SDE - 1

Traverse once through both the linked lists and find the longer one.
Connect the tail of the longer list to its head.
Now start from the head of the shorter linked list with two pointers as we did in the algorithm to find the cycle.
Find the point where both the pointers meet.
Reset one of the pointer to the head of the shorter linked list.
Now move both the pointers with the same speed (one step at a time).
The place where both the pointers meet is the merge point.

1. We will maintain an array called “islands” which will record the no. of islands visited.
2. Traverse through each element of the grid.
3. Check, if it is a land and it has not been visited yet i.e. grid[i][j] == 1, increment islands and change the value of that element to water i.e. 0.
4. Change the adjacent elements to water i.e. '0'.

the value of a node is the string length. We calculate the number of nodes in the recursion tree for string length=1, 2, ...., n respectively.

For example, when string length=4, the second layer of the recursion tree has three nodes where the string length is 3, 2 and 1 respectively. And the number of subtree rooted at these three nodes have been calculated when we do the mathmatical induction.

So time complexity is O(2^n).

BFS uses the indegrees of each node. We will first try to find a node with 0 indegree. If we fail to do so, there must be a cycle in the graph and we return false. Otherwise we set its indegree to be -1 to prevent from visiting it again and reduce the indegrees of its neighbors by 1. This process will be repeated for n (number of nodes) times.

Tip 1 : Clear out the requirements first.
Tip 2 : Convey your thoughts regularly and don't overthink the problem.
Tip 3 : Design small and then expand.