For every word we will do the following:
Let's say we are at word W(i). We compute its morse encoding and using a mapping from morseEncoding to number of apparitions, we will keep track of how many times a morse encoding appeared in the dictionary.
So at step i we will increment the number of apparitions of the morse encoding we obtained from W(i) where i=1...n. In order the manage this mapping we can use a hash table that maps strings to integers. We can then compute the answer by iterating through the values of the hash.
If there are no collisions, we print -1.
Let's say we are at word W(i). We compute its morse encoding and using a mapping from morseEncoding to number of apparitions, we will keep track of how many times a morse encoding appeared in the dictionary.
So at step i we will increment the number of apparitions of the morse encoding we obtained from W(i) where i=1...n. In order the manage this mapping we can use a hash table that maps strings to integers. We can then compute the answer by iterating through the values of the hash.
If there are no collisions, we print -1.
#include <iostream> #include <stdio.h> #include <unordered_map> #include <vector> using namespace std; int main() { int n, ans = 0; vector <string> d; unordered_map<char, string> repr; unordered_map<string, int> cnt; for(int i = 0; i < 26; ++ i) { char ch; string code; cin >> ch >> code; repr[ch] = code; } cin >> n; for(int i = 0; i < n; ++ i) { string word; cin >> word; string morse = ""; for(auto ch : word) morse = morse + repr[ch]; ans = max(ans, ++ cnt[morse]); } if(ans == 1) cout << -1 << '\n'; else cout << ans << '\n'; return 0; }
