Find and Replace Pattern#

Problem statement#

[1]You are provided with a list of strings named words and a string named pattern. Your task is to find the strings from words that match the given pattern. The order in which you return the answers does not matter.

A word is considered to match the pattern if there is a mapping p of the letters such that, when each letter x in the pattern is replaced with p(x), the word is formed.

Keep in mind that a permutation of letters is a one-to-one correspondence from letters to letters, where each letter is mapped to a distinct letter, and no two letters are mapped to the same letter.

Example 1#

Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
Output: ["mee","aqq"]
Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}. 
"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter.

Example 2#

Input: words = ["a","b","c"], pattern = "a"
Output: ["a","b","c"]

Constraints#

1 <= pattern.length <= 20.
1 <= words.length <= 50.
words[i].length == pattern.length.
pattern and words[i] are lowercase English letters.

Solution: Construct the bijection and check the condition#

Code#

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
vector<string> findAndReplacePattern(const vector<string>& words, const string& pattern) {
    vector<string> result;  
    // need two maps for the bijection
    unordered_map<char,char> w_to_p, p_to_w;
    int i;
    for (auto& w : words) {
        w_to_p.clear();
        p_to_w.clear();
        i = 0;
        while (i < w.length()) {
            if (w_to_p.find(w[i]) != w_to_p.end()) {
                // w[i] was mapped to some letter x
                // but x != pattern[i]
                if (w_to_p[w[i]] != pattern[i]) {
                    break;
                }
            } else {
                if (p_to_w.find(pattern[i]) != p_to_w.end()) {
                    // w[i] was not mapped to any letter yet
                    // but pattern[i] was already mapped to some letter
                    break;
                }
                // build the bijection w[i] <-> pattern[i]
                w_to_p[w[i]] = pattern[i];
                p_to_w[pattern[i]] = w[i];
            }
            i++;
        }
        if (i == w.length()) {
            result.push_back(w);
        }
    }
    return result;
}
void printResult(const vector<string>& result) {
    cout << "[";
    for (auto& s : result) {
        cout << s << ",";
    }
    cout << "]\n";
}
int main() {
    vector<string> words{"abc","deq","mee","aqq","dkd","ccc"};
    auto result = findAndReplacePattern(words, "abb");
    printResult(result);
    words = {"a", "b", "c"};
    result = findAndReplacePattern(words, "abb");
    printResult(result);
}

Output:
[mee,aqq,]
[a,b,c,]

Complexity#

Runtime: O(NL), where N = words.length and L = pattern.length.
Extra space: O(1). The maps w_to_p and p_to_w just map between 26 lowercase English letters.

Conclusion#

This solution efficiently finds and returns words from a vector of strings that match a given pattern in terms of character bijection. It uses two unordered maps to establish and maintain the bijection while iterating through the characters of the words and the pattern.