Question: Given a finite number, find out all the possible strings that can be generated by using encoding found on old telephone keypad.

Example: 223 => bad, bae, aae...etc. ( since 2 = abc , 3 = def ... )

Answer: This is a simple recursion problem. We should recurse the loop for each possible character of every digit in given number and append the character to the output. As soon as output length is equal to the input length process/print it.

A small trick is to properly choose function argument, which should allow us to check the current length of the o/p and also the number of characters for current digits.

Code: Ruby code

`#!usr/bin/ruby\$array = [" ", " ", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]def find_comb(input, length, index, out)if index == length    print out    print "\n"    returnendfor j in 0..(\$array[input[index].to_i].length-1)    out[index] = \$array[input[index].to_i][j]    find_comb(input, length, index+1, out)endendif ARGV[0].index('1') or ARGV[0].index('0')  p "0 and 1 are not allowed."  p "Syntax: ruby telephone.rb number"  exitendfind_comb(ARGV[0], ARGV[0].length, 0, "")`

Complexity:
If we assume that every number on keypad has 3 letters (actually 7 and 9 has 4 words each) then possible number of words are 3^n. Since we are processing every word just once, our algorithm complexity is also O(3^n).

PS: Follow-up to this question might be to remove duplicate in the output generated.
or
To show only the dictionary words in output.