Update 76_minimum_window_substring.md

Author: Joseph Luce

leetcode/hard/76_minimum_window_substring.md

@@ -1,7 +1,7 @@
 # 76. Minimum Window Substring
 
 ## Two Pointer with Map Solution
-- Runtime: O(S * T)
+- Runtime: O(S * T) but O(S * (S+T)) due to string slicing
 - Space: O(S)
 - S = Number of characters in string S
 - T = Number of unique characters in string T
@@ -21,13 +21,14 @@ If so, we can then try pruning with the left pointer all the way to the right po
 
 You may think that the run-time for this is exponential, especially when we are checking the two dictionaries.
 However, don't be mistaken, the comparison is actually a constant T run-time, it doesn't change based on S, but rather on T.
-There is one slight problem, python's implementation of string concatention is actually O(N).
+There is one slight problem, python's implementation of string concatention is actually **O(N)**.
 When the question wants the actual substring and not a count, even using a deque will not solve this problem.
-So this implementation is technically O(S * (S+T)) due to python, but in other languages
+So this implementation is technically **O(S * (S+T))** due to python.
 
 When implementing these type of two pointer questions.
 I recommend to avoid using indexes as much as possible and use iterators.
 It is very easy to get a one off error doing these and within a 30 minute timeframe, it is very risky.
+Just talk about using indexes instead and you will be fine.
 
 ```
 from collections import defaultdict
@@ -38,18 +39,18 @@ class Solution:
         all_ch_counts = Counter(t)
         ch_to_n_counts = defaultdict(int)
         str_builder, min_substr, found = '', s, False
-        for curr_right in s:
-            ch_to_n_counts[curr_right] += 1
-            str_builder += curr_right
+        for right_ch in s:
+            ch_to_n_counts[right_ch] += 1
+            str_builder += right_ch
             if chars_occur_ge(ch_to_n_counts, all_ch_counts):
-                for curr_left in str_builder:
+                for left_ch in str_builder:
                     if chars_occur_ge(ch_to_n_counts, all_ch_counts):
                         found = True
                         if len(str_builder) < len(min_substr):
                             min_substr = str_builder
                     else:
                         break
-                    ch_to_n_counts[curr_left] -= 1
+                    ch_to_n_counts[left_ch] -= 1
                     str_builder = str_builder[1:]
         return min_substr if found else ''