Tip 1 : Regarding DSA preparation, I have a theory. 20 percent of the questions will be asked in 80 percent of the interview and 80 percent of the questions will be asked in 20 percent of the interviews. In short, some questions have a very high chance of coming up during the interviews and some have very low chance. We should focus more on the questions that have more chance of coming up in the interview. You can find these questions on Striver SDE Sheet, InterviewBit, Leetcode 100 most liked, Leetcode 100 most important.
Tip 1: Make Sure that your resume is simple and also try to fit all the information on only one page.
Tip 2: Have at least 2 projects with the latest technologies, Github link of projects should be provided



You have been given two Strings “STR1” and “STR2” of characters. Your task is to find the length of the longest common subsequence.



The right view of a Binary Tree is a set of nodes visible when the tree is viewed from the Right side and the nodes are printed from top to bottom order.



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.
You have been given a binary matrix of size 'N' * 'M' where each element is either 0 or 1. You are also given a source and a destination cell, both of them lie within the matrix.



Consider 0-based indexing.
Consider the array 1, 2, 3, 4, 5, 6
We can Jump from index 0 to index 1
Then we jump from index 1 to index 2
Then finally make a jump of 3 to reach index N-1
There is also another path where
We can Jump from index 0 to index 1
Then we jump from index 1 to index 3
Then finally make a jump of 2 to reach index N-1
So multiple paths may exist but we need to return the minimum number of jumps in a path to end which here is 3.
You have been given an array 'ARR' of ‘N’ integers. You have to find the minimum number of jumps needed to reach the last index of the array i.e ‘N - 1’ if at any index ‘i’ we can jump to an index ‘i + k’ such that 1<= ‘k’ <= ARR[i] i.e the element you are currently at represents the maximum distance you can jump from the current element.



Approach is to store the value of the capacity in some variable whenever the capacity becomes less than zero.
In this case You have to traverse the array only once as compared to previous method which traverses the each index twice.

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
How do you remove whitespace from the start of a string?