Showing posts with label Dynamic programming. Show all posts
Showing posts with label Dynamic programming. Show all posts

December 23, 2020

Print different views of Tree like a Top, Bottom, Left, Right?

Question: Print different views of Tree like a Top, Bottom, Left, Right?

Problem link: https://www.hackerrank.com/challenges/tree-top-view/problem
Answer 
I am using Recursive approach to solve this Question, As it is easier to understand and explain, but it can lend up in stack overflow problem in case of a huge tree. It is good if we have some idea about the iterative approach. Followup Question may be asked that is the internal working of TreeMap and its data structure and its time complexity of inserting the data and traversing it.
1. TopView
 
To solve this Question, First, start thinking about how we can traverse the tree.
We can use recursion with the base condition that starts traversing left part of the tree and then right part. Here we got the hit that if we maintain horizontal distance from the root node that can give us the Top/Bottom view tree. 
To get the particular tree we use the level to get the specific Top/Bottom view tree.
Pair is a class that holds the two number that is node value and its level.
Time Complexity: O(n)
Space Complexity: O(n)
2. Bottom View
Only this below condition is changed.In order to get the bottom view of tree.
!map.containsKey(dist) || level >= map.get(dist).level
Time Complexity: O(n)
Space Complexity: O(n)
3. Left View 
now, this becomes easier as here we just need to maintain the max level till now while doing pre-order traversal.
Time Complexity: O(n)
Space Complexity: O(1)
4.Right View
now, this becomes easier as here we just need to maintain the max level till now while doing post-order traversal.
Time Complexity: O(n)
Space Complexity: O(1)
Code Solution :
Console 
6
1 2 5 3 6 4
Tree:  
  1
    \
     2
      \
       5
      /  \
     3    6
      \
       4
Top View: 
1 2 5 6 
Bottom View: 
1 3 4 6 
Left View: 
1 2 5 3 4 
Right View: 
1 2 5 6 4 

January 22, 2019

Fidelity Problem 4: More commonly it is known as the Longest String Chain Problem.

Description: You are provided a dictionary of words in a form of a list of size n.You have to find Longest String Chain Length as per following explanation:
From the dictionary, you pick up a word and in each step, you remove a single letter from it and check whether the remaining word is still in the dictionary. If the remaining word is present in the dictionary you increase the chain length by one. You perform the same operations until you are left with the last remaining word present in the dictionary.
Mind it, you have to remove every possible character in the string and check the remaining word for chain length.
Input :
a
b
ba
bca
bda
bdca
Output: 4