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Hashing problems

If some problem is not solvable without mental gymnastics, you can pretty much solve those with a hashtable or unordered_map<d_t, d_t> in C++.

Let's solve some of the hashing problems which may not be straight forward.

Find and Replace Pattern

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.


  • We can solve this problem with a hash table but the problem is one letter is mapped to only one letter. Let's take some examples see how this can cause some problems.
  • Let's say we have a pattern abb and we want to match from ["abc","deq","mee","aqq","dkd","ccc"]. Now if we make an unordered map with abb to for each string mapping it will show us that mee is matching, aqq is matching and also ccc is matching. You can only check if a is mapped to only c or not, not the other way around. So the test case acc will fail our algorithm. So let's try to map the reverse way round.
  • We'll map the string from the string array to the pattern string. Now the test case mee will have m->a mapping, e->b mapping and once more e->b mapping. Now let's take a look at the test case abc, a->a mapping then b->c mapping and then c->b mapping. So there is another error now. Once b is mapped to b then c can not be mapped to b. b can be only mapped to b.
  • So we've looked at both directions and none of the seems to work, what if we have a bi directional char to char mapping so that we can match one letter to exactly one letter?
  • So we need to implement a BiMap() class to solve the problem.

Below is the implementation for BiMap.

class BiMap {
    unordered_map<char, char> front;
    unordered_map<char, char> back;

    BiMap() {}

    void put(char key, char value) {
        front.insert({key, value});
        back.insert({value, key});

    char checkFront(char key) {
        if (front.find(key) == front.end()) return '~';
        return front[key];

    char checkBack(char value) {
        if (back.find(value) == back.end()) return '~';
        return back[value];

Now using that BiMap we'll solve the problem.

class Solution {
    bool findMatching(string word, string pattern) {

        if (word.size() != pattern.size()) return false;

        int index = 0;

        BiMap b = BiMap();

        while (index < pattern.size()) {
            char charFromWord = word[index];
            char charFromPattern = pattern[index];

            // if bimap returns ~ means it is not in the bimap
            // if both front and back returns ~ that means this is a new character
            if (b.checkFront(charFromWord) == '~' and b.checkBack(charFromPattern) == '~')
                b.put(charFromWord, charFromPattern);

            // else if checkFront and checkBack should return identical mapping
            // m->a and a->m
            // check if it is the same with the pattern?
            if (b.checkFront(charFromWord) != charFromPattern) {
                return false;

            if (b.checkBack(charFromPattern) != charFromWord) {
                return false;


        return true;

    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        vector<string> v;
        for (string str:words) {
            if (findMatching(str, pattern)) {

        return v;


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