This is what i had thought of too, btw, we need to update first in all
cases whether the char is same as a previous one or different.
int minD(char *pStr, int len, char a, char b)
{
if ((!pStr)|| (len <2)) return -1;
if (a == b) return 0;
bool foundA=false;
bool foundB=false;
int startIndex = -1;
int minD=INT_MAX;
while(pStr[i] !='\0')
{
if (*pStr == a)
{
if (foundA) //already found A
{ startIndex = i; }
else if (foundB)
{ minD = min(minD, i-startIndex-1); foundB = false;foundA = true;
startIndex = i; }
}
else (*pStr == b)
{
if (foundB) //already found B
{ startIndex = i; }
else if (foundA)
{ minD = min(minD, i-startIndex-1); foundA = false;foundB = true;
startIndex = i; }
}
i++;
}
if (minD >length) return -1;
return minD;
}
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Sun, Feb 5, 2012 at 11:13 PM, WgpShashank <[email protected]>wrote:
> @ashsih ..here is algo
>
> 1st Traverse array from left side and stop if either *a* or *b *are
> found. Store index of this first occurrence in a variable say first then
> Now traverse *array *after the index *first*. If the element at current
> index *i* matches with either x or y then check if it is different from *
> arr[first]* that we have already found. If it is different then update
> the minimum distance so far we have traversed the array . If it is same
> then update *first * i.e., make *first = i*.
> repeat same again .
>
> time complexity O(N) , space complexity O(1)
>
> Hope you can implement it :) Let me if other way exist to do the same or
> any flaw in approach ?
>
> *Thanks
> Shashank Mani Narayan
> Computer Science & Engineering
> Birla Institute of Technology,Mesra
> ** Founder Cracking The Code Lab "http://shashank7s.blogspot.com/"*
>
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