I gave an O(N) solution in a different thread by same author for this
question...
On Thu, Jul 21, 2011 at 6:08 PM, Abhi <abhi123khat...@gmail.com> wrote:
> My solution for this :
>
> #include<stdio.h>
> int max(int a,int b)
> {
> return a>b?a:b;
> }
>
> int main()
> {
> char str[] = "abcdab";
> int count=0,max1=0;
> int i=0,j,k;
> int hash[26];
> for(i=0;i<26;i++)
> hash[i]=-1;
> for(i=0;i<strlen(str);i++)
> {
> count=0;
> for(j=i;hash[str[j]-'a']==-1;j++)
> {
>
> hash[str[j]-'a'] = 1;
> count++;
> }
>
> max1=max(count,max1);
> for(k=0;k<26;k++)
> hash[k]=-1;
>
>
> }
> printf("%d ",max1);
> getch();
> return 0;
> }
>
> Worst case running time : O(n^2) when string is of the form "abcdeabcde".
>
> Does there exist an O(n) solution for this?
>
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--
Ankur Khurana
Computer Science
Netaji Subhas Institute Of Technology
Delhi.
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