LeetCode 1079. Letter Tile Possibilities

原题链接在这里：https://leetcode.com/problems/letter-tile-possibilities/

题目：

You have a set of tiles, where each tile has one letter tiles[i] printed on it.  Return the number of possible non-empty sequences of letters you can make.

Example 1:

Input: "AAB"
Output: 8
Explanation: The possible sequences are "A", "B", "AA", "AB", "BA", "AAB", "ABA", "BAA".

Example 2:

Input: "AAABBC"
Output: 188

Note:

1. 1 <= tiles.length <= 7
2. tiles consists of uppercase English letters.

题解：

Accumlate all the possibilities during the process of calculating permutations.

Time Complexity: expontential.

Space: O(expontential).

AC Java:

 class Solution {
     public int numTilePossibilities(String tiles) {
         if(tiles == null || tiles.length() == 0){
             return 0;
         }

         HashSet<String> hs = new HashSet<>();
         char [] arr = tiles.toCharArray();
         boolean [] used = new boolean[tiles.length()];
         dfs(arr, used, "", hs);
         return hs.size();
     }

     private void dfs(char [] arr, boolean [] used, String item, HashSet<String> hs){
         for(int i = 0; i<used.length; i++){
             if(!used[i]){
                 used[i] = true;
                 hs.add(item+arr[i]);
                 dfs(arr, used, item+arr[i], hs);
                 used[i] = false;
             }
         }
     }
 }

Count the frequency for letter in tiles.

AAB -> A:2, B:1.

Take one A sum++.

The rest is A:1, B:1. All the combinations of rest should be accumlated back to sum. sum += dfs(rest). A, AB, B, BA.

Time Complexity: O(expontential).

Space: O(n). n = tiles.length. n level stack space

AC Java:

 class Solution {
     public int numTilePossibilities(String tiles) {
         if(tiles == null || tiles.length() == 0){
             return 0;
         }

         int [] count = new int[26];
         for(int i = 0; i<tiles.length(); i++){
             count[tiles.charAt(i)-'A']++;
         }

         return dfs(count);
     }

     private int dfs(int [] count){
         int sum = 0;
         for(int i = 0; i<count.length; i++){
             if(count[i] == 0){
                 continue;
             }

             sum++;
             count[i]--;
             sum += dfs(count);
             count[i]++;
         }

         return sum;
     }
 }