my solution is in O(n) time and in O(1) space :D :D
On Tue, Aug 9, 2011 at 7:21 PM, sagar pareek <[email protected]> wrote:
> Simple ankit
>
>
>
>
>
> #include<stdio.h>
> #include<string.h>
>
> main()
> {
> int i,j=0,l,count;
> char str[100],tmp;
>
> printf("Please enter the string\n");
> fgets(str,100,stdin);
>
> l=strlen(str);
> tmp=str[0]; count=1;
>
> for(i=1;i<l;i++)
> {
> if(tmp==str[i])
> count++;
> else
> { if(count==1) str[j++]=tmp; else{ str[j++]=tmp;
> sprintf(&str[j++],"%d",count); } count=1; tmp=str[i]; }
> }
> //str[j++]=tmp; sprintf(&str[j++],"%d",count);
> str[j]='\0';
>
> printf("%s\n",str);
> }
>
>
> On Tue, Aug 9, 2011 at 7:05 PM, siddharam suresh
> <[email protected]>wrote:
>
>> #include<stdio.h>
>> #include<string.h>
>> #include<stdlib.h>
>> char * squeez(char *str)
>> {
>> char temp1[100];
>> char * temp;
>> temp=(char *) malloc(strlen(str)+1);
>> int start=0,cursor=0;
>> char mychar;
>> while(str[start]!='\0')
>> {
>> mychar=str[start];
>> while(str[start]==str[start+cursor]) cursor ++;
>> if(cursor==1)
>> sprintf(temp1,"%c",mychar);
>> else
>> sprintf(temp1,"%c%d",mychar,cursor);
>> strcat(temp,temp1);
>> start+=cursor;
>> cursor=0;
>> }
>> printf("%s\n",temp);
>> return (temp);
>> }
>>
>> int main()
>> {
>> char str[100];
>> printf("enter the string:");
>> scanf("%s",str);
>> printf("The squeezed string is: %s",squeez(str));
>> }
>> Thank you,
>> Siddharam
>>
>>
>>
>> On Tue, Aug 9, 2011 at 6:54 PM, Rohit Srivastava
>> <[email protected]>wrote:
>>
>>> google on RLE (run length encoding) its almost similar!!
>>>
>>>
>>> On Tue, Aug 9, 2011 at 6:46 PM, ankit sambyal <[email protected]>wrote:
>>>
>>>> @raghavan: ur approach uses O(n) space
>>>>
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>>>
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>
>
>
> --
> **Regards
> SAGAR PAREEK
> COMPUTER SCIENCE AND ENGINEERING
> NIT ALLAHABAD
>
>
--
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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