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*dst = (*src) & (-1 >>(8-nbit)); should be rewritten as *dst = (*src) & (-1 <<(8-nbit));

how cum u thought of such a good soln durin interview???

the code s very clean, modular, ..........

i got your point. the below soln will work for sure..try it!!!

algorithm that solves problem in linear time:

1.store the sum in an array from start till the point.
2.maintain the counts of sum in hash table
3.if there exists sum == 0 in the hash table, count of 0 = count of 0 + 1
4.for sum whose count == 2
no_of_sub-arrays = no_of_sub-arrays + 1;
else
for sum whose count > 2 ,let n = count
no_of_sub-arrays = no_of_sub-arrays + nC2 ,C denotes combination
5.repeat step 4 for all sums in hash table
6.print no_of_sub-arrays.

that's it!!!

jsdude, your idea of pushing null is awesome!!!

but there is a small correction. your code will get struck in an infinite loop because the queue will never become empty. after processing all elements the queue at end will always contain null.

so to fix the problem do the following:
before enqueuing null check if the queue is empty.
if (the queue is empty )
break from the loop
else
enqueue null and continue.

correct me if i am wrong..

@epic_coder instead of storing pointers to parent v can use a recursive soln which 'll be storing pointers in in-built stack . but again it requires o(log n) space for recursion but code will look simple with recursion.

4 and 6 are left child and right child of 3.

the above diagram was not clear so i posted again........

1
/ \
2 3
/ \
4 6
/
5

o/p should contain 4...

but acord., to the prgs posted in this site as well as in other sites o/p will not contain 4.

the best soln is to do bfs and keep track start and end nodes at each level...

please correct if i am wrong............

the above diagram was not clear so i posted again........

1
/ \
2 3
/ \
4 6
/
5

o/p should contain 4...

but acord., to the prgs posted in this site as well as in other sites o/p will not contain 4.

the best soln is to do bfs and keep track start and end nodes at each level...

please correct if i am wrong............

1
/ \
2 3
/ \
4 6
/
5

o/p should contain 4...

but acord., to the prgs posted in this site as well as in other sites o/p will not contain 4.

the best soln is to do bfs and keep track start and end nodes at each level...

please correct if i am wrong............

i think this question is asked by the interviewer to check if we are clarifying ambiguity regarding the question.

any person who s good in testing should get to know the requirements at first step.hence if such a qn s asked ask qns to interviewer to make the qn clear.

how will u represent 2^(n+1) - 1 nos with atmost n bits? explain.
with atmost n bits only 2^n nos can be represented..

great thinking, a small optimization........

after creating the 4 matrices(say br, bl, tr, tl), find min = min(br[i][j], bl[i][j], tr[i][j], tl[i][j]);

length of cross with centre at i,j = 2 * min - 1