Solution: Minimum Remove to Make Valid Parentheses

Leetcode Solutions (161 Part Series)

1 Solution: Next Permutation
2 Solution: Trim a Binary Search Tree
… 157 more parts…
3 Leetcode Solutions Index
4 Solution: Minimize Deviation in Array
5 Solution: Vertical Order Traversal of a Binary Tree
6 Solution: Count Ways to Make Array With Product
7 Solution: Smallest String With A Given Numeric Value
8 Solution: Linked List Cycle
9 Solution: Path With Minimum Effort
10 Solution: Concatenation of Consecutive Binary Numbers
11 Solution: Minimum Operations to Make a Subsequence
12 Solution: Longest Harmonious Subsequence
13 Solution: Simplify Path
14 Solution: Building Boxes
15 Solution: Decode XORed Permutation
16 Solution: Binary Tree Right Side View
17 Solution: Find Kth Largest XOR Coordinate Value
18 Solution: Change Minimum Characters to Satisfy One of Three Conditions
19 Solution: Shortest Distance to a Character
20 Solution: Peeking Iterator
21 Solution: Convert BST to Greater Tree
22 Solution: Copy List with Random Pointer
23 Solution: Valid Anagram
24 Solution: Number of Steps to Reduce a Number to Zero
25 Solution: Shortest Path in Binary Matrix
26 Solution: Is Graph Bipartite?
27 Solution: Maximum Score From Removing Substrings (ver. 1)
28 Solution: Maximum Score From Removing Substrings (ver. 2)
29 Solution: Sort the Matrix Diagonally
30 Solution: The K Weakest Rows in a Matrix (ver. 1)
31 Solution: The K Weakest Rows in a Matrix (ver. 2)
32 Solution: Letter Case Permutation
33 Solution: Container With Most Water
34 Solution: Arithmetic Slices
35 Solution: Minimum Remove to Make Valid Parentheses
36 Solution: Roman to Integer
37 Solution: Broken Calculator
38 Solution: Find the Most Competitive Subsequence
39 Solution: Longest Word in Dictionary through Deleting
40 Solution: Search a 2D Matrix II
41 Solution: Score of Parentheses
42 Solution: Shortest Unsorted Continuous Subarray
43 Solution: Validate Stack Sequences
44 Solution: Divide Two Integers (ver. 1)
45 Solution: Divide Two Integers (ver. 2)
46 Solution: Maximum Frequency Stack
47 Solution: Distribute Candies
48 Solution: Set Mismatch (ver. 1)
49 Solution: Set Mismatch (ver. 2)
50 Solution: Missing Number
51 Solution: Intersection of Two Linked Lists
52 Solution: Average of Levels in Binary Tree
53 Solution: Short Encoding of Words (ver. 1)
54 Solution: Design HashMap (ver. 1)
55 Solution: Short Encoding of Words (ver. 2)
56 Solution: Design HashMap (ver. 2)
57 Solution: Remove Palindromic Subsequences
58 Solution: Add One Row to Tree
59 Solution: Integer to Roman
60 Solution: Coin Change
61 Solution: Check If a String Contains All Binary Codes of Size K
62 Solution: Binary Trees With Factors
63 Solution: Swapping Nodes in a Linked List
64 Solution: Encode and Decode TinyURL
65 Solution: Best Time to Buy and Sell Stock with Transaction Fee
66 Solution: Generate Random Point in a Circle
67 Solution: Wiggle Subsequence
68 Solution: Keys and Rooms
69 Solution: Design Underground System
70 Solution: Reordered Power of 2
71 Solution: Vowel Spellchecker
72 Solution: 3Sum With Multiplicity
73 Solution: Advantage Shuffle
74 Solution: Pacific Atlantic Water Flow
75 Solution: Word Subsets
76 Solution: Palindromic Substrings
77 Solution: Reconstruct Original Digits from English
78 Solution: Flip Binary Tree To Match Preorder Traversal
79 Solution: Russian Doll Envelopes
80 Solution: Stamping The Sequence
81 Solution: Palindrome Linked List
82 Solution: Ones and Zeroes
83 Solution: Longest Valid Parentheses
84 Solution: Design Circular Queue
85 Solution: Global and Local Inversions
86 Solution: Minimum Operations to Make Array Equal
87 Solution: Determine if String Halves Are Alike
88 Solution: Letter Combinations of a Phone Number
89 Solution: Verifying an Alien Dictionary
90 Solution: Longest Increasing Path in a Matrix
91 Solution: Deepest Leaves Sum
92 Solution: Beautiful Arrangement II
93 Solution: Flatten Nested List Iterator
94 Solution: Partition List
95 Solution: Fibonacci Number
96 Solution: Remove All Adjacent Duplicates in String II
97 Solution: Number of Submatrices That Sum to Target
98 Solution: Remove Nth Node From End of List
99 Solution: Combination Sum IV
100 Solution: N-ary Tree Preorder Traversal
101 Solution: Triangle
102 Solution: Brick Wall
103 Solution: Count Binary Substrings
104 Solution: Critical Connections in a Network
105 Solution: Rotate Image
106 Solution: Furthest Building You Can Reach
107 Solution: Power of Three
108 Solution: Unique Paths II
109 Solution: Find First and Last Position of Element in Sorted Array
110 Solution: Powerful Integers
111 Solution: Prefix and Suffix Search
112 Solution: Course Schedule III
113 Solution: Running Sum of 1d Array
114 Solution: Non-decreasing Array
115 Solution: Jump Game II
116 Solution: Convert Sorted List to Binary Search Tree
117 Solution: Delete Operation for Two Strings
118 Solution: Super Palindromes
119 Solution: Construct Target Array With Multiple Sums
120 Solution: Count Primes
121 Solution: Maximum Points You Can Obtain from Cards
122 Solution: Range Sum Query 2D – Immutable
123 Solution: Ambiguous Coordinates
124 Solution: Flatten Binary Tree to Linked List
125 Solution: Valid Number
126 Solution: Binary Tree Cameras
127 Solution: Longest String Chain
128 Solution: Find Duplicate File in System
129 Solution: Minimum Moves to Equal Array Elements II
130 Solution: Binary Tree Level Order Traversal
131 Solution: Find and Replace Pattern
132 Solution: N-Queens
133 Solution: To Lower Case
134 Solution: Evaluate Reverse Polish Notation
135 Solution: Partitioning Into Minimum Number Of Deci-Binary Numbers
136 Solution: Maximum Product of Word Lengths
137 Solution: Maximum Erasure Value
138 Solution: N-Queens II
139 Solution: Maximum Gap
140 Solution: Search Suggestions System
141 Solution: Max Area of Island
142 Solution: Interleaving String
143 Solution: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts
144 Solution: Open the Lock
145 Solution: Maximum Performance of a Team
146 Solution: Longest Consecutive Sequence
147 Solution: Min Cost Climbing Stairs
148 Solution: Construct Binary Tree from Preorder and Inorder Traversal
149 Solution: Jump Game VI
150 Solution: My Calendar I
151 Solution: Stone Game VII
152 Solution: Minimum Number of Refueling Stops
153 Solution: Palindrome Pairs
154 Solution: Maximum Units on a Truck
155 Solution: Matchsticks to Square
156 Solution: Generate Parentheses
157 Solution: Number of Subarrays with Bounded Maximum
158 Solution: Swim in Rising Water
159 Solution: Pascal’s Triangle
160 Solution: Out of Boundary Paths
161 Solution: Redundant Connection

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode’s forums.

Leetcode Problem #1249 (Medium): Minimum Remove to Make Valid Parentheses

Description:

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.

Examples:

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.

Example 2:
Input:	s = “a)b(c)d”
Output:	“ab(c)d”

Example 3:
Input:	s = “))((“
Output:	“”
Explanation:	An empty string is also valid.

Example 4:
Input:	s = “(a(b(c)d)”
Output:	“a(b(c)d)”

Constraints:

1 <= s.length <= 10^5
s[i] is one of '(' , ')' and lowercase English letters.

Idea:

Valid parentheses follow the LIFO method (last in, first out), so we should automatically be thinking of some kind of stack solution.

To check for valid parentheses, you push any “(“ onto stack, then pop off the top stack element every time you find a matching “)”. If you find a “)” when stack is empty, that “)” must be invalid. At the end of S, any leftover “(“‘s left in stack must be invalid, as well. Since we’ll want to remove those “(“‘s by index at the end, stack should contain said indexes, rather than just the “(“.

Now that we’ve identified all the invalid parentheses, that leaves us with the problem of removing them from S. We could perform a lot of string slices and copies, but those are typically very slow and memory intensive, so we should probably find a data type that can be directly modified by index access and use that as an intermediary.

The most effective method varies by language, so I’ll discuss those in the Implementation section.

Then we can make our removals and re-form and return our answer.

Implementation:

Javascript has basic arrays, Python has lists, and Java has char arrays that will perform the job of a more flexible data type for this problem. C++ alone of the four languages has mutable strings, so we can just leave S as is.

While Java has stack/deque/list structures, they’re not always terribly efficient, so we can just use a more basic int[] with a length fixed to the size of S, along with an index variable (stIx).

Javascript conveniently allows us to directly delete an array element without screwing up our iteration, so we can use that on the initial pass for invalid “(“‘s. Python can’t do that, but we can easily replace each character we want to delete with an empty string, which effectively does the same thing once the string has been joined again.

Java and C++ won’t allow us to replace characters with empty strings, so we can just mark those characters with a character mask for later removal.

On the second pass through Javascript and Python can just repeat the same method while going through the remaining stack. Python is very fast with its appends and pops, so we can use that to our advantage.

For Java and C++, things are more difficult. We can’t change the length of the intermediary, but we can alter its contents by index assignment. That means we can use a two-pointer approach to rewrite the beginning portion of the intermediary before ultimately returning a subsection of it.

Since we want to iterate through stack in opposite order (FIFO) this time, we can just tag a -1 onto the end of the stack to avoid issues with going out-of-bounds, and then use stIx starting at 0.

Then, for every iteration, j will increment, but i will only increment if it’s not a character we want to remove (either by matching the character mask or the next stack entry), and we’ll overwrite the intermediary at i with j‘s value.

At the end, the substring between 0 and i will represent the “squeezed” string with all invalid parentheses removed, so we should return it.

Javascript Code:

var minRemoveToMakeValid = function(S) {
    S = S.split("")
    let len = S.length, stack = []
    for (let i = 0, c = S[0]; i < len; c = S[++i])
        if (c === ")")
            if (stack.length) stack.pop()
            else delete S[i]
        else if (c === "(") stack.push(i)
    for (let i = 0; i < stack.length; i++)
        delete S[stack[i]]
    return S.join("")
};

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Python Code:

class Solution:
    def minRemoveToMakeValid(self, S: str) -> str:
        S, stack = list(S), []
        for i, c in enumerate(S):
            if c == ")":
                if stack: stack.pop()
                else: S[i] = ""
            elif c == "(": stack.append(i)
        for i in stack: S[i] = ""
        return "".join(S)

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Java Code:

class Solution {
    public String minRemoveToMakeValid(String S) {
        char[] ans = S.toCharArray();
        int len = S.length(), stIx = 0, i = 0, j = 0;
        int[] stack = new int[len+1];
        for (; i < len; i++)
            if (ans[i] == ')')
                if (stIx > 0) stIx--;
                else ans[i] = '_';
            else if (ans[i] == '(') stack[stIx++] = i;
        for (i = 0, stack[stIx] = -1, stIx = 0; j < len; j++)
            if (j == stack[stIx]) stIx++;
            else if (ans[j] != '_') ans[i++] = ans[j];
        return new String(ans, 0, i);
    }
}

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C++ Code:

class Solution {
public:
    string minRemoveToMakeValid(string S) {
        int len = S.size(), i = 0, j = 0, stIx = 0;
        vector<int> stack;
        for (; i < len; i++)
            if (S[i] == ')')
                if (stack.size() > 0) stack.pop_back();
                else S[i] = '_';
            else if (S[i] == '(') stack.push_back(i);
        stack.push_back(-1);
        for (i = 0; j < len; j++)
            if (j == stack[stIx]) stIx++;
            else if (S[j] != '_') S[i++] = S[j];
        return S.substr(0, i);
    }
};

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Leetcode Solutions (161 Part Series)

原文链接：Solution: Minimum Remove to Make Valid Parentheses

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