SDE - 1

The problem can be solved using simple recursive traversal. We can keep track of level of a node by passing a parameter to all recursive calls. keep track of maximum level also. And traverse the tree in a manner that right subtree is visited before left subtree. Whenever we see a node whose level is more than maximum level so far, we print the node because this is the last node in its level.

Simple DP problem:
The state DP[i][j] will denote maximum value of ‘j-weight’ considering all values from ‘1 to ith’. 
So I considered ‘wi’ (weight in ‘ith’ row) we can fill it in all columns which have ‘weight values > wi’. Now two possibilities can take place: 

Fill ‘wi’ in the given column.
Do not fill ‘wi’ in the given column

Took maximum of these two possibilities

I didn't knew answer at first. Then interviewer told me to think in address direction and with their incremental help I was able to give perfect solution.
Initialize the elements of the array.
&a => This is the pointer to array which points at the same memory address as a.
&a + 1 => It points at the address after the end of the array.
*(a+1) => Dereferencing to *(&a + 1) gives the address after the end of the last element.
*(a+1)-a => Subtract the pointer to the first element to get the length of the array.
Print the size.

Tip 1 : Read Concepts of DBMS thoroughly
Tip 2 : Practice B+ tress from DS perspective as well

I had to construct n-ary tree myself.
Then I used a DFS traversal starting from the root
maximum value among all maximum path sums is the required answer.