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752. Open the Lock

Description

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.

 

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 cannot reach the target without getting stuck.

 

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.

Solutions

Solution 1

BFS.

Solution 2

This question can also use bidirectional BFS to optimize the search space to improve efficiency.

Bidirectional BFS is a common optimization method for BFS. The main implementation ideas are as follows:

1. Create two queues q1 and q2 for searches in the two directions of “starting point -> end point” and “end point -> starting point” respectively;
2. Create two hash tables m1 and m2 to record the visited nodes and the corresponding number of expansions (number of steps) respectively;
3. During each search, the queue with a smaller number of elements is given priority for search expansion. If during the expansion process, a node that has been visited in the other direction is searched, it means that the shortest path has been found;
4. As long as one of the queues is empty, it means that the search in the current direction can no longer proceed. It means that the starting point and the end point are not connected, and there is no need to continue searching.

while q1 and q2:
    if len(q1) <= len(q2):
        # 优先选择较少元素的队列进行扩展
        extend(m1, m2, q1)
    else:
        extend(m2, m1, q2)


def extend(m1, m2, q):
    # 新一轮扩展
    for _ in range(len(q)):
        p = q.popleft()
        step = m1[p]
        for t in next(p):
            if t in m1:
                # 此前已经访问过
                continue
            if t in m2:
                # 另一个方向已经搜索过，说明找到了一条最短的连通路径
                return step + 1 + m2[t]
            q.append(t)
            m1[t] = step + 1

Solution 3

The main idea of A* algorithm is as follows:

1. Convert the BFS queue to a priority queue (small root heap);
2. Each element in the queue is (dist[state] + f(state), state), dist[state] represents the distance from the starting point to the current state, and f(state) represents the estimated distance from the current state to the end point. The sum of these two distances is used as the basis for heap sorting;
3. When the end point is dequeued for the first time, it means that the shortest path from the starting point to the end point is found and the corresponding step is returned directly;
4. f(state) is the valuation function, and the valuation function must satisfy f(state) <= g(state), where g(state) represents the true distance from state to the end point;
5. The A* algorithm can only guarantee that when the end point is dequeued for the first time, it will find a minimum path from the starting point to the end point. It cannot guarantee that when other points are dequeued, it will also be the shortest path from the starting point to the current point.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      private String start;
      private String target;
      private Set<String> s = new HashSet<>();
  
      public int openLock(String[] deadends, String target) {
          if ("0000".equals(target)) {
              return 0;
          }
          start = "0000";
          this.target = target;
          for (String d : deadends) {
              s.add(d);
          }
          if (s.contains(start)) {
              return -1;
          }
          return bfs();
      }
  
      private int bfs() {
          Map<String, Integer> m1 = new HashMap<>();
          Map<String, Integer> m2 = new HashMap<>();
          Deque<String> q1 = new ArrayDeque<>();
          Deque<String> q2 = new ArrayDeque<>();
          m1.put(start, 0);
          m2.put(target, 0);
          q1.offer(start);
          q2.offer(target);
          while (!q1.isEmpty() && !q2.isEmpty()) {
              int t = q1.size() <= q2.size() ? extend(m1, m2, q1) : extend(m2, m1, q2);
              if (t != -1) {
                  return t;
              }
          }
          return -1;
      }
  
      private int extend(Map<String, Integer> m1, Map<String, Integer> m2, Deque<String> q) {
          for (int n = q.size(); n > 0; --n) {
              String p = q.poll();
              int step = m1.get(p);
              for (String t : next(p)) {
                  if (m1.containsKey(t) || s.contains(t)) {
                      continue;
                  }
                  if (m2.containsKey(t)) {
                      return step + 1 + m2.get(t);
                  }
                  m1.put(t, step + 1);
                  q.offer(t);
              }
          }
          return -1;
      }
  
      private List<String> next(String t) {
          List res = new ArrayList<>();
          char[] chars = t.toCharArray();
          for (int i = 0; i < 4; ++i) {
              char c = chars[i];
              chars[i] = c == '0' ? '9' : (char) (c - 1);
              res.add(String.valueOf(chars));
              chars[i] = c == '9' ? '0' : (char) (c + 1);
              res.add(String.valueOf(chars));
              chars[i] = c;
          }
          return res;
      }
  }
  
  
  // Solution 2
  class Solution {
      private String start;
      private String target;
      private Set<String> s = new HashSet<>();
  
      public int openLock(String[] deadends, String target) {
          if ("0000".equals(target)) {
              return 0;
          }
          start = "0000";
          this.target = target;
          for (String d : deadends) {
              s.add(d);
          }
          if (s.contains(start)) {
              return -1;
          }
          return bfs();
      }
  
      private int bfs() {
          Map<String, Integer> m1 = new HashMap<>();
          Map<String, Integer> m2 = new HashMap<>();
          Deque<String> q1 = new ArrayDeque<>();
          Deque<String> q2 = new ArrayDeque<>();
          m1.put(start, 0);
          m2.put(target, 0);
          q1.offer(start);
          q2.offer(target);
          while (!q1.isEmpty() && !q2.isEmpty()) {
              int t = q1.size() <= q2.size() ? extend(m1, m2, q1) : extend(m2, m1, q2);
              if (t != -1) {
                  return t;
              }
          }
          return -1;
      }
  
      private int extend(Map<String, Integer> m1, Map<String, Integer> m2, Deque<String> q) {
          for (int n = q.size(); n > 0; --n) {
              String p = q.poll();
              int step = m1.get(p);
              for (String t : next(p)) {
                  if (m1.containsKey(t) || s.contains(t)) {
                      continue;
                  }
                  if (m2.containsKey(t)) {
                      return step + 1 + m2.get(t);
                  }
                  m1.put(t, step + 1);
                  q.offer(t);
              }
          }
          return -1;
      }
  
      private List<String> next(String t) {
          List res = new ArrayList<>();
          char[] chars = t.toCharArray();
          for (int i = 0; i < 4; ++i) {
              char c = chars[i];
              chars[i] = c == '0' ? '9' : (char) (c - 1);
              res.add(String.valueOf(chars));
              chars[i] = c == '9' ? '0' : (char) (c + 1);
              res.add(String.valueOf(chars));
              chars[i] = c;
          }
          return res;
      }
  }
  
  
  
  // Solution 3
  class Solution {
      private String target;
  
      public int openLock(String[] deadends, String target) {
          if ("0000".equals(target)) {
              return 0;
          }
          String start = "0000";
          this.target = target;
          Set<String> s = new HashSet<>();
          for (String d : deadends) {
              s.add(d);
          }
          if (s.contains(start)) {
              return -1;
          }
          PriorityQueue<Pair<Integer, String>> q
              = new PriorityQueue<>(Comparator.comparingInt(Pair::getKey));
          q.offer(new Pair<>(f(start), start));
          Map<String, Integer> dist = new HashMap<>();
          dist.put(start, 0);
          while (!q.isEmpty()) {
              String state = q.poll().getValue();
              int step = dist.get(state);
              if (target.equals(state)) {
                  return step;
              }
              for (String t : next(state)) {
                  if (s.contains(t)) {
                      continue;
                  }
                  if (!dist.containsKey(t) || dist.get(t) > step + 1) {
                      dist.put(t, step + 1);
                      q.offer(new Pair<>(step + 1 + f(t), t));
                  }
              }
          }
          return -1;
      }
  
      private int f(String s) {
          int ans = 0;
          for (int i = 0; i < 4; ++i) {
              int a = s.charAt(i) - '0';
              int b = target.charAt(i) - '0';
              if (a > b) {
                  int t = a;
                  a = b;
                  b = a;
              }
              ans += Math.min(b - a, a + 10 - b);
          }
          return ans;
      }
  
      private List<String> next(String t) {
          List res = new ArrayList<>();
          char[] chars = t.toCharArray();
          for (int i = 0; i < 4; ++i) {
              char c = chars[i];
              chars[i] = c == '0' ? '9' : (char) (c - 1);
              res.add(String.valueOf(chars));
              chars[i] = c == '9' ? '0' : (char) (c + 1);
              res.add(String.valueOf(chars));
              chars[i] = c;
          }
          return res;
      }
  }
  
  

• class Solution {
  public:
      unordered_set<string> s;
      string start;
      string target;
  
      int openLock(vector<string>& deadends, string target) {
          if (target == "0000") return 0;
          for (auto d : deadends) s.insert(d);
          if (s.count("0000")) return -1;
          this->start = "0000";
          this->target = target;
          return bfs();
      }
  
      int bfs() {
          unordered_map<string, int> m1;
          unordered_map<string, int> m2;
          m1[start] = 0;
          m2[target] = 0;
          queue<string> q1{ {start} };
          queue<string> q2{ {target} };
          while (!q1.empty() && !q2.empty()) {
              int t = q1.size() <= q2.size() ? extend(m1, m2, q1) : extend(m2, m1, q2);
              if (t != -1) return t;
          }
          return -1;
      }
  
      int extend(unordered_map<string, int>& m1, unordered_map<string, int>& m2, queue<string>& q) {
          for (int n = q.size(); n > 0; --n) {
              string p = q.front();
              int step = m1[p];
              q.pop();
              for (string t : next(p)) {
                  if (s.count(t) || m1.count(t)) continue;
                  if (m2.count(t)) return step + 1 + m2[t];
                  m1[t] = step + 1;
                  q.push(t);
              }
          }
          return -1;
      }
  
      vector<string> next(string& t) {
          vector<string> res;
          for (int i = 0; i < 4; ++i) {
              char c = t[i];
              t[i] = c == '0' ? '9' : (char) (c - 1);
              res.push_back(t);
              t[i] = c == '9' ? '0' : (char) (c + 1);
              res.push_back(t);
              t[i] = c;
          }
          return res;
      }
  };
  
  
  // Solution 2
  class Solution {
  public:
      unordered_set<string> s;
      string start;
      string target;
  
      int openLock(vector<string>& deadends, string target) {
          if (target == "0000") return 0;
          for (auto d : deadends) s.insert(d);
          if (s.count("0000")) return -1;
          this->start = "0000";
          this->target = target;
          return bfs();
      }
  
      int bfs() {
          unordered_map<string, int> m1;
          unordered_map<string, int> m2;
          m1[start] = 0;
          m2[target] = 0;
          queue<string> q1{{start}};
          queue<string> q2{{target}};
          while (!q1.empty() && !q2.empty()) {
              int t = q1.size() <= q2.size() ? extend(m1, m2, q1) : extend(m2, m1, q2);
              if (t != -1) return t;
          }
          return -1;
      }
  
      int extend(unordered_map<string, int>& m1, unordered_map<string, int>& m2, queue<string>& q) {
          for (int n = q.size(); n > 0; --n) {
              string p = q.front();
              int step = m1[p];
              q.pop();
              for (string t : next(p)) {
                  if (s.count(t) || m1.count(t)) continue;
                  if (m2.count(t)) return step + 1 + m2[t];
                  m1[t] = step + 1;
                  q.push(t);
              }
          }
          return -1;
      }
  
      vector<string> next(string& t) {
          vector<string> res;
          for (int i = 0; i < 4; ++i) {
              char c = t[i];
              t[i] = c == '0' ? '9' : (char) (c - 1);
              res.push_back(t);
              t[i] = c == '9' ? '0' : (char) (c + 1);
              res.push_back(t);
              t[i] = c;
          }
          return res;
      }
  };
  
  
  
  // Solution 3
  class Solution {
  public:
      string target;
  
      int openLock(vector<string>& deadends, string target) {
          if (target == "0000") return 0;
          unordered_set<string> s(deadends.begin(), deadends.end());
          if (s.count("0000")) return -1;
          string start = "0000";
          this->target = target;
          typedef pair<int, string> PIS;
          priority_queue<PIS, vector<PIS>, greater<PIS>> q;
          unordered_map<string, int> dist;
          dist[start] = 0;
          q.push({f(start), start});
          while (!q.empty()) {
              PIS t = q.top();
              q.pop();
              string state = t.second;
              int step = dist[state];
              if (state == target) return step;
              for (string& t : next(state)) {
                  if (s.count(t)) continue;
                  if (!dist.count(t) || dist[t] > step + 1) {
                      dist[t] = step + 1;
                      q.push({step + 1 + f(t), t});
                  }
              }
          }
          return -1;
      }
  
      int f(string s) {
          int ans = 0;
          for (int i = 0; i < 4; ++i) {
              int a = s[i] - '0';
              int b = target[i] - '0';
              if (a < b) {
                  int t = a;
                  a = b;
                  b = t;
              }
              ans += min(b - a, a + 10 - b);
          }
          return ans;
      }
  
      vector<string> next(string& t) {
          vector<string> res;
          for (int i = 0; i < 4; ++i) {
              char c = t[i];
              t[i] = c == '0' ? '9' : (char) (c - 1);
              res.push_back(t);
              t[i] = c == '9' ? '0' : (char) (c + 1);
              res.push_back(t);
              t[i] = c;
          }
          return res;
      }
  };
  
  

• class Solution:
      def openLock(self, deadends: List[str], target: str) -> int:
          def next(s):
              res = []
              s = list(s)
              for i in range(4):
                  c = s[i]
                  s[i] = '9' if c == '0' else str(int(c) - 1)
                  res.append(''.join(s))
                  s[i] = '0' if c == '9' else str(int(c) + 1)
                  res.append(''.join(s))
                  s[i] = c
              return res
  
          def extend(m1, m2, q):
              for _ in range(len(q)):
                  p = q.popleft()
                  step = m1[p]
                  for t in next(p):
                      if t in s or t in m1:
                          continue
                      if t in m2:
                          return step + 1 + m2[t]
                      m1[t] = step + 1
                      q.append(t)
              return -1
  
          def bfs():
              m1, m2 = {"0000": 0}, {target: 0}
              q1, q2 = deque([('0000')]), deque([(target)])
              while q1 and q2:
                  t = extend(m1, m2, q1) if len(q1) <= len(q2) else extend(m2, m1, q2)
                  if t != -1:
                      return t
              return -1
  
          if target == '0000':
              return 0
          s = set(deadends)
          if '0000' in s:
              return -1
          return bfs()
  
  
  # Solution 2
  class Solution:
      def openLock(self, deadends: List[str], target: str) -> int:
          def next(s):
              res = []
              s = list(s)
              for i in range(4):
                  c = s[i]
                  s[i] = '9' if c == '0' else str(int(c) - 1)
                  res.append(''.join(s))
                  s[i] = '0' if c == '9' else str(int(c) + 1)
                  res.append(''.join(s))
                  s[i] = c
              return res
  
          def extend(m1, m2, q):
              for _ in range(len(q)):
                  p = q.popleft()
                  step = m1[p]
                  for t in next(p):
                      if t in s or t in m1:
                          continue
                      if t in m2:
                          return step + 1 + m2[t]
                      m1[t] = step + 1
                      q.append(t)
              return -1
  
          def bfs():
              m1, m2 = {"0000": 0}, {target: 0}
              q1, q2 = deque(['0000']), deque([target])
              while q1 and q2:
                  t = extend(m1, m2, q1) if len(q1) <= len(q2) else extend(m2, m1, q2)
                  if t != -1:
                      return t
              return -1
  
          if target == '0000':
              return 0
          s = set(deadends)
          if '0000' in s:
              return -1
          return bfs()
  
  
  
  # Solution 3
  class Solution:
      def openLock(self, deadends: List[str], target: str) -> int:
          def next(s):
              res = []
              s = list(s)
              for i in range(4):
                  c = s[i]
                  s[i] = '9' if c == '0' else str(int(c) - 1)
                  res.append(''.join(s))
                  s[i] = '0' if c == '9' else str(int(c) + 1)
                  res.append(''.join(s))
                  s[i] = c
              return res
  
          def f(s):
              ans = 0
              for i in range(4):
                  a = ord(s[i]) - ord('0')
                  b = ord(target[i]) - ord('0')
                  if a > b:
                      a, b = b, a
                  ans += min(b - a, a + 10 - b)
              return ans
  
          if target == '0000':
              return 0
          s = set(deadends)
          if '0000' in s:
              return -1
          start = '0000'
          q = [(f(start), start)]
          dist = {start: 0}
          while q:
              _, state = heappop(q)
              if state == target:
                  return dist[state]
              for t in next(state):
                  if t in s:
                      continue
                  if t not in dist or dist[t] > dist[state] + 1:
                      dist[t] = dist[state] + 1
                      heappush(q, (dist[t] + f(t), t))
          return -1
  
  

• func openLock(deadends []string, target string) int {
  	if target == "0000" {
  		return 0
  	}
  	s := make(map[string]bool)
  	for _, d := range deadends {
  		s[d] = true
  	}
  	if s["0000"] {
  		return -1
  	}
  	next := func(t string) []string {
  		s := []byte(t)
  		var res []string
  		for i, b := range s {
  			s[i] = b - 1
  			if s[i] < '0' {
  				s[i] = '9'
  			}
  			res = append(res, string(s))
  			s[i] = b + 1
  			if s[i] > '9' {
  				s[i] = '0'
  			}
  			res = append(res, string(s))
  			s[i] = b
  		}
  		return res
  	}
  
  	extend := func(m1, m2 map[string]int, q *[]string) int {
  		for n := len(*q); n > 0; n-- {
  			p := (*q)[0]
  			*q = (*q)[1:]
  			step, _ := m1[p]
  			for _, t := range next(p) {
  				if s[t] {
  					continue
  				}
  				if _, ok := m1[t]; ok {
  					continue
  				}
  				if v, ok := m2[t]; ok {
  					return step + 1 + v
  				}
  				m1[t] = step + 1
  				*q = append(*q, t)
  			}
  		}
  		return -1
  	}
  
  	bfs := func() int {
  		q1 := []string{"0000"}
  		q2 := []string{target}
  		m1 := map[string]int{"0000": 0}
  		m2 := map[string]int{target: 0}
  		for len(q1) > 0 && len(q2) > 0 {
  			t := -1
  			if len(q1) <= len(q2) {
  				t = extend(m1, m2, &q1)
  			} else {
  				t = extend(m2, m1, &q2)
  			}
  			if t != -1 {
  				return t
  			}
  		}
  		return -1
  	}
  
  	return bfs()
  }
  
  
  // Solution 2
  func openLock(deadends []string, target string) int {
  	if target == "0000" {
  		return 0
  	}
  	s := make(map[string]bool)
  	for _, d := range deadends {
  		s[d] = true
  	}
  	if s["0000"] {
  		return -1
  	}
  	next := func(t string) []string {
  		s := []byte(t)
  		var res []string
  		for i, b := range s {
  			s[i] = b - 1
  			if s[i] < '0' {
  				s[i] = '9'
  			}
  			res = append(res, string(s))
  			s[i] = b + 1
  			if s[i] > '9' {
  				s[i] = '0'
  			}
  			res = append(res, string(s))
  			s[i] = b
  		}
  		return res
  	}
  
  	extend := func(m1, m2 map[string]int, q *[]string) int {
  		for n := len(*q); n > 0; n-- {
  			p := (*q)[0]
  			*q = (*q)[1:]
  			step, _ := m1[p]
  			for _, t := range next(p) {
  				if s[t] {
  					continue
  				}
  				if _, ok := m1[t]; ok {
  					continue
  				}
  				if v, ok := m2[t]; ok {
  					return step + 1 + v
  				}
  				m1[t] = step + 1
  				*q = append(*q, t)
  			}
  		}
  		return -1
  	}
  
  	bfs := func() int {
  		q1 := []string{"0000"}
  		q2 := []string{target}
  		m1 := map[string]int{"0000": 0}
  		m2 := map[string]int{target: 0}
  		for len(q1) > 0 && len(q2) > 0 {
  			t := -1
  			if len(q1) <= len(q2) {
  				t = extend(m1, m2, &q1)
  			} else {
  				t = extend(m2, m1, &q2)
  			}
  			if t != -1 {
  				return t
  			}
  		}
  		return -1
  	}
  
  	return bfs()
  }
  
  

• function openLock(deadends: string[], target: string): number {
      const vis = Array<boolean>(10_000);
      const q = [[0, 0, 0, 0, 0]];
      const t = +target;
      deadends.forEach(s => (vis[+s] = true));
  
      for (const [a, b, c, d, ans] of q) {
          const key = a * 1000 + b * 100 + c * 10 + d;
          if (vis[key]) {
              continue;
          }
  
          vis[key] = true;
          if (key === t) {
              return ans;
          }
  
          for (let i = 0; i < 4; i++) {
              const next = [a, b, c, d, ans + 1];
              const prev = [a, b, c, d, ans + 1];
              next[i] = (next[i] + 11) % 10;
              prev[i] = (prev[i] + 9) % 10;
              q.push(prev, next);
          }
      }
  
      return -1;
  }
  
  

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