# All combinations of strings that can be used to dial a number

Given a number, print all possible combinations of strings that can be used to dial the given number in a phone with following specifications.

In the given phone, we can dial,
2 using A or B or C,
3 using D or E or F,
……………….
8 using T or U or V,
9 using W or X or Y or Z,
1 using only 1
0 using 0.

For example if 23, is the given phone number, the program should print AD, AE, AF, BD, BE, BF, CD, CE, CF

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to store digit to characters mapping in hash map. The map stores all characters that can be used dial a digit. We place every possible character for current digit and recur for remaining digits. Below is Java implementation of this idea.

 `// Java program to print all possible key strings ` `// that can be used to dial a phone number. ` `import java.util.HashMap; ` ` `  `class` `ConvertToString ` `{ ` `    ``// A Recursive function to print all combinations ` `    ``// that can be used to dial a given number. ` `    ``// phNo ==> Given Phone Number ` `    ``// i    ==> Current digit of phNo to be processed ` `    ``// hm   ==> Stores characters that can be used to ` `    ``//          to dial a digit. ` `    ``// str  ==> Current output string ` `    ``static` `void` `printStrings(String phNo, ``int` `i, ` `                    ``HashMap hm, ` `                    ``StringBuilder str) ` `    ``{ ` `        ``// If all digits are processed, print output ` `        ``// string ` `        ``if` `(i == phNo.length()) ` `        ``{ ` `            ``System.out.print(str + ``" "``); ` `            ``return``; ` `        ``} ` ` `  `        ``// Get current digit of phNo, and recur for all ` `        ``// characters that can be used to dial it. ` `        ``String s = hm.get(phNo.charAt(i)); ` `        ``for` `(``int` `j = 0; j < s.length(); j++) ` `        ``{ ` `            ``str.append(s.charAt(j)); ` `            ``printStrings(phNo, i+1, hm, str); ` `            ``str.deleteCharAt(str.length()-1); ` `        ``} ` `    ``} ` ` `  `    ``// Prints all possible combinations of strings that ` `    ``// can be used to dial c[]. ` `    ``static` `void` `printStringForNumber(String phNo) ` `    ``{ ` `        ``// Create a HashMap ` `        ``HashMap hm = ` `                    ``new` `HashMap(); ` ` `  `        ``// For every digit, store characters that can ` `        ``// be used to dial it. ` `        ``hm.put(``'2'``, ``"ABC"``); ` `        ``hm.put(``'3'``, ``"DEF"``); ` `        ``hm.put(``'4'``, ``"GHI"``); ` `        ``hm.put(``'5'``, ``"JKL"``); ` `        ``hm.put(``'6'``, ``"MNO"``); ` `        ``hm.put(``'7'``, ``"PQRS"``); ` `        ``hm.put(``'8'``, ``"TUV"``); ` `        ``hm.put(``'9'``, ``"WXYZ"``); ` `        ``hm.put(``'1'``, ``"1"``); ` `        ``hm.put(``'0'``, ``"0"``); ` ` `  `        ``// Create a string to store a particular output ` `        ``// string ` `        ``StringBuilder str = ``new` `StringBuilder(); ` ` `  `        ``// Call recursive function ` `        ``printStrings(phNo, 0, hm, str); ` `    ``} ` ` `  `    ``// Driver code to test above methods ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``// Prints ` `        ``printStringForNumber(``"23"``); ` `    ``} ` `} `

Output :

`AD AE AF BD BE BF CD CE CF `

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