LeetCode 1554. Strings Differ by One Character Solution in Java, C++, Python & Go | Explanation + Code

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.

Solutions

Solution 1

PythonJavaC++Go

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)