Cracking The Coding Interview 3rd -- 1.5*

Implement a method to perform basic string compression using the counts of repeated characters. For example, the string "aabcccccaaa" would become a2b1c5a3. If the "compressed" string would not become smaller than the original string, your method should return the original string.

Analysis: This problem only asks for compressing in such way that we don't need to worry about combining "a2" and "a3" together. As required, it would be pretty straight forward that we simply scan original string and create a new string, then compare. 

Solution 1:

Scan every character in string and count its occurrences.

Solution 2:

Further compression, combine "a2" and "a3" into "a5", as a result the order of characters will change. In this algorithm, we use an integer array of size 256 to store the counts of each character as discussed in previous questions.

Solution 3 ... 

According to book solution, string concatenation takes much longer time, as a result we should use stringbuffer (Java). I will post c++ version later.

 Header file

#ifndef __QUESTION_1_5_H_CHONG__
#define __QUESTION_1_5_H_CHONG__

#include <string>

using namespace std;

class Question1_5 {
public:
  string stringCompression(string target);
  string stringCompression_2(string target);
  string stringCompression_3(string target);
  int run();
};
#endif // __QUESTION_1_5_H_CHONG__

Source file

#include "stdafx.h"
#include "Question1_5.h"
#include <iostream>

using namespace std;

string Question1_5::stringCompression(string target)
{
  if (target.size()==0) {
    return target;
  }

  string res = "";  // new string storing results
  char c = target[0];   // previous different character
  int cnt = 1;    // counts on each character
  for (int i=1; i<target.size(); ++i) {
    if (target[i]==c) {
      cnt++;
    }
    else {
      res += c;
      res += char(cnt+'0');
      cnt = 1;
      c = target[i];
    }
  }
  // The loop always misses the last character in string, so we add them back
  res += c;
  res += char(cnt+'0');

  if (res.size()<target.size()) {
    return res;
  }
  else 
    return target;
}

// use ASCII 256 array to storage counts on each character
string Question1_5::stringCompression_2(string target)
{
  if (target.size()==0) {
    return target;
  }

  string res = "";
  int counts[256] = {0};
  for (int i=0; i<target.size(); ++i) {
    counts[int(target[i])]++;
  }
  for (int i=0; i<256; ++i) {
    if (counts[i]!=0) {
      res += char(i);
      res += char(counts[i]+'0');
    }
  }

  if (res.size()<target.size()) {
    return res;
  }
  else return target;
}

string Question1_5::stringCompression_3(string target) 
{
	...
}

int Question1_5::run() 
{
  // assume there's no spaces in string
  string input[] = {"abcdabcd", "aaabbbcccddddaaabbbcccdddd"};
  for (int i=0; i<2; ++i) {
    cout << input[i] << " after compression_1 is " << stringCompression(input[i]) 
      << " after compression_2 is " << stringCompression_2(input[i]) << endl;
  }

  return 0;
}