SDE - 1

You have been given the preorder and inorder traversal of a binary tree. Your task is to construct a binary tree using the given inorder and preorder traversals.

Note:

You may assume that duplicates do not exist in the given traversals.

For example :

For the preorder sequence = [1, 2, 4, 7, 3] and the inorder sequence = [4, 2, 7, 1, 3], we get the following binary tree.

You're given a string 'S' consisting of "{", "}", "(", ")", "[" and "]" .

Return true if the given string 'S' is balanced, else return false.

For example:

'S' = "{}()".

There is always an opening brace before a closing brace i.e. '{' before '}', '(' before ').
So the 'S' is Balanced.

Design and implement a data structure for Least Recently Used (LRU) cache to support the following operations:

1. get(key) - Return the value of the key if the key exists in the cache, otherwise return -1.

2. put(key, value), Insert the value in the cache if the key is not already present or update the value of the given key if the key is already present. When the cache reaches its capacity, it should invalidate the least recently used item before inserting the new item.

You will be given ‘Q’ queries. Each query will belong to one of these two types:

Type 0: for get(key) operation.
Type 1: for put(key, value) operation.

Note :

1. The cache is initialized with a capacity (the maximum number of unique keys it can hold at a time).

2. Access to an item or key is defined as a get or a put operation on the key. The least recently used key is the one with the oldest access time.

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

1. Push(num): Push the given number in the stack.
2. Pop: Remove and return the top element from the stack if present, else return -1.
3. Top: return the top element of the stack if present, else return -1.
4. getMin: Returns minimum element of the stack if present, else return -1.

For Example:

For the following input: 
1
5
1 1
1 2
4
2
3

For the first two operations, we will just insert 1 and then 2 into the stack which was empty earlier. So now the stack is => [2,1]
In the third operation, we need to return the minimum element of the stack, i.e., 1. So now the stack is => [2,1]
For the fourth operation, we need to pop the topmost element of the stack, i.e., 2. Now the stack is => [1]
In the fifth operation, we return the top element of the stack, i.e. 1 as it has one element. Now the stack is => [1]

So, the final output will be: 
1 2 1

Simple location preference, any other offers in hand, etc.