If abcdef is changed to a1b1c1d1e1f1,
then we need to allocate memory dynamically.
Because length is increased,I think this has no practical
implementation.As "abcdef " serves the same purpose.
On Sunday, 3 June 2012 09:36:25 UTC+5:30, utsav sharma wrote:
>
> @ashish:-algo given in link wiil fail for "abcdef"
>
> @navin:- output of "abcdef" should be "1a1b1c1d1e1f"
>
> On Sun, May 27, 2012 at 3:24 PM, Ashish Goel <ashg...@gmail.com> wrote:
>
>> Will fail for the sing having say 257characters all same
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>> On Sat, May 26, 2012 at 12:26 PM, Navin Gupta <navin.nit...@gmail.com>wrote:
>>
>>> This is called Run-Length-Encoding (RLE) of a string.
>>> Its purpose is to save space.So in case of abcdef,I think the output
>>> needed is abcdef (1 is implicit).
>>> The added benefit is it makes the solution in-place.
>>>
>>> Approach:- (In-place and Linear Time)
>>> Start from the left of string and PREVIOUS_CHAR = str[0]
>>> move forward till u match the CURRENT_CHAR with PREVIOUS_CHAR and keep
>>> count of PREVIOUS_CHAR
>>> At any point if (PREVIOUS_CHAR!=CURRENT_CHAR)
>>> put the count of prev_char next to the start position of the previous
>>> character.
>>>
>>> Below is the working code :-
>>> void torle(char *str)
>>> { int i=0,j=0,k=0,cnt=1;
>>> char cur_char=str[0],num[100];
>>> while(str[j+1])
>>> {
>>> cnt=1;
>>> while(str[j+1]==cur_char && str[j]!='\0'){
>>> j++;
>>> cnt++;
>>> }
>>> str[i++]=cur_char;
>>> if( cnt>9 ){
>>> itoa(cnt,num);
>>> k=0;
>>> while(num[k]) str[i++]=num[k++];
>>> }
>>> else if( cnt>1 && cnt<10 )
>>> str[i++]= cnt+'0';
>>> j++;
>>> if(str[j]) cur_char=str[j];
>>> }
>>> if(i!=0){
>>> if(cnt==1)
>>> str[i++]=cur_char;
>>> str[i]='\0';
>>>
>>> }
>>> }
>>>
>>>
>>> On Saturday, 26 May 2012 04:32:35 UTC+5:30, utsav sharma wrote:
>>>>
>>>> Implement a method to perform basic string compression using the counts
>>>> of repeated characters.(inplace)
>>>>
>>>> eg:- input: "aaabbbbbcdef"
>>>> output:"3a5b1c1d1e1f".
>>>>
>>>> what should be my approach to this problem
>>>>
>>>> if i calculate the size of array required to store the output string
>>>> and start from the last of the array then i wldn't get the right answer
>>>> of above input case.
>>>>
>>>> and if start from front then i wldn't get the right answer of this
>>>> input case
>>>> eg:- input: "abcdef"
>>>> output: "1a1b1c1d1e1f"
>>>>
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