Question

1554. Strings Differ by One Character

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Description

Given a list of strings dict where all the strings are of the same length.

Return true if there are 2 strings that only differ by 1 character in the same index, otherwise return false.

 

Example 1:

Input: dict = ["abcd","acbd", "aacd"]
Output: true
Explanation: Strings "abcd" and "aacd" differ only by one character in the index 1.

Example 2:

Input: dict = ["ab","cd","yz"]
Output: false

Example 3:

Input: dict = ["abcd","cccc","abyd","abab"]
Output: true

 

Constraints:

• The number of characters in dict <= 105
• dict[i].length == dict[j].length
• dict[i] should be unique.
• dict[i] contains only lowercase English letters.

 

Follow up: Could you solve this problem in O(n * m) where n is the length of dict and m is the length of each string.

Solutions

Solution 1

class Solution:
  def differByOne(self, dict: List[str]) -> bool:
    s = set()
    for word in dict:
      for i in range(len(word)):
        t = word[:i] + "*" + word[i + 1 :]
        if t in s:
          return True
        s.add(t)
    return False

class Solution {
  public boolean differByOne(String[] dict) {
    Set<String> s = new HashSet<>();
    for (String word : dict) {
      for (int i = 0; i < word.length(); ++i) {
        String t = word.substring(0, i) + "*" + word.substring(i + 1);
        if (s.contains(t)) {
          return true;
        }
        s.add(t);
      }
    }
    return false;
  }
}

class Solution {
public:
  bool differByOne(vector<string>& dict) {
    unordered_set<string> s;
    for (auto word : dict) {
      for (int i = 0; i < word.size(); ++i) {
        auto t = word;
        t[i] = '*';
        if (s.count(t)) return true;
        s.insert(t);
      }
    }
    return false;
  }
};

func differByOne(dict []string) bool {
  s := make(map[string]bool)
  for _, word := range dict {
    for i := range word {
      t := word[:i] + "*" + word[i+1:]
      if s[t] {
        return true
      }
      s[t] = true
    }
  }
  return false
}