update

Author: Aaron Liu

Meta/LC1249 Minimum Removeto Make Valid Parentheses.py

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+'''
+Given a string s of '(' , ')' and lowercase English characters.
+
+Your task is to remove the minimum number of parentheses ( '(' or ')', in any positions ) so that the resulting parentheses string is valid and return any valid string.
+
+Formally, a parentheses string is valid if and only if:
+
+It is the empty string, contains only lowercase characters, or
+It can be written as AB (A concatenated with B), where A and B are valid strings, or
+It can be written as (A), where A is a valid string.
+
+Example 1:
+
+Input: s = "lee(t(c)o)de)"
+Output: "lee(t(c)o)de"
+Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.
+'''
+
+'''
+solution 1: two pass
+只有左右括号匹配 才算balanced
+first pass to count number of '('
+second pass:
+遍历过程中 记录左括号个数left. 当前字符char 分类讨论:
+1. char == ')'. 
+  - if left>0 -> 左边有对应的匹配. char加入ret, left--.
+  - else 左边没有‘（’与之匹配. right-- 避免右边的'('跟自己匹配.
+2. char == '('.
+  - 
+
+'''
+
+def minRemoveToMakeValid0(s: str) -> str:
+    if not s:
+        return ""
+    
+    right = 0
+    for char in s:
+        if char == ')':
+            right += 1
+    
+    ret = []
+    left = 0
+    for char in s:
+        if char == ')':
+            if left > 0:
+                ret.append(char)
+                left -= 1 # 与当前')'配对
+            else:
+                right -= 1 # 防止与后面的'('配对
+        elif char == '(':
+            if right > 0:
+                ret.append(char)
+                left += 1 # 必须+=1
+                right -= 1 # 与当前的配对
+        else:
+            ret.append(char)
+    return "".join(ret)
+
+
+    # Pass 1: Remove all invalid ")"
+    first_pass_chars = []
+    balance = 0
+    open_seen = 0
+    for c in s:
+        if c == "(":
+            balance += 1
+            open_seen += 1
+        if c == ")":
+            if balance == 0:
+                continue
+            balance -= 1
+        first_pass_chars.append(c)
+
+    # Pass 2: Remove the rightmost "("
+    result = []
+    open_to_keep = open_seen - balance
+    for c in first_pass_chars:
+        if c == "(":
+            open_to_keep -= 1
+            if open_to_keep < 0:
+                continue
+        result.append(c)
+
+    return "".join(result)
+
+
+def minRemoveToMakeValid1(self, s: str) -> str:
+    # Pass 1: Remove all invalid ")"
+    first_pass_chars = []
+    balance = 0
+    open_seen = 0
+    for c in s:
+        if c == "(":
+            balance += 1
+            open_seen += 1
+        if c == ")":
+            if balance == 0:
+                continue
+            balance -= 1
+        first_pass_chars.append(c)
+
+    # Pass 2: Remove the rightmost "("
+    result = []
+    open_to_keep = open_seen - balance
+    for c in first_pass_chars:
+        if c == "(":
+            open_to_keep -= 1
+            if open_to_keep < 0:
+                continue
+        result.append(c)
+
+    return "".join(result)