Tip 1 : Practice as many questions as you can of DS algo
Tip 2 : Do at least 2-3 Development Projects.
Tip 1 : Have some projects on resume.
Tip 2 : Do not put false things on resume.
Test consisted of 10-15 MCQ questions based on OOPs , DS algo , DBMS and 2 coding questions



Suppose given input is "abacb", then the length of the longest substring without repeating characters will be 3 ("acb").



This round was technical round -1 . Interviewer ask questions from resume , and gave coding questions to solve.



1. Delete a character
2. Replace a character with another one
3. Insert a character
Strings don't contain spaces in between.
The idea is to process all characters one by one starting from either from left or right sides of both strings.
Let us traverse from right corner, there are two possibilities for every pair of character being traversed.
m: Length of str1 (first string)
n: Length of str2 (second string)
If last characters of two strings are same, nothing much to do. Ignore last characters and get count for remaining strings. So we recur for lengths m-1 and n-1.
Else (If last characters are not same), we consider all operations on ‘str1’, consider all three operations on last character of first string, recursively compute minimum cost for all three operations and take minimum of three values.
Insert: Recur for m and n-1
Remove: Recur for m-1 and n
Replace: Recur for m-1 and n-1

1. Both the strings contain only lowercase alphabets and can contain duplicates.
2. Return the uncommon characters in lexicographically sorted order.
Concept of hashing can be applied to solve this problem.
1. Create a hash table with a size of 26 for all lowercase characters.
2. Mark each character as '0' (indicating it is not present in both strings).
3. Mark the presence of each character in the 1st string as a '1' in the hash table, denoting the 1st string.
4. Next, traverse the 2nd string. For every character in string 2, check whether its presence in the hash table is '1' or not. If it is a 1, you should mark its presence as a -1 (indicating that the character is common to both strings), otherwise you should mark it as a 2 (indicating the 2nd string).
explain your project.
explain recursion to me.
difference between RDMS and DBMS
SQL program to find salary of a particular person in the office .
what are interfaces in java.
This was technical round -2. This round was tougher than previous round. The interviewer starts by asking questions from DS algo basics , DBMS , OS






Let S = “abdd” and X = “bd”.
The windows in S which contain all the characters in X are: 'abdd', 'abd', 'bdd', 'bd'.
Out of these, the smallest substring in S which contains all the characters present in X is 'bd'.
All the other substring have a length larger than 'bd'.
First check if the length of the string is less than the length of the given pattern, if yes then “no such window can exist “.
Store the occurrence of characters of the given pattern in a hash_pat[].
we will be using two pointer technique basically
Start matching the characters of pattern with the characters of string i.e. increment count if a character matches.
Check if (count == length of pattern ) this means a window is found.
If such a window found, try to minimize it by removing extra characters from the beginning of the current window.
delete one character from first and again find this deleted key at right, once found apply step 5 .
Update min_length.
Print the minimum length window.

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?