Problem of the day
I practiced a lot of problems related to Data structures on the Coding ninjas platform and other coding platforms. If you are stuck somewhere in a problem and want to see reference, first, try to understand the algorithm and then move on to code. Just see the algorithm that is how to approach first and don't try to understand from the code right off the bat, code has finer implementation variation that can be done on your own if you are clear with the algorithm. Also for big product companies, prepare enough problems of graphs and trees as these are the most important topics for interviews.
Mention a good level of projects in your resume, and also give a brief detail about each and every project.
This round consisted of two coding questions which were of medium level.
You have given a m*n matrix filled with "1" and "0"."1" means you can use the cell and "0" means the cell is blocked. You can move in 4 directions left, right, top, bottom. Every time you move upward direction or downward it costs you 1$. Moving left or right will not be added to cost. So you have to write a code to check whether one can reach to destination index (x,y) starting from (0,0). If ...
You are given a bag of size 'W' kg and provided with the costs of packets with different weights of oranges as a list/array with the name 'cost'. Every i-th position in the cost denotes the price of 'i+1' kg packet of oranges.
If at any point in time the i-th cost is -1, it means that 'i+1' kg packet of orange is unavailable.
You are required to find the minimum total cost to buy ex...
It was a subjective pen and paper coding round. Interviewer gave me one coding question that is:
Given a snake and ladder board, find the minimum number of dice throws required to reach the destination or last cell from source or 1st cell. Basically, the player has total control over the outcome of dice throw and wants to find out the minimum number of throws required to reach the last cell.
If the player reaches a cell which is base of a ladder, the player has to climb ...