SDE - 1

This problem can be solved using DFS approach.
We maintain a visited 2D array to check if the particular cell is visited or not.We iterate through the whole grid, now we check if it's a 1 and if it's not visited, then we mark visited of this cell as 1 and then call for DFS on all the four sides of the cell because essentially we need to find the number of components in the whole grid. Now we go to next 1 and check for the same condition, so we also keep increasing the count as and when we explore a new component. So finally we return the count of islands.

The optimal solution for this problem is to use Binary Search. We first fix a value for the number of pages as mid of current minimum and maximum. We initialize minimum and maximum as 0 and sum-of-all-pages respectively. If a current mid can be a solution, then we search on the lower half, else we search in higher half. So we need to check if the midvalue can be possible or not. We have to check if we can assign pages to all students in a way that the maximum number doesn’t exceed current value. To do this, we sequentially assign pages to every student, while the current number of assigned pages doesn’t exceed the value. In this process, if the number of students becomes more than m, then the solution is not feasible. If it's not more than m students, then we can allocate so it's feasible.

First we create the copy of node 1 and insert it between node 1 and node 2 in original Linked List, then we create the copy of 2 and insert it between 2 and 3. We continue the same process and add the copy of last after the last node. Next we copy the random links also in the same fashion. Next we restore the original and copy linked lists in this fashion in a single loop. Then finally the next pointer of the last node has to be NULL.

We compare the root's value with both the values given. If root's value is bigger then both the values, then this means that the LCA lies on the left part of root's sub-tree. If both the values are smaller than the gives values, then LCA lies on the right part of root's sub-tree. Suppose if root's value is bigger than one value and smaller than the other value, then root itself is the LCA since both the values lie on different parts of the sub-tree and the common element will be the root itself.

I have explained the importance of OOPS like code reusability and how OOPS is beneficial if there are large projects. Next, I have explained about Polymorphism, Inheritance, Abstraction, and Encapsulation in detail, with an example for each of them.

Tip 1 : Mention about your college and current degree
Tip 2 : Mention about your interests in Computer Science
Tip 3 : Why do you want to join Citrix

Tip 1 : First tell about what do you want to do in the first 2 years as you will be a fresher
Tip 2 : Next mention about the next 3 years since you will be promoted to SDE 2
Tip 3 : How do you improve on skills other than technical skills

Tip 1 : Be very thorough with your project. You should know in and out about your project
Tip 2 : You should be able to explain them clearly, if possible screen share, and explain all the components
Tip 3 : Mention about the challenges you faced and how did you overcome then