Microsoft IDC Interview Experience
Test:
- Find first non-repeating character in the string.
Input: “aabcbd”
Output: c
- K-reverse linked list
Input: 1 2 3 4 5 -1 k = 3
Output: 3 2 1 5 4
- Cut short binary search tree in the range of given integers.
Input: 7 5 10 4 6 8 -1 2 -1 -1 -1 -1 9 1 3 -1 -1 -1 -1 -1 -1 (level order input)
Lower range = 4 upper range = 8
Output:
7: 5, 8
5: 4, 6
8: -1, 9
4: -1, -1
6: -1, -1
9: -1, -1
Around 200-300 people gave this round and 80 were able to clear it.
Group fly round:
- Remove and replace the character ‘c’ from a given input string by double characters “**”
Input: “calcic”
Output: **al**i**
- Given a binary tree whose structure is as below
Class BSTspecial{
public:
BSTspecial* parent = NULL;
BSTspecial* left = NULL;
BSTspecial* right = NULL;
int data;
BSTspecial(int data){
this->data = data;
}
};
Given a node (note it can be either of nodes of the tree whether it is root or not) you need to find its immediate right sibling/cousin if any or return NULL if not present.
Input:
1 2 3 4 5 -1 6 -1 -1 -1 -1 -1 -1(level order input)
For node ‘5’ answer is ‘6’
For node ‘4’ answer is ‘5’
For node ‘6’ answer is -1
For node ‘1’ answer is -1
Out of 80 people, 14 were shortlisted for interviews.
Interview:
Round #1:
- What do you understand by time complexity? And what was your time complexity for the question which were asked in group fly?
- You are given n strings and a string joining function which takes two arguments (both strings) its time complexity is such than it is the sum of lengths of both the strings.
If s1 is k units long and s2 is l units long T.C = O(k + l)
Now, you are required to generate an algorithm such that minimal time is taken to join n strings.
Strings: s1, s2, s3, s4, ……………………………, sn.
Lengths: l1, l2, l3, l4, ……………………………., ln.
Hint: O(log(n)*(l1 + l2 + l3 + …….. + ln)) might not be the best way, this only works for strings with almost equal length.
- Find whether a given linked list is palindrome or not.
Without extra space in O(n) and not breaking any links.
Note: recursion’s stack space will be ignored.
Round #2:
Due to late night Round #3 and HR were clubbed we were only 4 people, but I was asked to leave.
Last Updated :
25 Oct, 2018
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