1
2             3   is not a BST and its pre-order traversal is 1 2 3, pre
order of other is 3 2 1 .



On Wed, Jul 4, 2012 at 5:17 PM, Bhupendra Dubey <[email protected]>wrote:

> 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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