SDE - Intern

You have been given a Binary Tree of integers. You are supposed to return the level order traversal of the given tree.

For example:

For the given binary tree

The level order traversal will be {1,2,3,4,5,6,7}.

The lowest common ancestor (LCA) is a concept in graph theory and computer science.

Let ‘T’ be a rooted tree with ‘N’ nodes. The lowest common ancestor is defined between two nodes, ‘u’ and ‘v’, as the lowest node in ‘T’ that has both ‘u’ and ‘v’ as descendants (where we allow a node to be a descendant of itself). - Wikipedia

For the given tree, The LCA of nodes 5 and 8 will be node 2, as node 2 is the first node that lies in the path from node 5 to root node 1 and from node 8 to root node 1.

Path from node 5 to root node looks like 5 → 2 → 1.

Path from node 8 to root node looks like 8 → 6 → 2 → 1.

Since 2 is the first node that lies in both paths. Hence LCA will be 2.

Given any two nodes ‘u’ and ‘v’, find the LCA for the two nodes in the given Tree ‘T’.

Note: For each test case, the tree is rooted at node 1.

Given an array of ‘N’ integers and ‘Q’ queries. The query is defined as below :-

Given 2 integers ‘l’ and ‘r’ (‘l’ >= 0 and ‘r’ < N) find the maximum subarray sum between ‘l’ to ‘r’ (Inclusive).

Query( ‘l’ ,’r’) = max(arr[i] + arr[i+1] +...arr[j].( i >= ‘l’ and j <= ‘r’)

You have been given a grid containing some oranges. Each cell of this grid has one of the three integers values:

Value 0 - representing an empty cell.

Value 1 - representing a fresh orange.

Value 2 - representing a rotten orange.

Every second, any fresh orange that is adjacent(4-directionally) to a rotten orange becomes rotten.

Your task is to find out the minimum time after which no cell has a fresh orange. If it's impossible to rot all the fresh oranges then print -1.

Note:

1. The grid has 0-based indexing.
2. A rotten orange can affect the adjacent oranges 4 directionally i.e. Up, Down, Left, Right.

Given a singly linked list of integers. Your task is to return the head of the reversed linked list.

For example:

The given linked list is 1 -> 2 -> 3 -> 4-> NULL. Then the reverse linked list is 4 -> 3 -> 2 -> 1 -> NULL and the head of the reversed linked list will be 4.

Follow Up :

Can you solve this problem in O(N) time and O(1) space complexity?

You have been given a permutation of ‘N’ integers. A sequence of ‘N’ integers is called a permutation if it contains all integers from 1 to ‘N’ exactly once. Your task is to rearrange the numbers and generate the lexicographically next greater permutation.

To determine which of the two permutations is lexicographically smaller, we compare their first elements of both permutations. If they are equal — compare the second, and so on. If we have two permutations X and Y, then X is lexicographically smaller if X[i] < Y[i], where ‘i’ is the first index in which the permutations X and Y differ.

For example, [2, 1, 3, 4] is lexicographically smaller than [2, 1, 4, 3].