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1215. Stepping Numbers 👍

Approach 1: BFS

  • Time: $O(|\texttt{ans}|)$
  • Space: $O(|\texttt{ans}|)$
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class Solution {
 public:
  vector<int> countSteppingNumbers(int low, int high) {
    vector<int> ans;
    if (low == 0)
      ans.push_back(0);

    queue<long> q;

    for (int i = 1; i <= 9; ++i)
      q.push(i);

    while (!q.empty()) {
      const long curr = q.front();
      q.pop();
      if (curr > high)
        continue;
      if (curr >= low)
        ans.push_back(curr);
      const int lastDigit = curr % 10;
      if (lastDigit > 0)
        q.push(curr * 10 + lastDigit - 1);
      if (lastDigit < 9)
        q.push(curr * 10 + lastDigit + 1);
    }

    return ans;
  }
};

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class Solution {
  public List<Integer> countSteppingNumbers(int low, int high) {
    List<Integer> ans = new ArrayList<>();
    if (low == 0)
      ans.add(0);

    Queue<Long> q = new ArrayDeque<>();

    for (long i = 1; i <= 9; ++i)
      q.offer(i);

    while (!q.isEmpty()) {
      final long curr = q.poll();
      if (curr > high)
        continue;
      if (curr >= low)
        ans.add((int) curr);
      final long lastDigit = curr % 10;
      if (lastDigit > 0)
        q.offer(curr * 10 + lastDigit - 1);
      if (lastDigit < 9)
        q.offer(curr * 10 + lastDigit + 1);
    }

    return ans;
  }
}

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class Solution:
  def countSteppingNumbers(self, low: int, high: int) -> list[int]:
    ans = [0] if low == 0 else []
    q = collections.deque(list(range(1, 10)))

    while q:
      curr = q.popleft()
      if curr > high:
        continue
      if curr >= low:
        ans.append(curr)
      lastDigit = curr % 10
      if lastDigit > 0:
        q.append(curr * 10 + lastDigit - 1)
      if lastDigit < 9:
        q.append(curr * 10 + lastDigit + 1)

    return ans

Approach 2: DFS

  • Time: $O(\texttt{sort}(\texttt{ans}))$
  • Space: $O(|\texttt{ans}|)$
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class Solution {
 public:
  vector<int> countSteppingNumbers(int low, int high) {
    vector<int> ans;
    if (low == 0)
      ans.push_back(0);

    for (int i = 1; i <= 9; ++i)
      dfs(i, low, high, ans);

    ranges::sort(ans);
    return ans;
  }

 private:
  void dfs(long curr, int low, int high, vector<int>& ans) {
    if (curr > high)
      return;
    if (curr >= low)
      ans.push_back(curr);

    const int lastDigit = curr % 10;
    if (lastDigit > 0)
      dfs(curr * 10 + lastDigit - 1, low, high, ans);
    if (lastDigit < 9)
      dfs(curr * 10 + lastDigit + 1, low, high, ans);
  }
};

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class Solution {
  public List<Integer> countSteppingNumbers(int low, int high) {
    List<Integer> ans = new ArrayList<>();
    if (low == 0)
      ans.add(0);

    for (long i = 1; i <= 9; ++i)
      dfs(i, low, high, ans);

    Collections.sort(ans);
    return ans;
  }

  private void dfs(long curr, int low, int high, List<Integer> ans) {
    if (curr > high)
      return;
    if (curr >= low)
      ans.add((int) curr);

    final long lastDigit = curr % 10;
    if (lastDigit > 0)
      dfs(curr * 10 + lastDigit - 1, low, high, ans);
    if (lastDigit < 9)
      dfs(curr * 10 + lastDigit + 1, low, high, ans);
  }
}

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class Solution:
  def countSteppingNumbers(self, low: int, high: int) -> list[int]:
    ans = [0] if low == 0 else []

    def dfs(curr: int) -> None:
      if curr > high:
        return
      if curr >= low:
        ans.append(curr)

      lastDigit = curr % 10
      if lastDigit > 0:
        dfs(curr * 10 + lastDigit - 1)
      if lastDigit < 9:
        dfs(curr * 10 + lastDigit + 1)

    for i in range(1, 9 + 1):
      dfs(i)

    ans.sort()
    return ans