Solution Explanation:

This problem asks to determine if any two strings in a given list differ by only one character at the same index. The optimal approach uses a hash table (or set) to efficiently track modified strings.

Algorithm:

1. Initialization: Create a set s to store modified strings.

2. Iteration: Iterate through each string word in the input list dict.

3. Inner Iteration: For each word, iterate through each character index i.

4. Modification: Create a modified string t by replacing the character at index i with a special character (e.g., '*'). This represents the wildcard – any character could be at that position.

5. Check in Set: Check if t exists in the set s. If it does, it means we've encountered another string that differs only at this index; return true.

6. Add to Set: If t is not in s, add it to the set.

7. Return False: If the loop completes without finding any such pair, return false.

Time Complexity: O(N*M), where N is the number of strings and M is the length of each string. We iterate through each string once and for each string, we iterate through each character.

Space Complexity: O(N*M). In the worst case, we might store almost all modified strings in the set. The space complexity is proportional to the size of the modified strings which is linear with respect to the number of strings and the length of the strings.

Code Explanation (Python):

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 :] #create modified string with *
                if t in s:
                    return True # Found a pair that differ by one
                s.add(t) #add modified string to the set
        return False #No such pair found
 

The Python code directly implements the algorithm. It utilizes Python's built-in string slicing and set operations for efficiency.

Code Explanation (Java):

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;
    }
}

The Java code is very similar to the Python version, using Java's String manipulation and HashSet for the same purpose.

Code Explanation (C++):

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] = '*'; // Modify the string in place
                if (s.count(t)) return true;
                s.insert(t);
            }
        }
        return false;
    }
};

The C++ code efficiently modifies the string in place using t[i] = '*'; It leverages unordered_set for fast lookups.

Code Explanation (Go):

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
}

The Go code mirrors the logic of the other solutions, using Go's string manipulation and map for efficient set operations.

All four code snippets implement the same fundamental algorithm and achieve the same time and space complexity. The choice of language depends on personal preference and the specific context of the application.