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I used the Kadane algorithm to solve this question and wrote a clean code with which could work on negative numbers also.



If the input tree is as depicted in the picture:
The Left View of the tree will be: 2 35 2
This was again a very standard question and solved it using recursion.



I told the interviewer dynamic programming based solution on either including the current element to current sum or excluding the current element and then he asked me to write its code and I gave properly commented code for this.



Down: (row+1,col)
Right: (row, col+1)
Down right diagonal: (row+1, col+1)
It was a standard DP question



V is the number of vertices present in graph G and vertices are numbered from 0 to V-1.
E is the number of edges present in graph G.
The Graph may not be connected i.e there may exist multiple components in a graph.
As I have to find a number of connected components in an undirected graph so I simply use Depth-first search to solve this question by visiting each element and mapped it to the number of component in which it belongs. At last, I print the vertices of each component.



A subtree of a node X is X, plus every node that is a descendant of X.
Look at the below example to see a Binary Tree pruning.
Input: [1, 1, 1, 0, 1, 0, 1, 0, 0, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1]
Output: [1, 1, 1, -1, 1, -1, 1, -1, -1, -1, -1]
For example, the input for the tree depicted in the below image would be :

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null (-1).
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

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