
+*   `1 <= m, n <= 105`
+*   `s` and `t` consist of uppercase and lowercase English letters.
+
+**Follow up:** Could you find an algorithm that runs in `O(m + n)` time?
+
+https://leetcode.com/problems/minimum-window-substring

0076_minimum-window-substring/python3/brute_force_tle.py

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+from collections import Counter
+
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        tcounts = Counter(t)
+        
+        # If chars of t is not included in the entire
+        # s string, then we can just return right away
+        if not tcounts <= Counter(s):
+            return ""
+        
+        tlen = len(t)
+        min_window = ""
+        
+        for start in range(len(s)):
+            end = start + tlen - 1
+            
+            # We can all the substrings starting from `start` which is every
+            # letter of s
+            while end < len(s):
+                curr_window = s[start:end + 1]
+                curr_window_counts = Counter(curr_window)
+                
+                # <= is overloaded for Counters to mean inclusion
+                #
+                # i.e, counter_a -is-included-in- counter_b
+                #
+                if tcounts <= curr_window_counts:
+                    # If no min_window has been computed yet or if current
+                    # window is shorter, then we assign that as the new
+                    # min_window
+                    if min_window == "" or len(curr_window) < len(min_window):
+                        min_window = curr_window
+                        
+                    # Since at this point, if we keep iterating in the inner loop,
+                    # we'll only get longer substrings so we just break and proceed
+                    # to find the substrings starting from the next letter (start + 1)
+                    break
+
+                end += 1
+        
+        return min_window