Tip 1 : You need a strong aptitude for knowledge as a first tip
Tip 2 : Solve questions from previous coding questions
Tip 3 : Make sure your resume includes at least two projects
Tip 1 : Have at least 2 good projects mentioned in your resume with a link
Tip 2 : Focus on skills, internships, projects, and experiences.
Tip 3 : Make it simple, crisp, and one page
This was technical round in which they gave me two coding question which need to be solved in one hour.



Do not allocate extra space for another array. You need to do this by modifying the given input array in place with O(1) extra memory.
'n' = 5, 'arr' = [1 2 2 2 3].
The new array will be [1 2 3].
So our answer is 3.
Time complexity = O(n) + time taken by erase function.
Space complexity = O(1)
class Solution {
public:
int removeDuplicates(vector& nums) {
int cnt = 1;
for(int i=1; i if(nums[i]!=nums[i-1]){
cnt++;
}
else{
nums.erase(nums.begin()+i);
i=i-1; // to keep i in the same place after performing the deletion.
}
}
return cnt;
}
};



1. There might be duplicates present in the array.
2. The order of the permutations in the output does not matter.
3. Do not use any kind of in-built library functions to find the answer.
Basic backtrack solution
class Solution {
List> result;
public List> permute(int[] nums) {
result = new ArrayList<>();
backtrack(nums, new ArrayList());
return result;
}
public void backtrack(int[] nums, List curr) {
if (curr.size() == nums.length) {
result.add(new ArrayList(curr));
return;
}
for (int i = 0; i < nums.length; i++) {
if (!curr.contains(nums[i])) {
curr.add(nums[i]);
backtrack(nums, curr);
curr.remove(curr.size() - 1);
}
}
}
}
This was technical round taken by the SDE1 at scaler based on problem solving and data structures




Solution in Java
class Solution {
public List spiralOrder(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
int up = 0, down = m-1, left = 0, right = n-1;
List res = new LinkedList<>();
while (res.size() < m * n) {
// top: left to right
if (up <= down) {
for (int i = left; i <= right; i++) {
res.add(matrix[up][i]);
}
up++;
}
// right: up to down
if (left <= right) {
for (int i = up; i <= down; i++) {
res.add(matrix[i][right]);
}
right--;
}
// bottom: right to left
if (up <= down) {
for (int i = right; i >= left; i--) {
res.add(matrix[down][i]);
}
down--;
}
// left: down to up
if (left <= right) {
for (int i = down; i >= up; i--) {
res.add(matrix[i][left]);
}
left++;
}
}
return res;
}
}



At first I thought it was as trivial as using regular expressions but then I thought about it and it's actually easier than that.
Just convert string to float and check whether it passes.
Every programming language already has a function that verifies whether a string is a valid number!
If this is an interview question, it would test your creativity much more than your coding skills.
class Solution:
def isNumber(self, s: str) -> bool:
if 'inf' in s.lower():
return False
try:
float(s)
return True
except:
return False
This was technical + HR round taken by senior Manager and HR of the Scaler
Tell me about yourself
Tell me about the most challenging project
What do you know about scaler
Tell me about the basic pillars of OOPs
How will you merge the sorted linkedlist
what are acid properties
Am I ready to relocate to any location in India
Why should we hire you
How will you resolve conflict with your manager
Will you join us if you got offer from Amazon ?
Tip 1: Make sure your resume is up-to-date
Tip 2: Be confident and focus on your communication
Tip 3: Prepare for the behavioral questions

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
Which keyword is used for inheritance?