given a number as a string write a algorithm to map to its oral description. I.e.

"1" - "11" //this can be thought of as there is one one.
"11" - "21" // there are two ones
"21" - "1211" // there is one two and one one
"1211" - "111221" ect. //there is one one, one two and two ones

Question

given a number as a string write a algorithm to map to its oral description. I.e.

"1" -> "11" //this can be thought of as there is one one.
"11" -> "21" // there are two ones
"21" -> "1211" // there is one two and one one
"1211" -> "111221" ect. //there is one one, one two and two ones

Bharath · Accepted Answer

Rio, your code breaks on your own provided example. The following code solves this:

f = string[0]

for i in range(0, len(string)):
    if string[i] == f:
        count = 1
        if i == len(string)-1:
            sys.stdout.write("%d%s" %(count, f))
            break
        while (string[i+1] == string[i]):
            count += 1
            i += 1
            if (i+1) >= len(string):
                break
        sys.stdout.write("%d%s" %(count, f))
        if (i+1) >= len(string):
            break
        f = string[i+1]
print ""

I might have some extra unnecessary code and could definitely clean this up a little, but it's a working version.

Rio · Answer

The above implementation is incorrect. Firstly, for 11 it will return 2 for 1 because it counts the original. The following implementation in Python does the job nicely.

def oralDescription(string):
	firstchar = string[0]
	
	count = 0
	all = ""
	for i in range(0, len(string), 2):
		print "There are", string[i], "of", string[i+1]
		
oralDescription("112111")

J · Answer

In PHP:

function lookAndSay($number) {
  $len = strlen($number);
  $count = 1;
  $output = "";
  
  for($i = 1; $i < $len; $i++) {
    if ($number{$i} === $number{$i-1}) {
      $count++;
    }
    else {
      $output .= $count.$number{$i-1};
      $count = 1;
    }
  }
  
  $output .= $count.$number{$i-1};
  
  return $output;
}

nilkn · Answer

Probably the simplest Python solution:

from itertools import groupby

def oral(num):
    return ''.join([str(len(list(j))) + str(i) for i, j in groupby(num)])

kenneth · Answer

std::string oralDescription(std::string input) {
    if (input.size() == 0) {
        return '';
    }

    std::string output;
    char c = input[0];
    int pos = 0;
    int count = 1;

    while (pos < input.size()) {
        ++pos;
        while (pos < input.size() && input[pos] == c) {
            ++pos;
            ++count;
        }
        output += std::to_string(count) + c;

        if (pos < input.size()) {
            c = input[pos];
            count = 1;
        }
    } 

    return output;
}

Sigiloso · Answer

public static String expand(String input){
		String output = "";
		char prev = input.charAt(0);
		int count = 0;
		for(int i = 0; i = input.length())
				output += count + "" + cur;
		}
		return output;
	}

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