Given two strings s and t, you want to transform string s into string t using the following operation any number of times:

• Choose a non-empty substring in s and sort it in-place so the characters are in ascending order.

For example, applying the operation on the underlined substring in "14234" results in "12344".

Return true if it is possible to transform string s into string t. Otherwise, return false.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "84532", t = "34852"
Output: true
Explanation: You can transform s into t using the following sort operations:
"84532" (from index 2 to 3) -> "84352"
"84352" (from index 0 to 2) -> "34852"

Example 2:

Input: s = "34521", t = "23415"
Output: true
Explanation: You can transform s into t using the following sort operations:
"34521" -> "23451"
"23451" -> "23415"

Example 3:

Input: s = "12345", t = "12435"
Output: false

Example 4:

Input: s = "1", t = "2"
Output: false

Constraints:

• s.length == t.length
• 1 <= s.length <= 105
• s and t only contain digits from '0' to '9'.

Solution: Queue

We can move a smaller digit from right to left by sorting two adjacent digits.
e.g. 18572 -> 18527 -> 18257 -> 12857, but we can not move a larger to the left of a smaller one.

Thus, for each digit in the target string, we find the first occurrence of it in s, and try to move it to the front by checking if there is any smaller one in front of it.

Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
  bool isTransformable(string s, string t) {
    vector<deque<int>> idx(10);
    for (int i = 0; i < s.length(); ++i)
      idx[s[i] - '0'].push_back(i);
    for (char c : t) {
      const int d  = c - '0';
      if (idx[d].empty()) return false;
      for (int i = 0; i < d; ++i)
        if (!idx[i].empty() && idx[i].front() < idx[d].front())
          return false;
      idx[d].pop_front();      
    }
    return true;
  }
};

class Solution:
  def isTransformable(self, s: str, t: str) -> bool:
    idx = defaultdict(deque)
    for i, c in enumerate(s):
      idx[int(c)].append(i)
    for c in t:
      d = int(c)
      if not idx[d]: return False
      for i in range(d):
        if idx[i] and idx[i][0] < idx[d][0]: return False
      idx[d].popleft()
    return True

The post 花花酱 LeetCode 1585. Check If String Is Transformable With Substring Sort Operations appeared first on Huahua's Tech Road.

Problem:

Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False

// Author: Huahua
class MagicDictionary {
public:
    /** Initialize your data structure here. */
    MagicDictionary() {
        dict_.clear();
    }
    
    /** Build a dictionary through a list of words */
    void buildDict(vector<string> dict) {
        for(string& word: dict) {
            for(int i = 0; i < word.length(); ++i) {
                char c = word[i];
                word[i] = '*';
                dict_[word].insert(c);
                word[i] = c;
            }
        }
    }
    
    /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
    bool search(string word) {
        for(int i = 0; i < word.length(); ++i) {
            char c = word[i];
            word[i] = '*';
            if (dict_.count(word)) {
                const auto& char_set = dict_[word];
                if (!char_set.count(c) || char_set.size() > 1)
                    return true;
            }
            word[i] = c;
        }
        return false;
    }
private:
    unordered_map<string, unordered_set<char>> dict_;
};

/**
 * Your MagicDictionary object will be instantiated and called as such:
 * MagicDictionary obj = new MagicDictionary();
 * obj.buildDict(dict);
 * bool param_2 = obj.search(word);
 */

Java / Trie

class MagicDictionary {
    private Trie root;    
    
    public MagicDictionary() {
        root = new Trie();
    } 
    
    public void buildDict(String[] dict) {
        for (String word : dict) {
            Trie curr = root;
            for (char c : word.toCharArray()) {                
                int index = c - 'a';
                if (curr.children[index] == null)
                    curr.children[index] = new Trie();
                curr = curr.children[index];
            }
            curr.ends = true;
        }
    }  
    
    public boolean search(String word) {
        char[] s = word.toCharArray();
        for(int i = 0; i < s.length; i++) {
            for (char c = 'a'; c <= 'z'; ++c) {
                if (s[i] == c) continue;
                char tmp = s[i];
                s[i] = c;
                if (contains(s)) return true;
                s[i] = tmp;
            }
        }
        return false;
    }
    
    private boolean contains(char[] word) {        
        Trie curr = root;
        for (int i = 0; i < word.length; ++i) {
            curr = curr.children[word[i] - 'a'];
            if (curr == null) return false;            
        }
        return curr.ends;
    }
    
    class Trie {
        private boolean ends;
        private Trie[] children;
        public Trie() {
            children = new Trie[26];
            ends = false;
        }
    }
}

Java / Trie v2

class MagicDictionary {
    private Trie root;    
    
    public MagicDictionary() {
        root = new Trie();
    } 
    
    public void buildDict(String[] dict) {
        for (String word : dict) {
            Trie curr = root;
            for (char c : word.toCharArray()) {                
                int index = c - 'a';
                if (curr.children[index] == null)
                    curr.children[index] = new Trie();
                curr = curr.children[index];
            }
            curr.ends = true;
        }
    }  
    
    public boolean search(String word) {
        char[] s = word.toCharArray();
        Trie prefix = root;
        for(int i = 0; i < s.length; i++) {
            for (char c = 'a'; c <= 'z'; ++c) {
                if (s[i] == c) continue;
                char tmp = s[i];
                s[i] = c;
                if (contains(s, prefix, i)) return true;
                s[i] = tmp;                
            }
            prefix = prefix.children[s[i] - 'a'];
            if (prefix == null) return false;
        }
        return false;
    }
    
    private boolean contains(char[] word, Trie prefix, int s) {        
        Trie curr = prefix;
        for (int i = s; i < word.length; ++i) {
            curr = curr.children[word[i] - 'a'];
            if (curr == null) return false;
        }
        return curr.ends;
    }
    
    class Trie {
        private boolean ends;
        private Trie[] children;
        public Trie() {
            children = new Trie[26];
            ends = false;
        }
    }
}

The post 花花酱 LeetCode 676. Implement Magic Dictionary appeared first on Huahua's Tech Road.