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Associate machine learning engineer

Flix Stock
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4 rounds | 4 Coding problems

Interview preparation journey

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Preparation
Duration: 6 months
Topics: Machine learning, deep learning, Computer vision, image processing, AI
Tip
Tip

Tip 1 : focus on maths
Tip 2 : Focus on intution

Application process
Where: Other
Eligibility: published Research paper, Good ML projects, 7 cgpa
Resume Tip
Resume tip

Tip 1 : include project links
Tip 2 : must include research work and github link

Interview rounds

01
Round
Medium
Assignment
Duration2 Days
Interview date10 Jun 2022
Coding problem1

1. Assignment

Garments in the fashion domain can be of multiple shapes, sizes, and colors. Finding garments similar to each other is an important feature used by e-commerce websites to show recommendations to its users. We would like to find visually similar garments for any input garment from within a given dataset of garment images. Write a python solution to find the 10 most similar images to a query image from within a database of images

02
Round
Medium
Video Call
Duration60 Minutes
Interview date20 Jul 2022
Coding problem2

1. Integer To Roman Numeral

Easy
10m average time
90% success
0/40
Asked in companies
AdobeFacebookPayPal

Given an integer ‘N’, the task is to find its corresponding Roman numeral.

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol      Value
  I           1
  V           5
  X           10
  L           50
  C           100
  D           500
  M           1000

Example :

2 is written as II in the roman numeral, just two one’s added together. 
12 is written as XII, which is simply X(ten) + II(one+one). 
The number 27 is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. 
However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four.
The same principle applies to the number nine, which is written as IX.

There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Problem approach

how we can take each place in the decimal system (times a power of 10) and add the numbers for our total number, similarly we can separately evaluate and then concatenate THOUS + HUNDS + TENS + ONES for the total roman numeral.

Furthermore, for any place X*2 = XX)
If the place is 4, the term is UNIT + HALF (Ex. 4 -> I + V = IV)
If the place is 5 or more, the term is HALF + UNIT (place-5) (Ex. 800 -> D + C * (8-5) = DCCC)
If the place is 9, the term is UNIT + NEXT_UNIT(Ex. 90 -> X + C = XC)
So, for all relevant powers of 10, we can go through the above 4 considerations, only changing our definitions of UNIT, HALF, and NEXT_UNIT. This reduces our written code while maintaining legibility.]

Try solving now

2. Populating Next Right Pointers In Each Node

Moderate
25m average time
75% success
0/80
Asked in companies
AmazonMathworksMorgan Stanley

You have been given a complete binary tree of ‘N’ nodes. The nodes are numbered 1 to ‘N’.

You need to find the ‘next’ node that is immediately right in the level order form for each node in the given tree.

Note :

1. A complete binary tree is a binary tree in which nodes at all levels except the last level have two children and nodes at the last level have 0 children.
2. Node ‘U’ is said to be the next node of ‘V’ if and only if  ‘U’ is just next to ‘V’ in tree representation.
3. Particularly root node and rightmost nodes have ‘next’ node equal to ‘Null’ 
4. Each node of the binary tree has three-pointers, ‘left’, ‘right’, and ‘next’. Here ‘left’ and ‘right’ are the children of node and ‘next’ is one extra pointer that we need to update.

For Example :

1

The‘next’ node for ‘1’ is ‘Null’, ‘2’ has ‘next’ node ‘6’, ‘5’ has ‘next’ node ‘3’, For the rest of the nodes check below.

1

Problem approach

I used the level order traversal algorithm as my base and modifying it. The last if condition in the code handles the linking of nodes.

class Solution {
public Node connect(Node root) {
if (root == null) return root;

Queue queue = new ArrayDeque<>();
queue.add(root);

while (!queue.isEmpty()) {
int size = queue.size();

for (int i = 0; i < size; i++) {
Node curr = queue.remove();

if (curr.left != null) {
queue.add(curr.left);
}
if (curr.right != null) {
queue.add(curr.right);
}

if (i < size - 1) {
curr.next = (queue.peek() != null) ? queue.peek() : null;
}
}
}
return root;
}
}

Try solving now
03
Round
Easy
Telephonic
Duration90 Minutes
Interview date22 Jul 2022
Coding problem0

In depth Questions regarding Machine learning, Image processing and other fields in AI. What is image processing. How do you know about its accuracy.

04
Round
Easy
Video Call
Duration60 Minutes
Interview date22 Jul 2022
Coding problem1

Questions regarding your resume, work experience, non tech skills.

1. Basic HR Questions

What keeps you motivated?

Why should we hire you?

Here's your problem of the day

Solving this problem will increase your chance to get selected in this company

Skill covered: Programming

What is recursion?

Choose another skill to practice
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