Tip 1 : Have atleast one internship in Data science or ML domain...Will help you to gain exposure
Tip 2 : Do good practice of advanced data structures like Tries,graphs etc.
Tip 3 : Be good in your communication
Tip 1 : Keep your resume up to date and mention three or four good level projects which will give a good impression to the interviewer
Tip 2 : You should be well aware and knowledgeable about all the things that are mentioned in your resume.



If N = 4, K = 2 and subjects = {50,100,300,400}
Assignment of problems can be done in the following ways among the two friends.
{} and {50,100,300,400}. Time required = max(0, 50+100+300+400) = max(0, 850) = 850
{50} and {100,300,400}. Time required = max(50, 100+300+400) = max(50, 800) = 800
{50, 100} and {300,400}. Time required = max(50+100, 300+400) = max(150, 700) = 700
{50,100,300} and {400}. Time required = max(50+100+300, 400) = max(450, 400) = 400
{50,100,300, 400} and {}. Time required = max(50+100+300+400, 0) = max(850, 0) = 850
So, out of all the above following ways, 400 is the lowest possible time.



If A = [3, 2, 3], and K = 2.
Then max of [3, 2] = 3 and max of [2, 3] = 3
So, the answer will be [3, 3]
If A = [3, 2, 3, 5, 1, 7] and K = 3.
Then max of [3, 2, 3] = 3
Then max of [2, 3, 5] = 5
Then max of [3, 5, 1] = 5
Then max of [5, 1, 7] = 7
So the answer will be [3, 5, 5, 7]
Can you solve the problem in O(N) time complexity and O(K) space complexity?
1. Create a queue to store k elements.
2. Run a loop and insert till index less than k(window size) in the queue.
3. Then for an index greater than equal to k, check the greater element of the current queue and output it. And then to slide the window, dequeue the first element of the queue, and enqueue new element of the current index in the queue.
4. And when the loop completes, check the greater element of the queue outside the loop for the last window



1. Coordinates of the cells are given in 0-based indexing.
2. You can move in 4 directions (Up, Down, Left, Right) from a cell.
3. The length of the path is the number of 1s lying in the path.
4. The source cell is always filled with 1.
1 0 1
1 1 1
1 1 1
For the given binary matrix and source cell(0,0) and destination cell(0,2). Few valid paths consisting of only 1s are
X 0 X X 0 X
X X X X 1 X
1 1 1 X X X
The length of the shortest path is 5.
1. Initialize distances of all vertices as infinite.
2. Create an empty priority_queue pq. Every item of pq is a pair (weight, vertex). Weight (or distance) is used as the first item of pair as the first item is by default used to compare two pairs
3. Insert source vertex into pq and make its distance as 0. While either pq doesn't become empty
a) Extract minimum distance vertex from pq. Let the extracted vertex be u.
b) Loop through all adjacent of u and do following for every vertex v.
If there is a shorter path to v through u.
If dist[v] > dist[u] + weight(u, v)
(i) Update distance of v, i.e., do dist[v] = dist[u] + weight(u, v)
(ii) Insert v into the pq (Even if v is already there)
Introduce yourself, why do you want to join the company, what do you know about the company etc.

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
To make an AI less repetitive in a long paragraph, you should increase: