Practo Software Developer Intern Interview Questions and Answers

Easy to medium level of leet code

Q1. 1. Fibonacci series with recursion and without recursion 2. String Palindrome 3. Count all repeated numbers from the array
Ans. 
Answering questions related to Fibonacci series, string palindrome, and counting repeated numbers in an array.
• For Fibonacci series with recursion, write a function that calls itself to calculate the next number in the series.

• For Fibonacci series without recursion, use a loop to calculate the series.

• For string palindrome, compare characters from start and end of the string.

• To count all repeated numbers from the array, u
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Q1. 1.Given an array(0-based indexing), you have to find the max sum of i*A[i] where A[i] is the element at index i in the array. The only operation allowed is to rotate(clock-wise or counter clock-wise) the ...read more
Ans. 
Rotate array to find max sum of i*A[i]
• Rotate array to bring maximum element to front

• Calculate sum of i*A[i] for each rotation

• Keep track of maximum sum found
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Q1. 1.Oops concepts 2.Oops code 2.About Java Spring 3. Array stack questions 4.Project discussion
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Q1. It was based on dynamic programming
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Round duration - 90 mintues
Round difficulty - Easy

Nice environment everything went good.

Q1. Sum of LCM
You are given an integer ‘N’ , calculate and print the sum of :
LCM(1,N) + LCM(2,N) + .. + LCM(N,N) 
where LCM(i,n) denotes the Least Common Multiple of the integers ‘i’ and ‘N’.
Input Form...read more
Ans. Brute force approach
 
Using the mathematical formula, A*B = LCM(A,B) * GCD(A,B) where ‘A’ and ‘B’ are the two integers, and GCD(A,B) denotes the Greatest common divisor of ‘A’ and ‘B’.
For example, suppose 
 
A = 20 and B = 30
Factors of A = {1,2,4,5,10,20}
Factors of B = {1,2,3,5,6,10,15,30}
 
The greatest common factor is 10, hence GCD(20,30) = 10.
 
So LCM(A,B) = A*B / GCD(A,B)
 
All we need to fin...read more
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Q2. Vertical Order Traversal
Given a binary tree, return the vertical order traversal of the values of the nodes of the given tree.
For each node at position (X, Y), (X-1, Y-1) will be its left child positio...read more
Ans. Iterative Approach
Our very basic intuition is that while traversing the given binary tree, we need to keep a track of the horizontal distance of all the nodes of the given binary tree with respect to the root node.
 
We initially pass the horizontal distance as 0 for root. For the left subtree, we pass the ‘HD’ as the Horizontal distance of root minus 1. For the right subtree, we pass the ‘HD’ as Horizontal Distance...read more
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Round duration - 90 minutes
Round difficulty - Hard

It was a tough round focus on the key concepts

Q1. Technical Questions
Derive merge sort complexity.
Different methods to obtain fibonacci series , Space complexity of fibonacci recursion.
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Q2. Design a stack that supports getMin() in O(1) time and O(1) extra space
Implement a SpecialStack Data Structure that supports getMin() in O(1) time and O(1) extra space along with push(), pop(), top(), isE...read more
Ans. Optimal Solution
• You need to make two separate stacks for solving the problem.
• The first stack would have the actual number and the second stack would contain the minimum number present in the current stack.
• Now, when we need to push a number in the stack, we first need to check if the stack is empty or not. If the stack is empty, we simply push the integer in both the stacks. Otherwise, we push the integer in the first s...read more
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Round duration - 90 mintues
Round difficulty - Medium

Goes good no problem at all exam is between 10 to 11:30.

Q1. Game of Dominoes
Rafiq loves to play with dominoes. On his birthday his father gifted him ‘N’ piles of dominoes, where each pile contains a positive number of dominoes stacked above.
Rafiq loves to play ...read more
Ans. Brute Force
Let's reduce our question to a fixed array instead of every subarray of size ‘K’ and try to find an answer specific to that array/list ‘height’, so we could do the same thing for each window, now, as mentioned in the hint, we need to calculate the sum of absolute difference of every element from its median, so after getting the new array, next step would be to calculate the cost of that window, so first we w...read more
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Q2. Pair Sum
You are given an integer array 'ARR' of size 'N' and an integer 'S'. Your task is to return the list of all pairs of elements such that each sum of elements of each pair eq...read more
Ans. Brute Force
• Initialize a list to store our results.
• For each element in the array 'ARR[i]', check if ('ARR[i]' + ‘ARR[j]’), equals to given sum or not, where ‘i’ < ‘j’ < ‘N’.
• If the condition matches, add the pair('ARR[i]', ‘ARR[j]’) to the list. Sort the list of pairs as per the given output format and return this list.
Space Complexity: O(1)Explanation:
O(1).
 
Constant extra space is required.
Time Complexity: O...read more
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Round duration - 90 minutes
Round difficulty - Hard

Goes good no problem at all exam at same time between 10 to 11:30.

Q1. Technical Question
He started asking what problem i faced in my project and how i deal with it.
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Q2. Ninja and Sorted Arrays
Ninja has been given two sorted integer arrays/lists ‘ARR1’ and ‘ARR2’ of size ‘M’ and ‘N’. Ninja has to merge these sorted arrays/lists into ‘ARR1’ as one sorted array. You may hav...read more
Ans. Brute Force
Our main task is that after merging ‘ARR2’ into ‘ARR1’. The resultant ‘ARR1’ is also sorted. So first we simply add all the elements of ‘ARR2’ into ‘ARR1’. Then we can apply any sorting algorithm to sort ‘ARR1’. 
 
Algorithm:
1. We run a loop for ‘i’ = 0 to ‘N’:
   • ‘ARR1[M + i]’ = ‘ARR2[i]’
2. Sort the array/list ‘ARR1’.
3. Finally, return ‘ARR1’.
Space Complexity: O(1)Explanation:
O(1) .
 
Because we are not...read more
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Round duration - 1 hour
Round difficulty - Medium

It was a 1 hour round on coding and problem solving. 

Q1. Search In Rotated Sorted Array
Aahad and Harshit always have fun by solving problems. Harshit took a sorted array and rotated it clockwise by an unknown amount. For example, he took a sorted array = [1, 2,...read more
Ans. 
Step 1 : Initially I applied bubble sort.
Step 2 : The interviewer asked me to optimize the solution
Step 3 : I had to find a pivot and then sort .
Step 4 : I applied Binary search to find the pivot and then performed sort.
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Q2. Merge Two Sorted Linked Lists
You are given two sorted linked lists. You have to merge them to produce a combined sorted linked list. You need to return the head of the final linked list.
Note:
The give...read more
Ans. 
Step 1 : I first merged the linked list and applied sorting.
Step 2 : The interviewer asked me to optimize the solution
Step 3: I then took an empty linked list and starting adding elements by comparing the elements( as they are sorted so no need to compare one element with the complete list) of both the given linked list.
Step 4 : Then interviewer asked me to optimize space complexity.
Step 5: I took the first linked list...read more
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