How to solve Leetcode 890. Find and Replace Pattern

How to solve Leetcode 890. Find and Replace Pattern

How to create a bijection map

Problem statement

Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order.

A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the desired word.

Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map 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"]


  • 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


#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
    vector<string> result;  
    // need two maps for the bijection
    unordered_map<char,char> w_to_p, p_to_w;
    int i;
    for (string& w : words) {
        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]) {
            } 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
                // build the bijection w[i] <-> pattern[i]
                w_to_p[w[i]] = pattern[i];
                p_to_w[pattern[i]] = w[i];
        if (i == w.length()) {
    return result;
void printResult(const vector<string>& result) {
    cout << "[";
    for (const string& s : result) {
        cout << s << ",";
    cout << "]\n";
int main() {
    vector<string> words{"abc","deq","mee","aqq","dkd","ccc"};
    auto result = findAndReplacePattern(words, "abb");
    words = {"a", "b", "c"};
    result = findAndReplacePattern(words, "abb");


  • Runtime: O(N*L), where N = words.length and L = pattern.length.

  • Extra space: O(1) if N or L is very larger than 26. The maps w_to_p and p_to_w just map between 26 lowercase English letters.