Consider trees:
1 3
2 3 2
1
pre order traversal is same still (1 , 2 ,3) even though ist tree doesn't
satisfy the criteria . So I don't think it can be determined.
On Wed, Jul 4, 2012 at 4:16 PM, Ashish Goel <[email protected]> wrote:
> Q4
>
>
> vector<String> prefix;
> prefix[0]=NULL;
> prefixCount =1;
> for (int i=0;i<n;i++)
> for (int j=0;j<n;j++)
> for (int k=0; k<prefixCount;k++)
> {
> if (visited[i][j]) continue;
> visited[i][j] = true;
> String s=prefix[k]+a[i][j];
> if (isWord(s) { printWord(s); prefix[k]=s; continue;}
> else if isPrefix(s) prefix[prefixCount++] = s;
> else removePrefix(prefix[k], prefixCount);
> prefix[prefixCount++] = String(a[i][j];
> }
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Wed, Jul 4, 2012 at 4:13 PM, Ashish Goel <[email protected]> wrote:
>
>> 1. inverted hasp map
>> 2. not clear
>> 3. VLR, how do you identify end of L and start of R, question incomplete
>> 4. One problem: consider
>>
>> .......
>> a b...
>> c d...
>> .......
>>
>> if ab is a prefix, can aba be another prefix, i would assume so. But if
>> that is true, i am not sure if this program will come to an end.
>> vector<String> prefix;
>> prefix[0]=NULL;
>> prefixCount =1;
>> for (int i=0;i<n;i++)
>> for (int j=0;j<n;j++)
>> for (int k=0; k<prefixCount;k++)
>> {
>> String s=prefix[k]+a[i][j];
>> if (isWord(s) { printWord(s); prefix[k]=s; continue;}
>> else if isPrefix(s) prefix[prefixCount++] = s;
>> else removePrefix(prefix[k], prefixCount);
>> prefix[prefixCount++] = String(a[i][j];
>> }
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>>
>> On Wed, Jul 4, 2012 at 12:22 PM, Decipher <[email protected]> wrote:
>>
>>> Find the next higher number in set of permutations of a given number
>>
>>
>>
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Thanks & regards
Bhupendra
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