Create 076_minimum_window_substring.md

Author: Joseph Luce

leetcode/hard/076_minimum_window_substring.md

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+# 76. Minimum Window Substring
+
+## Sliding Window with Dictionary Solution
+- Runtime: O(N)
+- Space: O(N)
+- N = Number of characters in S
+
+Lets focus on how to figure out if a sub-string has all characters of T.
+You can do this with a dictionary counter, keeping occurances of each character of T.
+To avoid checking if each character in the dictionary is less than or equal to zero occurances, we can keep a separate variable as the remmaining characters needed to be found.
+
+Next is the idea of a sliding window, if we iterate from left to right, we can find all the sub-strings containing T using the above soluiton.
+Once that sub-string is found, then its the matter of decrementing the left most character of the sub-string until we need to find another character.
+With that, we have to decrement the dictionary and the number of remaining characters accordingly.
+
+```
+from collections import Counter
+
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        char_counter = Counter(t)
+        left_idx, n_chars_needed = 0, len(t)
+        result = (-1, -1) # left and right result indexes
+        
+        for right_idx, ch in enumerate(s):
+            if ch in char_counter:
+                char_counter[ch] -= 1
+                if char_counter[ch] >= 0:
+                    n_chars_needed -= 1
+            
+            while n_chars_needed == 0:
+                if result[0] == -1 or result[1]-result[0] > right_idx-left_idx:
+                    result = (left_idx, right_idx)
+                left_ch = s[left_idx]
+                if left_ch in char_counter:
+                    char_counter[left_ch] += 1
+                    if char_counter[left_ch] == 1:
+                        n_chars_needed += 1
+                left_idx += 1
+            
+        return s[result[0]:result[1]+1] if result[0] != -1 else ''
+```