# Phone Number Combinations - Amazon Top Interview Questions

### Problem Statement :

```Given a string digits containing 2 to 9 inclusive, return in sorted lexicographic order all possible strings it could represent when mapping to letters on a phone dialpad.

These are the mappings on a phone dialpad:

| 2 | abc  |
| 3 | def  |
| 4 | ghi  |
| 5 | jkl  |
| 6 | mno  |
| 7 | pqrs |
| 8 | tuv  |
| 9 | wxyz |

Example 1

Input

digits = "23"

Output

["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]```

### Solution :

```                        ```Solution in C++ :

vector<string> keypad = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

vector<string> solve(string digits) {
if (digits.empty()) return {""};
for (auto first : keypad[digits[0] - '0']) {
for (auto rest : solve(digits.substr(1))) {
}
}
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public String[] solve(String digits) {
Map<Character, char[]> digitToChars = new HashMap<>();
digitToChars.put('2', new char[] {'a', 'b', 'c'});
digitToChars.put('3', new char[] {'d', 'e', 'f'});
digitToChars.put('4', new char[] {'g', 'h', 'i'});
digitToChars.put('5', new char[] {'j', 'k', 'l'});
digitToChars.put('6', new char[] {'m', 'n', 'o'});
digitToChars.put('7', new char[] {'p', 'q', 'r', 's'});
digitToChars.put('8', new char[] {'t', 'u', 'v'});
digitToChars.put('9', new char[] {'w', 'x', 'y', 'z'});

List<String> collector = new ArrayList<>();
backtrack(digits, 0, digitToChars, new StringBuilder(), collector);

return collector.toArray(new String[collector.size()]);
}

private void backtrack(String digits, int idx, Map<Character, char[]> digitToChars,
StringBuilder sb, List<String> collector) {
if (digits.length() == idx) {
} else {
for (char c : digitToChars.get(digits.charAt(idx))) {
sb.append(c);
backtrack(digits, idx + 1, digitToChars, sb, collector);
sb.deleteCharAt(sb.length() - 1);
}
}
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, digits):
phone = {}
phone["2"] = list("abc")
phone["3"] = list("def")
phone["4"] = list("ghi")
phone["5"] = list("jkl")
phone["6"] = list("mno")
phone["7"] = list("pqrs")
phone["8"] = list("tuv")
phone["9"] = list("wxyz")

result = []
curr = []

def build(d, curr):
nonlocal phone, result

if len(d) == 0:
result.append("".join(curr))
return

for i in phone[d[0]]:
curr.append(i)
build(d[1:], curr)
curr.pop()

build(list(digits), [])

return result```
```

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