Solution: Open the Lock

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 #752 (Medium): Open the Lock

Description:

(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'. The wheels can rotate freely and wrap around: for example we can turn '9' to be '0', or '0' to be '9'. Each move consists of turning one wheel one slot.

The lock initially starts at '0000', a string representing the state of the 4 wheels.

You are given a list of deadends dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it.

Given a target representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible.

Examples:

Example 1:

Input: deadends = [“0201″,”0101″,”0102″,”1212″,”2002”], target = “0202”

Output: 6

Explanation: A sequence of valid moves would be “0000” -> “1000” -> “1100” -> “1200” -> “1201” -> “1202” -> “0202”.
Note that a sequence like “0000” -> “0001” -> “0002” -> “0102” -> “0202” would be invalid, because the wheels of the lock become stuck after the display becomes the dead end “0102”.

Example 2:

Input: deadends = [“8888”], target = “0009”

Output: 1

Explanation: We can turn the last wheel in reverse to move from “0000” -> “0009”.

Example 3:

Input: deadends = [“8887″,”8889″,”8878″,”8898″,”8788″,”8988″,”7888″,”9888”], target = “8888”

Output: -1

Explanation: We can’t reach the target without getting stuck.

Example 4:

Input: deadends = [“0000”], target = “8888”

Output: -1

Example 1:
Input:	deadends = [“0201″,”0101″,”0102″,”1212″,”2002”], target = “0202”
Output:	6
Explanation:	A sequence of valid moves would be “0000” -> “1000” -> “1100” -> “1200” -> “1201” -> “1202” -> “0202”. Note that a sequence like “0000” -> “0001” -> “0002” -> “0102” -> “0202” would be invalid, because the wheels of the lock become stuck after the display becomes the dead end “0102”.

Example 2:
Input:	deadends = [“8888”], target = “0009”
Output:	1
Explanation:	We can turn the last wheel in reverse to move from “0000” -> “0009”.

Example 3:
Input:	deadends = [“8887″,”8889″,”8878″,”8898″,”8788″,”8988″,”7888″,”9888”], target = “8888”
Output:	-1
Explanation:	We can’t reach the target without getting stuck.

Example 4:
Input:	deadends = [“0000”], target = “8888”
Output:	-1

Constraints:

1 <= deadends.length <= 500

deadends[i].length == 4

target.length == 4

target will not be in the list deadends.

target and deadends[i] consist of digits only.

Idea:

(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

There are 10^4 combinations for the lock, and we can think of each one as a node on a graph. We then have to find the shortest path from “0000” to the target combination without going through one of the deadends.

In a normal problem dealing with a shortest path on a graph, we keep track of previously visited nodes in a boolean array of combinations (seen), so we can just go ahead and add all of the deadends into seen by converting the strings to numbers.

Then, we can solve the shortest path problem with a standard queue. We’ll have an outer loop to keep track of the number of turns we’ve taken, while the inner loop will run the length of the current turn (qlen).

On each turn, we’ll take the current queue entry (curr), then we’ll iterate through the four digits and create both a mask for that digit as well as a masked version of curr. (For example, if curr = 4213 and we’re on the 2nd digit, mask would be 1 and masked would be 4203.) This way we can change the mask and add it back to masked to form the next combination. For each digit, we’ll also have to attempt both the forward and backward move, so we can add 1 and then 9 to the mask, before applying modulo 10, to get the new values.

For each next combination, if it’s our target we should return turns, and if it’s been seen, we should continue to the next iteration. Otherwise, we should consider it seen and add it to the queue. If we ever completely empty the queue, then there are no more possible moves, so we should return -1.

We also need to remember to account for edge cases where “0000” is either a deadend or the target.

Time Complexity: O(1e4) or O(1) because there are always a maximum of 1e4 possible combinations
Space Complexity: O(2e4) or O(1) for seen and the maximum length of the queue

Javascript Code:

(Jump to: Problem Description || Solution Idea)

var openLock = function(deadends, target) {
    if (target === "0000") return 0
    let queue = [0], seen = new Uint8Array(10000)
    for (let d of deadends)
        seen[~~d] = 1
    target = ~~target
    if (seen[0]) return -1
    for (let turns = 1; queue.length; turns++) {
        let qlen = queue.length
        for (let i = 0; i < qlen; i++) {
            let curr = queue.shift()
            for (let j = 1; j < 10000; j *= 10) {
                let mask = ~~(curr % (j * 10) / j),
                    masked = curr - (mask * j)
                for (let k = 1; k < 10; k += 8) {
                    let next = masked + (mask + k) % 10 * j
                    if (seen[next]) continue
                    if (next === target) return turns
                    seen[next] = 1
                    queue.push(next)
                }
            }
        }
    }
    return -1
};

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Python Code:

(Jump to: Problem Description || Solution Idea)

class Solution:
    def openLock(self, deadends: List[str], target: str) -> int:
        if target == "0000": return 0
        queue, target = deque([0]), int(target)
        seen, turns = [0] * 10000, 1
        for d in deadends: seen[int(d)] = 1
        if seen[0]: return -1
        while len(queue):
            qlen = len(queue)
            for i in range(qlen):
                curr, j = queue.popleft(), 1
                while j < 10000:
                    mask = curr % (j * 10) // j
                    masked = curr - (mask * j)
                    for k in range(1,10,8):
                        nxt = masked + (mask + k) % 10 * j
                        if seen[nxt]: continue
                        if nxt == target: return turns
                        seen[nxt] = 1
                        queue.append(nxt)
                    j *= 10
            turns += 1
        return -1

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Java Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
    public int openLock(String[] deadends, String target) {
        if (target.equals("0000")) return 0;
        Queue<Integer> queue = new LinkedList<>();
        queue.add(0);
        boolean[] seen = new boolean[10000];
        for (String el : deadends)
            seen[Integer.parseInt(el)] = true;
        int targ = Integer.parseInt(target);
        if (seen[0]) return -1;
        for (int turns = 1; !queue.isEmpty(); turns++) {
            int qlen = queue.size();
            for (int i = 0; i < qlen; i++) {
                int curr = queue.poll();
                for (int j = 1; j < 10000; j *= 10) {
                    int mask = curr % (j * 10) / j,
                        masked = curr - (mask * j);
                    for (int k = 1; k < 10; k += 8) {
                        int next = masked + (mask + k) % 10 * j;
                        if (seen[next]) continue;
                        if (next == targ) return turns;
                        seen[next] = true;
                        queue.add(next);
                    }
                }
            }
        }
        return -1;
    }
}

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C++ Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
public:
    int openLock(vector<string>& deadends, string target) {
        if (target == "0000") return 0;
        queue<int> queue;
        queue.push(0);
        bool seen[10000]{false};
        for (auto& d : deadends)
            seen[stoi(d)] = true;
        int targ = stoi(target);
        if (seen[0]) return -1;
        for (int turns = 1; queue.size(); turns++) {
            int qlen = queue.size();
            for (int i = 0; i < qlen; i++) {
                int curr = queue.front();
                queue.pop();
                for (int j = 1; j < 10000; j *= 10) {
                    int mask = curr % (j * 10) / j,
                        masked = curr - (mask * j);
                    for (int k = 1; k < 10; k += 8) {
                        int next = masked + (mask + k) % 10 * j;
                        if (seen[next]) continue;
                        if (next == targ) return turns;
                        seen[next] = true;
                        queue.push(next);
                    }
                }
            }
        }
        return -1;
    }
};

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Leetcode Solutions (161 Part Series)

原文链接：Solution: Open the Lock

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