Tip 1 : Practice Atleast 250 Questions
Tip 2 : Practice Atleast 250 Questions
Tip 3 : Take time to study DSA
Tip 1 : Have some projects on resume.
Tip 2 : Do not put false things on resume.
The test consists of Aptitude questions , English Questions , Technical . The technical questions were from DBMS DS algo . and 3 coding questions Those who clear the first test would have a second test of Psychometric duration 60 Mins on the same day of the test.



1. Choose an index i (1 <= i <= N - 1) and swap A[i] and A[i+1].
2. Choose an index i (1 <= i <= N - K + 1) and if A[i], A[i+1],. . . . , A[i+K-1] all are equal to some character x (x != ‘z’), then you can replace each one with the next character (x + 1) , i.e. ‘a’ is replaced by ‘b’, ‘b’ is replaced by ‘c’ and so on.
You are allowed to perform any operation any number of times(possibly zero) only on string 'A'.
If the given strings are A = ‘xbbx’ and B = ‘xddx’ and K is given as 2. Then it is possible to convert string A into B by applying the second operation two times on index 2 (1 based indexing).
1) Loop from 1st to last element and take count is one
2) And store 1st alphabet in output array and go to next
3) Increase the count if alphabet of that index is same as previous else first store the count as string in array and then insert new alphabet in output array and again go to next element and do the same



The width of each bar is the same and is equal to 1.
Input: ‘n’ = 6, ‘arr’ = [3, 0, 0, 2, 0, 4].
Output: 10
Explanation: Refer to the image for better comprehension:

You don't need to print anything. It has already been taken care of. Just implement the given function.
Traverse every array element and find the highest bars on the left and right sides. Take the smaller of two heights. The difference between the smaller height and the height of the current element is the amount of water that can be stored in this array element.
Follow the steps mentioned below to implement the idea:
Traverse the array from start to end:
For every element:
Traverse the array from start to that index and find the maximum height (a) and
Traverse the array from the current index to the end, and find the maximum height (b).
The amount of water that will be stored in this column is min(a,b) – array[i], add this value to the total amount of water stored
Print the total amount of water stored.



String 'S' is NOT case sensitive.
Let S = “c1 O$d@eeD o1c”.
If we ignore the special characters, whitespaces and convert all uppercase letters to lowercase, we get S = “c1odeedo1c”, which is a palindrome. Hence, the given string is also a palindrome.
Find length of str. Let length be n.
Initialize low and high indexes as 0 and n-1 respectively.
Do following while low index ‘l’ is smaller than high index ‘h’.
If str[l] is not same as str[h], then return false.
Increment l and decrement h, i.e., do l++ and h--
If we reach here, it means we didn’t find a miss
I was asked questions from DS algo and DBMS. I was also given coding questions to solve.



The start time of one chosen meeting can’t be equal to the end time of the other chosen meeting.
'N' = 3, Start = [1, 3, 6], End = [4, 8, 7].
You can organize a maximum of 2 meetings. Meeting number 1 from 1 to 4, Meeting number 3 from 6 to 7.
Idea is to solve the problem using the greedy approach which is the same as Activity Selection Problem.
Sort all pairs(Meetings) in increasing order of second number(Finish time) of each pair.
Select first meeting of sorted pair as the first Meeting in the room and push it into result vector and set a variable time_limit(say) with the second value(Finishing time) of the first selected meeting.
Iterate from the second pair to last pair of the array and if the value of the first element(Starting time of meeting) of the current pair is greater then previously selected pair finish time (time_limit) then select the current pair and update the result vector (push selected meeting number into vector) and variable time_limit.
Print the Order of meeting from vector.



A binary search tree is a binary tree data structure, with the following properties :
a. The left subtree of any node contains nodes with a value less than the node’s value.
b. The right subtree of any node contains nodes with a value equal to or greater than the node’s value.
c. Right, and left subtrees are also binary search trees.
It is guaranteed that,
d. All nodes in the given tree are distinct positive integers.
e. The given BST does not contain any node with a given integer value.


A new key is always inserted at the leaf. Start searching a key from the root until we hit a leaf node. Once a leaf node is found, the new node is added as a child of the leaf node.
Can we overload the constructors
Yes, the constructors can be overloaded by changing the number of arguments accepted by the constructor or by changing the data type of the parame
Is multiple inheritance possible in Java. If so, how?
When the child class extends from more than one superclass, it is known as multiple inheritance. However, Java does not support multiple inheritance. To achieve multiple inheritance in Java, we must use the interface.

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?