simple problem in c language

Question

1.00/5 (4 votes)

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i have given two input strings like

HELLO
OPEN

then string1 (HELLO) compares with string2(OPEN) then same characters are deleted from string1.

output is

HLL

please tell me if any one knows.

Thanks in adv.

Posted 3-Jun-12 19:05pm

sankar071237

Updated 4-Jun-12 10:30am

Aescleal

v2

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Comments

Prasad_Kulkarni 4-Jun-12 1:17am

Homework??

sankar071237 4-Jun-12 2:28am

ya.

you know the answer????

Prasad_Kulkarni 4-Jun-12 2:35am

Yes, Have you tried anything so far? If yes, then I will help you to move on..

6 solutions

Solution 5

As I criticised three of the other solutions up there I thought I'd stick my money where my mouth was and write one of me own:

C++

/***/

int is_capital_letter( char to_test )
{
    return to_test <= 'Z' && to_test >= 'A';
}

/***/

unsigned capital_letter_to_index( char letter )
{
    return letter - 'A';
}

/***/

void mark_letter_to_be_eliminated(
    char elimination_table[ 26 ],
    char letter_to_eliminate )
{
    elimination_table[ capital_letter_to_index( letter_to_eliminate ) ] = 1;
}

/***/

int is_letter_to_be_kept( const char elimination_table[ 26 ], char letter_to_test )
{
    return !elimination_table[ capital_letter_to_index( letter_to_test ) ];
}

/***/

void build_elimination_table(
    const char *to_eliminate,
    char elimination_table[26] )
{
    for( ; *to_eliminate; to_eliminate++ )
    {
        if( is_capital_letter( *to_eliminate ) )
        {
            mark_letter_to_be_eliminated( elimination_table, *to_eliminate );
        }
    }
}

/***/

void remove_duplicates_using_elimination_table(
    const char *source,
    const char elimination_table[26],
    char *destination,
    const char *end_of_destination )
{
    while( *source && destination < end_of_destination )
    {
        if( is_capital_letter( *source ) )
        {
            if( is_letter_to_be_kept( elimination_table, *source ) )
            {
                *destination++ = *source;
            }

            source++;
        }
    }

    *destination = 0;
}

/***/

void remove_duplicates(
    const char *source,
    const char *to_eliminate,
    char *destination,
    const char *end_of_destination )
{
    char elimination_table[26] = { 0 };

    build_elimination_table( to_eliminate, elimination_table );

    remove_duplicates_using_elimination_table(
            source,
            elimination_table,
            destination,
            end_of_destination );
}

This lot works by creating a table of letters that are to be eliminated from the source string as it's copied to the destination string. The functions are all pretty safe and won't crash with invalid input they'll just not copy characters that aren't capital letters. And it'll still work if the source and destination are aliased.

Unlike some of the approaches above which have a complexity of O(n*m) where n is the size of the elimination string and m is the size of the string to be copied, this approach has a complexity of O(n+m).

Anyway you'd call it something like this:

C++

char source[] = "HELLO";
char to_eliminate[] = "OPEN";

remove_duplicates( source, to_eliminate, source, source + sizeof source );

which will show it still works when source and destination refer to the same memory.

Final points: it's broken up into so many functions because...
- I have trouble reading functions with more than one loop or one conditional
- It's easier to unit test small functions

Posted 4-Jun-12 10:23am

Aescleal

Comments

nv3 4-Jun-12 17:01pm

When the strings are long this is the absolute winner. The approach via an elimination table is very professional. What I like is the simplicity of the code; what I don't like is that it's a bit lengthy -- but better length and right than short and wrong :-)

Aescleal 5-Jun-12 3:30am

The reason there are loads of functions is to make unit testing far easier - the functions are defined almost in the order I wrote them. I didn't include the tests as they rely on a C++ unit test framework and defeated the idea of the question being for C.

TRK3 4-Jun-12 19:49pm

OP said "same characters" not "same letters". The way I read that, he means any characters... Not just letters.

So, we'll have to backup and define "characters" -- if we are dealing with ascii - then your solution gets a lot simpler with a 256 byte elimination table (and no check for capital letters).

If we are dealing with unicode characters then it's a whole different issue.

Aescleal 5-Jun-12 3:25am

The original poster only supplied one test case and it just had capitals in it. I suppose I could have gone the extreme programming route and just copied "HLL" to the destination but that would have been rather trivial.

Handling UCS 2 characters could be done in the same way, the full character set only occupies 128K which isn't a lot by modern standards, especially on 64 bit machine.

TRK3 5-Jun-12 13:35pm

Yeah, as long as we aren't including chinese pictographs...

Clearly, we are having more fun with this problem than OP. :^)

michaelmel 4-Jun-12 20:47pm

There is a massive flaw in logic - what does it have to do with CAPITAL letters? And, in fact, with letters in general - it should work on any characters. Using embedded constants -why 26? - at least #define it somewhere. Not parenthesizing e.g.: return to_test <= 'Z' && to_test >= 'A' - whilst precedence is correct coding style is bad in my opinion. There is nothing wrong with functions with more than one loop even if they make you dizzy. And finally, your solution is overcomplicated and bulky, especially in the context of the question.

Aescleal 5-Jun-12 3:28am

Why 26? How soon are the number of characters in the roman alphabet going to change? They haven't changed sine the middle ages as far as I know... It's a bit like #define ONE 1, completely pointless. Maybe the variable name is incorrect - perhaps it should be capital_letter_elimination table or I could have used a typedef to make it clearer...

For the reasons there are loads of functions see the original description and the reply to nv3.

Solution 6

Old School Solution...

C++

void DeleteCommonChars(char * szString1, char * szString2)
{
	char * pchar = szString1;

	while (pchar = strpbrk(pchar, szString2))
		memmove(pchar, pchar + 1, strlen(pchar));
}

Or by blocks as Ascleal's solution demonstrates...

C++

void DeleteCommonCharsByBlocks(char * szString1, char * szString2)
{
	char * pFindChar = szString1;
	char * pDestChar = szString1;
	char * pNextStart = szString1;

	while (pFindChar = strpbrk(pNextStart, szString2))
	{
		size_t MoveSize = (size_t) (pFindChar - pNextStart);

		if (MoveSize)
		{
			memmove(pDestChar, pNextStart, MoveSize);

			pDestChar  += MoveSize;
		}

		pNextStart = pFindChar + 1;
	}

	*pDestChar = '\0';
	strcpy(pDestChar, pNextStart);
}

Posted 4-Jun-12 11:14am

JackDingler

Updated 4-Jun-12 11:57am

v5

Comments

JackDingler 4-Jun-12 17:19pm

An optimization might use a reverse find...

Another might optimize the strlen() out of the loop...

Solution 1

Here is the solution

Program :-

VB

#include<stdio.h>
#include<string.h>

main()
{
   char a[100], b[100];

   printf("Enter the first string\n");
   gets(a);

   printf("Enter the second string\n");
   gets(b);

   if( strcmp(a,b) == 0 )
      printf("Entered strings are equal.\n");
   else
      printf("Entered strings are not equal.\n");

   return 0;
}

For More :-
http://www.programmingsimplified.com/c-program-compare-two-strings[^]

Thanks

Posted 3-Jun-12 19:23pm

Anurag Sarkar

Comments

sankar071237 4-Jun-12 2:20am

your just comparing.

i have given
HELLO is the first string and
OPEN is the second string then

HLL is my out put.

it means both string contains "E","O" same characters then delete those characters in first string and display.

[no name] 4-Jun-12 2:36am

Hey ! i am not going to do your homework Ok .

[no name] 4-Jun-12 2:37am

Just compare & get the function by yourself for deleting it .

Anurag Sarkar 4-Jun-12 2:41am

You are going yo use one more IF

CPallini 4-Jun-12 3:46am

What has it to do with OP question?

Aescleal 4-Jun-12 6:04am

Someone's missed the point here...

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michaelmel · Accepted Answer · 2012-06-03T20:22:00

Solution 2

Brutal force approach, connoisseurs should forgive (hopefully)

C++

C#

void StringTransform()
{
	const int size = 100;
	char a[size], b[size], output[size];
	
	// zeroise
	memset(a, 0, size);
	memset(b, 0, size);
	memset(output, 0, size);

	// set to initial content
	strcpy(a, "HELLO");
	strcpy(b, "OPEN");

	BOOL found;
	int k = 0;

	for (int i = 0 ; i < size ; i++)
	{
		found = FALSE;
		for (int j = 0 ; j < size ; j++)
		{
			if (a[i] == b[j])
			{
				found = TRUE;
				break;
			}
		}
		// only copy from a to output if not found in b
		if (!found)
		{
			output[k++] = a[i];
		}
	}
}

Posted 3-Jun-12 20:22pm

michaelmel

Updated 3-Jun-12 20:38pm

v2

Comments

Aescleal 4-Jun-12 5:59am

Doesn't strike me as being that brute force and quite a direct method. However (which is why I've given it a 3)...

- It's not written in C
- I'd eliminate the variable found, it just complicates things needlessly
- Give the function four parameters and allow aliasing of the destination array with the first input string
- You've made it a variable number arguments function for no reason I can see!

michaelmel 4-Jun-12 20:32pm

1) Droplets of C++ won't hurt. Makes it more compact. The only non ANSI-C thing in the code is BOOL type and initialization for i, j, k variables. Something for the poster to ponder on (after all, we are doing his homework, aren't we?)
2) It is written intentionally in a way that a student can understand hence "found" variable. It is HOMEWORK, remember? Besides, I cannot see how it can be eliminated without changing the solution substantially.
3)
It is not a "variable number of arguments function" - where did you get that idea?

CPallini · Accepted Answer · 2012-06-03T21:45:00

Try:

C++

#include <stdio.h>
#include <stdlib.h>
int main()
{
  char s[] = "HELLO";
  char t[] = "OPEN";

  size_t LS = sizeof(s) / sizeof(s[0])-1;
  size_t LT = sizeof(t) / sizeof(t[0])-1;

  size_t i, j, k, deleted;

  deleted = 0;

  i = LS;
  while (i--)
  {
    for(j=0; j<LT; j++)
    {
      if (s[i] == t[j])
      {
        deleted++;
        for (k=i; k<LS-deleted-1; k++)
        {
          s[k] = s[k+1];
        }
        break;
      }
    }
  }
  s[LS-deleted] = '\0';

  printf("%s \n", s);

  return 0;
}

gameloftyou · Accepted Answer · 2012-06-03T20:44:00

C#

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
    int i = 0,j=0,c = 0;
    int hasFind = 0;
    char str1[100],str2[100],str3[100];
    puts("please input string1(len<100):");
    gets(str1);
    puts("please input string2(len<100):");
    gets(str2);
    for(i=0;i<strlen(str1);++i)
    {
        hasFind = 0;
        for(j=0;j<strlen(str2);++j)
        {
            if(str1[i]==str2[j])
            {
                hasFind = 1;
                break;
            }
        }
        if(!hasFind)
        {
            str3[c++] = str1[i];
        }
    }
    str3[c] = '\0';
    printf("string1 deleted the same characters with string2 is:\n%s\n",str3);
    system("pause");
    return 0;
}

simple problem in c language

6 solutions

Solution 5

Solution 2

Solution 4

Solution 6

Solution 1

Solution 3

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