Programming Interview Questions

sahaj

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Comment hidden because of low score. Click to expand.

of 0 vote

With binary elimination it will take log(n) time.

- sahaj October 11, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 16 vote

bool is_power_of_2 (int num) {
     return ((num & (num-1)) == 0);
}

e.g 4 in binary is 100
3 in binary is 011
4&3 000

- sahaj September 27, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

By traversing using two pointers, in every iteration, first pointer will advance by one node and the, second will advance by two nodes. When two pointers matches (i.e. pointing to the same node), you can say their is a loop.

- sahaj September 27, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

when array is not sorted, create an array (which will act like a hash map, index as key). If you get number for which the array[number] is already filled, you found a duplicate. Now, you can stop right there and using summation function for 1toN number, you can find missing number.
sum_of_n = (N (N+1))/2
i.e. missing_number = sum_of_n - (sum_of_array_elements - duplicate_number_found)

If array is sorted, traverse the array, if A[i] != i+1 then i is a duplicate number and again use above formula for missing number.

- sahaj September 27, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 3 vote

Use hash map. In worst case, you will get answer after visiting first (N/2 + 2) numbers.

- sahaj September 26, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

int ptr1incr = 1;
int ptr2incr = 2; //this is for 2x
// You can substitute it with 3, 4, 5 etc for 3x, 4x, 5x respectively

int count = 0;
int ptr1pos = 0;
int ptr2pos = 0;
int N = 3; // Total number of nodes
do {
  count++;
  ptr1pos = (ptr1pos + ptr1incr) % N;
  ptr2pos = (ptr2pos + ptr2incr) % N
} while (ptr1pos != ptr2pos);

float total_cycles = (float) count/N;

- sahaj September 25, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Consider this case:
bus1 10 11
bus2 11 12

Shouldn't you take care of this case? To handle this case, for same pair of arrival and departure times, don't increment count.

- sahaj September 25, 2012 | Flag Reply

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