C interview question---run-length coding of strings [closed] - c

Question

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.
Closed 11 years ago.
A friend of mine was asked the following question a Yahoo interview:
Given a string of the form "abbccc" print "a1b2c3". Write a function that takes a string and return a string. Take care of all special cases.
How would you experts code it?
Thanks a lot

Answer 1

if (0==strcmp(s, "abbccc"))
return "a1b2c3";
else
tip_the_interviewer(50);
Taken care of.

Answer 2

There's more than one way to do it, but I'd probably run over the input string twice: once to count how many bytes are required for the output, then allocate the output buffer and go again to actually generate the output.
Another possibility is to allocate up front twice the number of bytes in the input string (plus one), and write the output into that. This keeps the code simpler, but is potentially very wasteful of memory. Since the operation looks like a rudimentary compression (RLE), perhaps it's best that the first implementation doesn't have the output occupy double the memory of the input.
Another possibility is to take a single pass, and reallocate the output string as necessary, perhaps increasing the size exponentially to ensure O(N) overall performance. This is quite fiddly in C, so probably not the initial implementation of the function, especially in interview conditions. It's also not necessarily any faster than my first version.
However it's done, the obvious "special case" is an empty input string, because the obvious (to me) implementation will start by storing the first character, then enter a loop. It's also easy to write something where the output may be ambiguous: "1122" is the output for the input "122", but perhaps it is also the output for the input consisting of 122 1 characters. So you might want to limit run lengths to at most 9 characters (assuming base 10 representation) to prevent ambiguity. It depends what the function is for - conjuring a complete function specification from a single example input and output is not possible.
There's also more than one way to design the interface: the question says "returns a string", so presumably that's a NUL-terminated string in a buffer newly-allocated with malloc. In the long run, though, that's not always a great way to write all your string APIs. In a real project I would prefer to design a function that takes as input the string to process, together with a pointer to an output buffer and the length of that buffer. It returns either the number of bytes written, or if the output buffer isn't big enough it returns the number which would have been written. Implementing the stated function using this new function is easy:
char *stated_function(const char *in) {
size_t sz = new_function(in, NULL, 0);
char *buf = malloc(sz);
if (buf) new_function(in, buf, sz);
return buf;
}
I'm also confused what "print" means in the question - other answerers have taken it to mean "write to stdout", meaning that no allocation is necessary. Does the interviewer want a function that prints the encoded string and returns it? Prints and returns something else? Just returns a string, and is using "print" when they don't really mean it?

Answer 3

Follow the following algo and implement it.
Run a loop for all the letters in
string.
Store the first character in a temp
char variable.
For each change in character
initialize a counter with 1 and
print the count of previous
character and then the new letter.

Answer 4

This smells like a homework question, but the code was just too much fun to write.
The key ideas:
A string is a (possibly empty) sequence of nonempty runs of identical characters.
Pointer first always points to the first in a run of identical characters.
After the inner while loop, pointer beyond points one past the end of a run of identical characters.
If the first character of a run is a zero, we've reached the end of the string. The empty string falls out as an instance of the more general problem.
The space required for a decimal numeral is always at most the length of a run, so the result needs at most double the memory. The code works fine with a run length of 53: valgrind reports no memory errors.
Pointer arithmetic is beautiful.
The code:
char *runcode(const char *s) {
char *t = malloc(2 * strlen(s) + 1); // eventual answer
assert(t);
char *w = t; // writes into t;
const char *first, *beyond; // mark limits of a run in s
for (first = s; *first; first = beyond) { // for each run do...
beyond = first+1;
while (*beyond == *first) beyond++; // move to end of run
*w++ = *first; // write char
w += sprintf(w, "%d", beyond-first); // and length of run
}
*w = '\0';
return t;
}
Things I like:
No auxiliary variable for the character whose run we're currently scanning.
No auxiliary variable for the count.
Reasonably sparing use of other local variables.

Answer 5

As others have pointed out, the spec is ambiguous. I think that's fine for an interview question: the point may well be to see what the job applicant does in an ambiguous situation.
Here's my take on the code. I've made some assumptions (since I can't very well ask the interviewer in this case):
This is a simple form of run-length encoding.
Output is of the form {character}{count}.
To avoid ambiguity, the count is 1..9.
Runs of the same character longer than 9 are split into multiple counts.
No dynamic allocation is done. In C, it's usually better to let caller take care of that. We return true/false to indicate if there was enough space.
I hope the code is clear enough to stand on its own. I've included a test harness and some test cases.
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static void append(char **output, size_t *max, int c)
{
if (*max > 0) {
**output = c;
*output += 1;
*max -= 1;
}
}
static void encode(char **output, size_t *max, int c, int count)
{
while (count > 9) {
append(output, max, c);
append(output, max, '0' + 9);
count -= 9;
}
append(output, max, c);
append(output, max, '0' + count);
}
static bool rle(const char *input, char *output, size_t max)
{
char prev;
int count;
prev = '\0';
count = 0;
while (*input != '\0') {
if (*input == prev) {
count++;
} else {
if (count > 0)
encode(&output, &max, prev, count);
prev = *input;
count = 1;
}
++input;
}
if (count > 0)
encode(&output, &max, prev, count);
if (max == 0)
return false;
*output = '\0';
return true;
}
int main(void)
{
struct {
const char *input;
const char *facit;
} tests[] = {
{ "", "" },
{ "a", "a1" },
{ "aa", "a2" },
{ "ab", "a1b1" },
{ "abaabbaaabbb", "a1b1a2b2a3b3" },
{ "abbccc", "a1b2c3" },
{ "1", "11" },
{ "12", "1121" },
{ "1111111111", "1911" },
{ "aaaaaaaaaa", "a9a1" },
};
bool errors;
errors = false;
for (int i = 0; i < sizeof(tests) / sizeof(tests[0]); ++i) {
char buf[1024];
bool ok;
ok = rle(tests[i].input, buf, sizeof buf);
if (!ok || strcmp(tests[i].facit, buf) != 0) {
printf("FAIL: i=%d input=<%s> facit=<%s> buf=<%s>\n",
i, tests[i].input, tests[i].facit, buf);
errors = true;
}
}
if (errors)
return EXIT_FAILURE;
return 0;
}

Answer 6

int priya_homework(char *input_str, char *output_str, int out_len)
{
char pc,c;
int count=0,used=0;
/* Check for NULL and empty inputs here and return*/
*output_str='\0';
pc=*input_str;
do
{
c=*input_str++;
if (c==pc)
{
pc=c;
count++;
}
else
{
used=snprintf(output_str,out_len,"%c%d",pc,count);
if (used>=out_len)
{
/* Output string too short */
return -1;
}
output_str+=used;
out_len-=used;
pc=c;
count=1;
}
} while (c!='\0' && (out_len>0));
return 0;
}

Answer 7

Damn, thought you said C#, not C. Here is my C# implementation for interest's sake.
private string Question(string input)
{
var output = new StringBuilder();
while (!string.IsNullOrEmpty(input))
{
var first = input[0];
var count = 1;
while (count < input.Length && input[count] == first)
{
count++;
}
if (count > input.Length)
{
input = null;
}
else
{
input = input.Substring(count);
}
output.AppendFormat("{0}{1}", first, count);
}
return output.ToString();
}

Answer 8

Something like this:
void so(char s[])
{
int i,count;
char cur,prev;
i = count = prev = 0;
while(cur=s[i++])
{
if(!prev)
{
prev = cur;
count++;
}
else
{
if(cur != prev)
{
printf("%c%d",prev,count);
prev = cur;
count = 1;
}
else
count++;
}
}
if(count)
printf("%c%d",prev,count);
printf("\n");
}