[프로그래머스] [C] Lv.1 대충 만든 자판

문제 설명
휴대폰의 자판은 컴퓨터 키보드 자판과는 다르게 하나의 키에 여러 개의 문자가 할당될 수 있습니다. 키 하나에 여러 문자가 할당된 경우, 동일한 키를 연속해서 빠르게 누르면 할당된 순서대로 문자가 바뀝니다.

예를 들어, 1번 키에 "A", "B", "C" 순서대로 문자가 할당되어 있다면 1번 키를 한 번 누르면 "A", 두 번 누르면 "B", 세 번 누르면 "C"가 되는 식입니다.

같은 규칙을 적용해 아무렇게나 만든 휴대폰 자판이 있습니다. 이 휴대폰 자판은 키의 개수가 1개부터 최대 100개까지 있을 수 있으며, 특정 키를 눌렀을 때 입력되는 문자들도 무작위로 배열되어 있습니다. 또, 같은 문자가 자판 전체에 여러 번 할당된 경우도 있고, 키 하나에 같은 문자가 여러 번 할당된 경우도 있습니다. 심지어 아예 할당되지 않은 경우도 있습니다. 따라서 몇몇 문자열은 작성할 수 없을 수도 있습니다.

이 휴대폰 자판을 이용해 특정 문자열을 작성할 때, 키를 최소 몇 번 눌러야 그 문자열을 작성할 수 있는지 알아보고자 합니다.

1번 키부터 차례대로 할당된 문자들이 순서대로 담긴 문자열배열 keymap과 입력하려는 문자열들이 담긴 문자열 배열 targets가 주어질 때, 각 문자열을 작성하기 위해 키를 최소 몇 번씩 눌러야 하는지 순서대로 배열에 담아 return 하는 solution 함수를 완성해 주세요.

단, 목표 문자열을 작성할 수 없을 때는 -1을 저장합니다.

제한사항
1 ≤ keymap의 길이 ≤ 100
1 ≤ keymap의 원소의 길이 ≤ 100
keymap[i]는 i + 1번 키를 눌렀을 때 순서대로 바뀌는 문자를 의미합니다.
예를 들어 keymap[0] = "ABACD" 인 경우 1번 키를 한 번 누르면 A, 두 번 누르면 B, 세 번 누르면 A 가 됩니다.
keymap의 원소의 길이는 서로 다를 수 있습니다.
keymap의 원소는 알파벳 대문자로만 이루어져 있습니다.
1 ≤ targets의 길이 ≤ 100
1 ≤ targets의 원소의 길이 ≤ 100
targets의 원소는 알파벳 대문자로만 이루어져 있습니다.

 

 

 

#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>

typedef struct {
    char key;
    int count;
} KeyPress;

KeyPress* calculateKeyPresses(const char* keymap[], size_t keymap_len, size_t* keypress_len) {
    int minPresses[26];
    for (int i = 0; i < 26; i++) {
        minPresses[i] = -1;
    }

    for (size_t i = 0; i < keymap_len; i++) {
        for (size_t j = 0; j < strlen(keymap[i]); j++) {
            int charIndex = keymap[i][j] - 'A';
            if (minPresses[charIndex] == -1 || minPresses[charIndex] > (int)j) {
                minPresses[charIndex] = j + 1; 
            }
        }
    }

    *keypress_len = 0;
    KeyPress* keypresses = (KeyPress*)malloc(26 * sizeof(KeyPress));
    for (int i = 0; i < 26; i++) {
        if (minPresses[i] != -1) {
            keypresses[*keypress_len].key = 'A' + i;
            keypresses[*keypress_len].count = minPresses[i];
            (*keypress_len)++;
        }
    }

    return keypresses;
}

int* solution(const char* keymap[], size_t keymap_len, const char* targets[], size_t targets_len) {
    int* answer = (int*)malloc(targets_len * sizeof(int));
    size_t keypress_len = 0;
    KeyPress* keypresses = calculateKeyPresses(keymap, keymap_len, &keypress_len);
    /*
    for(int i=0; i<26; i++){
        printf("%c %d\n",keypresses[i].key,keypresses[i].count);
    }
    */
    for (size_t i = 0; i < targets_len; i++) {
        int pressCount = 0;
        bool canType = true;

        for (size_t j = 0; j < strlen(targets[i]); j++) {
            bool found = false;

            for (size_t k = 0; k < keypress_len; k++) {
                if (keypresses[k].key == targets[i][j]) {
                    pressCount += keypresses[k].count;
                    found = true;
                    break;
                }
            }

            if (!found) {
                canType = false;
                break;
            }
        }

        answer[i] = canType ? pressCount : -1;
    }
    
    return answer;
}