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1585. Check If String Is Transformable With Substring Sort Operations 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  bool isTransformable(string s, string t) {
    if (getCount(s) != getCount(t))
      return false;

    vector<queue<int>> positions(10);

    for (int i = 0; i < s.length(); ++i)
      positions[s[i] - '0'].push(i);

    // For each digit in `t`, check if we can put this digit in `s` at the same
    // position as `t`. Ensure that all the left digits are equal to or greater
    // than it. This is because the only operation we can perform is sorting in
    // ascending order. If there is a digit to the left that is smaller than it,
    // we can never move it to the same position as in `t`. However, if all the
    // digits to its left are equal to or greater than it, we can move it one
    // position to the left until it reaches the same position as in `t`.
    for (const char c : t) {
      const int d = c - '0';
      const int front = positions[d].front();
      positions[d].pop();
      for (int smaller = 0; smaller < d; ++smaller)
        if (!positions[smaller].empty() && positions[smaller].front() < front)
          return false;
    }
    return true;
  }

 private:
  vector<int> getCount(const string& s) {
    vector<int> count(10);
    for (const char c : s)
      ++count[c - '0'];
    return count;
  }
};

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class Solution {
  public boolean isTransformable(String s, String t) {
    if (!Arrays.equals(getCount(s), getCount(t)))
      return false;

    Queue<Integer>[] positions = new Queue[10];
    for (int i = 0; i < 10; i++)
      positions[i] = new LinkedList<>();

    for (int i = 0; i < s.length(); i++)
      positions[s.charAt(i) - '0'].offer(i);

    // For each digit in `t`, check if we can put this digit in `s` at the same
    // position as `t`. Ensure that all the left digits are equal to or greater
    // than it. This is because the only operation we can perform is sorting in
    // ascending order. If there is a digit to the left that is smaller than it,
    // we can never move it to the same position as in `t`. However, if all the
    // digits to its left are equal to or greater than it, we can move it one
    // position to the left until it reaches the same position as in `t`.
    for (final char c : t.toCharArray()) {
      final int d = c - '0';
      final int front = positions[d].poll();
      for (int smaller = 0; smaller < d; ++smaller)
        if (!positions[smaller].isEmpty() && positions[smaller].peek() < front)
          return false;
    }

    return true;
  }

  private int[] getCount(String s) {
    int[] count = new int[10];
    for (char c : s.toCharArray())
      count[c - '0']++;
    return count;
  }
}

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class Solution:
  def isTransformable(self, s: str, t: str) -> bool:
    if collections.Counter(s) != collections.Counter(t):
      return False

    positions = [collections.deque() for _ in range(10)]

    for i, c in enumerate(s):
      positions[int(c)].append(i)

    # For each digit in `t`, check if we can put this digit in `s` at the same
    # position as `t`. Ensure that all the left digits are equal to or greater
    # than it. This is because the only operation we can perform is sorting in
    # ascending order. If there is a digit to the left that is smaller than it,
    # we can never move it to the same position as in `t`. However, if all the
    # digits to its left are equal to or greater than it, we can move it one
    # position to the left until it reaches the same position as in `t`.
    for c in t:
      d = int(c)
      front = positions[d].popleft()
      for smaller in range(d):
        if positions[smaller] and positions[smaller][0] < front:
          return False

    return True