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2 Coding questions
1. Subtract two numbers given in form of linkedlist
2. LCA application



If the first linked list is 1 -> 2 -> 3 -> 4 -> 5 -> NULL and the second linked list is 4 -> 5 -> NULL.
The two numbers represented by these two lists are 12345 and 45, respectively. So, adding these two numbers gives 12390.
So, the linked list representation of this number is 1 -> 2 -> 3 -> 9 -> 0 -> NULL.
Calculate sizes of given two linked lists.
If sizes are not the same, then append zeros in the smaller linked list.
If the size is the same, then follow the below steps:
Find the smaller valued linked list.
One by one subtract nodes of the smaller-sized linked list from the larger size. Keep track of borrow while subtracting.



Let ‘TREE’ be a binary tree. The lowest common ancestor of two nodes, ‘N1’ and ‘N2’, is defined as the lowest node in ‘TREE’ with ‘N1’ and ‘N2’ as descendants (where we allow a node to be a descendant of itself).
Find a path from the root to n1 and store it in a vector or array.
Find a path from the root to n2 and store it in another vector or array.
Traverse both paths till the values in arrays are the same. Return the common element just before the mismatch.
2 coding questions
1. Backtracking(Similar to rat in a maze)
2. graph(Application of Dijkstra



Here, sorted paths mean that the expected output should be in alphabetical order.
Given a square matrix of size 4*4 (i.e. here 'N' = 4):
1 0 0 0
1 1 0 0
1 1 0 0
0 1 1 1
Expected Output:
DDRDRR DRDDRR
i.e. Path-1: DDRDRR and Path-2: DRDDRR
The rat can reach the destination at (3, 3) from (0, 0) by two paths, i.e. DRDDRR and DDRDRR when printed in sorted order, we get DDRDRR DRDDRR.
1. Create a solution matrix, initially filled with 0’s.
2. Create a recursive function, which takes initial matrix, output matrix and position of rat (i, j).
if the position is out of the matrix or the position is not valid then return.
3. Mark the position output[i][j] as 1 and check if the current position is destination or not. If destination is reached print the output matrix and return.
4. Recursively call for position (i-1,j), (I,j-1), (i+1, j) and (i, j+1).
Unmark position (i, j), i.e output[i][j] = 0.



Input:
4 5
0 1 5
0 2 8
1 2 9
1 3 2
2 3 6

In the given input, the number of vertices is 4, and the number of edges is 5.
In the input, following the number of vertices and edges, three numbers are given. The first number denotes node ‘X’, the second number denotes node ‘Y’ and the third number denotes the distance between node ‘X’ and ‘Y’.
As per the input, there is an edge between node 0 and node 1 and the distance between them is 5.
The vertices 0 and 2 have an edge between them and the distance between them is 8.
The vertices 1 and 2 have an edge between them and the distance between them is 9.
The vertices 1 and 3 have an edge between them and the distance between them is 2.
The vertices 2 and 3 have an edge between them and the distance between them is 6.
1. There are no self-loops(an edge connecting the vertex to itself) in the given graph.
2. There can be parallel edges i.e. two vertices can be directly connected by more than 1 edge.
1) Create a set sptSet (shortest path tree set) that keeps track of vertices included in the shortest-path tree, i.e., whose minimum distance from the source is calculated and finalized. Initially, this set is empty.
2) Assign a distance value to all vertices in the input graph. Initialize all distance values as INFINITE. Assign distance value as 0 for the source vertex so that it is picked first.
3) While sptSet doesn’t include all vertices
….a) Pick a vertex u which is not there in sptSet and has a minimum distance value.
….b) Include u to sptSet.
….c) Update distance value of all adjacent vertices of u. To update the distance values, iterate through all adjacent vertices. For every adjacent vertex v, if the sum of distance value of u (from source) and weight of edge u-v, is less than the distance value of v, then update the distance value of v.

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