# 17. Letter Combinations of a Phone Number

![](/files/-MdX1S9X1Vzi8ZLjt-jY)

![](/files/-MdX1WEflWArSpDoEZs2)

time : 

![](/files/-MdX2VX6gLZJyWdFmNnP)

space: O(n)

```java
class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<>();
        if (digits == null || digits.length() == 0) return res;
        
        helper(res, "", digits, 0);
        return res;
    }
    private String[] mapping = new String[]{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    
    private void helper(List<String> res, String s, String digits, int index) {
        if (s.length() == digits.length()) {
            res.add(s);
            return;
        }
        String letter = mapping[digits.charAt(index) - '0'];
        for (int i = 0; i < letter.length(); i++) {
            helper(res, s + letter.charAt(i), digits, index + 1);
        }
    }
}
```

```java
class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList<>();
        
        if (digits == null || "".equals(digits)) return result;
        
        Map<Character, String> map = new HashMap<>();
        map.put('2', "abc");
        map.put('3', "def");
        map.put('4', "ghi");
        map.put('5', "jkl");
        map.put('6', "mno");
        map.put('7', "pqrs");
        map.put('8', "tuv");
        map.put('9', "wxyz");
        search("", digits, result, map, 0);
        return result;

    }
    private void search(String s, String digits, List<String> result, Map<Character, String> map, int i) {
        
        //terminator
        if (i == digits.length()) {
            result.add(s);
            return;
        }
        String combinations = map.get(digits.charAt(i)); // get first digits' letters
        
        //process
        for (int j = 0; j < combinations.length(); j++) {
            
            // drill down
            String letter = combinations.substring(j, j+1); //use substring is faster
            search(s + letter, digits, result, map, i+1); // i is recursion level
        }
    }
}
```

for example

![](/files/-M1UCBk30HmMrl2uaYke)

## new version

T: O(4^n\*n), at most 4 digits, n is digits length, last \*n => is String concat..

S: O(2\*4 + n), dfs stack is n

```java
class Solution {
    public List<String> letterCombinations(String digits) {
        List<String>  result = new ArrayList<>();
        if ("".equals(digits)) {
            return result;
        }

        Map<Character, List<Character>> map = new HashMap<>();
        map.put('2', Arrays.asList('a', 'b', 'c'));
        map.put('3', Arrays.asList('d', 'e', 'f'));
        map.put('4', Arrays.asList('g', 'h', 'i'));
        map.put('5', Arrays.asList('j', 'k', 'l'));
        map.put('6', Arrays.asList('m', 'n', 'o'));
        map.put('7', Arrays.asList('p', 'q', 'r', 's'));
        map.put('8', Arrays.asList('t', 'u', 'v'));
        map.put('9', Arrays.asList('w', 'x', 'y', 'z'));
        
        dfs(digits, 0, "", map,result);
        return result;
    }
    private void dfs(String digits, int start, String str, Map<Character, List<Character>> map, List<String> result) {
        if (str.length() == digits.length()) {
            result.add(str);
            return;
        }
        char cur = digits.charAt(start);
        for (char c : map.get(cur)) {
            dfs(digits, start+1, str + String.valueOf(c), map, result);
        }
    }
}

/*

23

2 -> abc
3 -> def

  2 a   b    c
  /     |     \ 
def    def    def
*/  

```

## better one I think

T: O(4^n\*n), at last concat String with n times , n is digits length

S: O(n), recursion stack&#x20;

```java
class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList<>();
        if ("".equals(digits)) {
            return result;
        }
        String mappings[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        dfs(digits, 0, mappings, result, new StringBuilder()); // sb will do backtracking to original "", so always use same sb
        return result;
    }
    private void dfs(String digits, int start, String mappings[], List<String> result, StringBuilder sb) {
        if (start == digits.length()) {
            result.add(sb.toString());
            return;
        }
        
        String mapping = mappings[digits.charAt(start) - '0'];
        for (char m : mapping.toCharArray()) {
            sb.append(m);
            dfs(digits, start + 1, mappings, result, sb);
            sb.deleteCharAt(sb.length()-1);
        }
    }
}

/*
  2 [abc]
  //\
a  b c
3 [def]
/   \    \ backtracking
def  def def -> reach result length add to result(List<String>)


4^4
*/
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://timmybeeflin.gitbook.io/cracking-leetcode/dfs-and-bfs/dfs/tips/17.-letter-combinations-of-a-phone-number.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.