Snapdeal Software Developer Interview Questions, Process, and Tips

Round duration - 60 minutes
Round difficulty - Medium

It’ll range from the very very basics of programming to the toughest of DPs. In between questions were being popped up on your projects. If you’ve some worth-discussing development projects in your resume(like I’ve my BTP and an Android game), substantial amount of time goes in discussing that.

Q1. 
Nth Fibonacci Problem Statement
Calculate the Nth term of the Fibonacci series, denoted as F(n), using the formula: F(n) = F(n-1) + F(n-2) where F(1) = 1 and F(2) = 1.
Input:
The first line of each test...read more
Ans. 
The recursive approach involves direct implementation of mathematical recurrence formula. 
F(n) = F(n-1)+F(n-2)
 
Pseudocode :
fibonacci(n):
	if(n<=1)
		return n;
	return fibonacci(n-1) + fibonacci(n-2)
 
This is an exponential approach. 
It can be optimized using dynamic programming. Maintain an array that stores all the calculated fibonacci numbers so far and return the nth fibonacci number at last...read more
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Q2. 
Inorder Traversal of Binary Tree Without Recursion
Given a Binary Tree consisting of 'N' nodes with integer values, your task is to perform an In-Order traversal of the tree without using recursion.
Inpu...read more
Ans. 
Inorder traversal requires that we print the leftmost node first and the right most node at the end. 
So basically for each node we need to go as far as down and left as possible and then we need to come back and go right. So the steps would be : 
1. Start with the root node. 
2. Push the node in the stack and visit it's left child.
3. Repeat step 2 while node is not NULL, if it's NULL then pop the topmost n...read more
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Q3. 
Largest Rectangular Area In A Histogram
Given an array HEIGHTS of length N, where each element represents the height of a histogram bar and the width of each bar is 1, determine the area of the largest rec...read more
Ans. 
Traverse all bars from left to right, maintain a stack of bars. Every bar is pushed to stack once. A bar is popped from stack when a bar of smaller height is seen. When a bar is popped, calculate the area with the popped bar as smallest bar. The current index is the ‘right index’ and index of previous item in stack is the ‘left index’. 
Following is the complete algorithm.

1) Create an empty stack.
 
2) Start fro...read more
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Round duration - 60 minutes
Round difficulty - Hard

Director (search) had come along with the recruitment team this time. So, he was only taking the rounds for all the candidates in this round. Summary, two questions were asked.
After these two, he asked me if I’d any questions for him. I asked about the work culture and what kind of people he was looking for. It’ll give you an idea whether you’re selected or not. Be eager to hear the answer he gives and feel attracted to the prospects he puts forward about the company! AGAIN IMPORTANT, DON’T GIVE UP ANY QUESTION. UNTIL HE DECIDES TO MOVE ON TO THE NEXT. If you’re not able to come up with a solution. Don’t panic. Show your fighting spirit.

Q1. 
Search in a Row-wise and Column-wise Sorted Matrix Problem Statement
You are given an N * N matrix of integers where each row and each column is sorted in increasing order. Your task is to find the positi...read more
Ans. 
The brute force solution is to traverse the array and to search elements one by one. Run a nested loop, outer loop for row and inner loop for the column
Check every element with x and if the element is found then print “element found”. If the element is not found, then print “element not found”.
 
Efficient Approach : The idea here is to remove a row or column in each comparison until an element is found. Start searc...read more
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Q2. 
Closest Pair of Points Problem Statement
Given an array containing 'N' points in a plane, your task is to find the distance between the closest pair of points.
Explanation:
The distance between two poin...read more
Ans. 
Algorithm:
1) Sort all points according to x coordinates.
2) Divide all points in two halves.
3) Recursively find the smallest distances in both subarrays.
4) Take the minimum of two smallest distances. Let the minimum be d. 
5) Create an array that stores all points which are at most d distance away from the middle line dividing the two sets.
6) Find the smallest distance in the created array.
7) Return the minimum of d...read more
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Round duration - 30 minutes
Round difficulty - Easy

Most Important here is do not fake your personality here. They’re HR guys, they’re trained to catch the fake ones. So, be genuine.

Round duration - 60 minutes
Round difficulty - Medium

Technical Interview round with 2 DSA based questions.

Q1. 
Find Pair with Given Sum in BST
You are provided with a Binary Search Tree (BST) and a target value 'K'. Your task is to determine if there exist two unique elements in the BST such that their sum equals ...read more
Ans. 
This problem can be solved using hashing. The idea is to traverse the tree in an inorder fashion and insert every node’s value into a set. Also check if, for any node, the difference between the given sum and node’s value is found in the set, then the pair with the given sum exists in the tree.
Time Complexity : O(n).
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Q2. 
Zig Zag Tree Traversal Problem Statement
Given a binary tree, compute the zigzag level order traversal of the nodes' values. In a zigzag traversal, start at the root node and traverse from left to right a...read more
Ans. 
This problem can be solved using two stacks.
Assume the two stacks are current : current level and next level. We would also need a variable to keep track of the current level order(whether it is left to right or right to left). We pop from the current level stack and print the nodes value. Whenever the current level order is from left to right, push the nodes left child, then its right child to the stack next level.&nb...read more
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Round duration - 60 minutes
Round difficulty - Easy

Technical round with questions on OS and DSA.

Q1. 
Subtree Check in Binary Trees
You are given two binary trees, T and S. Your task is to determine whether tree S is a subtree of tree T, meaning S must match the structure and node values of some subtree i...read more
Ans. 
The recursive approach would be to traverse the tree T in preorder fashion. For every visited node in the traversal, see if the subtree rooted with this node is identical to S.
Time worst-case complexity is O(mn) where m and n are number of nodes in given two trees. 
Auxiliary space : O(n)

The above approach can be optimised to O(n) time complexity. The idea is based on the fact that inorder and preorder/postorder un...read more
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Round duration - 30 minutes
Round difficulty - Easy

HR round with typical behavioral problems.

Round duration - 60 minutes
Round difficulty - Easy

Technical Interview round with questions based on DSA.

Q1. 
LCA of Binary Tree Problem Statement
You are given a binary tree consisting of distinct integers and two nodes, X and Y. Your task is to find and return the Lowest Common Ancestor (LCA) of these two nodes...read more
Ans. 
The recursive approach is to traverse the tree in a depth-first manner. The moment you encounter either of the nodes node1 or node2, return the node. The least common ancestor would then be the node for which both the subtree recursions return a non-NULL node. It can also be the node which itself is one of node1 or node2 and for which one of the subtree recursions returns that particular node.
Pseudo code :
LowestCommonA...read more
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Q2. 
Reverse a Linked List Problem Statement
You are given a Singly Linked List of integers. Your task is to reverse the Linked List by changing the links between nodes.
Input:
The first line of input contai...read more
Ans. 
This can be solved both: recursively and iteratively.
The recursive approach is more intuitive. First reverse all the nodes after head. Then we need to set head to be the final node in the reversed list. We simply set its next node in the original list (head -> next) to point to it and sets its next to NULL. The recursive approach has a O(N) time complexity and auxiliary space complexity.
For solving the question is c...read more
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Q3. 
Linked List Cycle Detection
Determine if a given singly linked list of integers forms a cycle.
Explanation:
A cycle in a linked list occurs when a node's next reference points back to a previous node in...read more
Ans. 
Floyd's algorithm can be used to solve this question.
Define two pointers slow and fast. Both point to the head node, fast is twice as fast as slow. There will be no cycle if it reaches the end. Otherwise, it will eventually catch up to the slow pointer somewhere in the cycle.
Let X be the distance from the first node to the node where the cycle begins, and let X+Y be the distance the slow pointer travels. To catch up, t...read more
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Round duration - 45 minutes
Round difficulty - Medium

Technical round with questions on DSA and DBMS.

Q1. What is an Inner Join?
Ans. 
It is a type of join operation in SQL. Inner join is an operation that returns combined tuples between two or more tables where at least one attribute is in common. If there is no attribute in common between tables then it will return nothing. 

Syntax: 

select * 
from table1 INNER JOIN table2
on table1.column_name = table2.column_name;
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Q2. What is an outer join?
Ans. 
It is a type of Join operation in SQL. Outer join is an operation that returns combined tuples from a specified table even if the join condition fails. There are three types of outer join in SQL i.e. 

Left Outer Join
Right Outer Join
Full Outer Join

Syntax of Left Outer Join: 
select *
from table1 LEFT OUTER JOIN table2
on table1.column_name = table2.column_name;
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Q3. 
Count Unique Rectangles in Grid
Given a grid with 'M' rows and 'N' columns, calculate the total number of unique rectangles that can be formed within the grid using its rows and columns.
Input:
The firs...read more
Ans. 
If the grid is 1×1, there is 1 rectangle. 
If the grid is 2×1, there will be 2 + 1 = 3 rectangles 
If it grid is 3×1, there will be 3 + 2 + 1 = 6 rectangles. 
we can say that for N*1 there will be N + (N-1) + (n-2) … + 1 = (N)(N+1)/2 rectangles
If we add one more column to N×1, firstly we will have as many rectangles in the 2nd column as the first, 
and then we have that same number of 2×M rectangles.&nb...read more
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Round duration - 40 minutes
Round difficulty - Easy

Technical round with questions on DSA and Puzzles.

Q1. You have three jars, each mislabeled. One jar contains only apples, another contains only oranges, and the third contains both apples and oranges. You can pick one fruit from one jar to determine its conte...read more
Ans. 
1 pick of an eatable is required to correctly label the Jars.

Solution :
You have to pick only one eatable from jar C. Suppose the eatable is a candy, then the jar C contains candies only(because all the jars were mislabeled).
Now, since the jar C has candies only, Jar B can contain sweets or mixture. But, jar B can contain only the mixture because its label reads “sweets” which is wrong.

Therefore, Jar A contains sweets. ...read more
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Q2. How can you measure 45 minutes using two identical wires?
Ans. 
If we light a stick, it takes 60 minutes to burn completely. What if we light the stick from both sides? It will take exactly half the original time, i.e. 30 minutes to burn completely.

0 minutes – Light stick 1 on both sides and stick 2 on one side.

30 minutes – Stick 1 will be burnt out. Light the other end of stick 2.

45 minutes – Stick 2 will be burnt out. Thus 45 minutes is completely measured.
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Q3. 
Find All Anagrams in a String
Given a string STR and a non-empty string PTR, your task is to identify all starting indices of PTR’s anagrams in STR.
Explanation:
An anagram of a string is another string...read more
Ans. 
Algorithm : 
Define a map m, n := size of s, set left := 0, right := 0, counter := size of p
define an array ans
store the frequency of characters in p into the map m
for right := 0 to n – 1
if m has s[right] and m[s[right]] is non zero, then decrease m[s[right]] by 1, decrease counter by 1 and if counter = 0, then insert left into ans
otherwise
while left < right{
   if s[left] is not present in m, then incr...read more
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