Microsoft SDE Intern Interview Experience
💼 LTIMindtree Interview Experience (On-Campus) | Fresher | 2026
Salesforce SMTS | Interview Experience | Rejected
JPMC | SDE2 (Associate) - Java Backend - Interview Experience + Compensation
Microsoft - SDE2 - Coding Round
Atlassian | Karat
Summary
I had a Karat interview for Atlassian which consisted of two sections: scenario-based design questions and Data Structures & Algorithms (DSA) problems. I was able to implement one DSA question and provide an approach for the second, though it needed more time.
Full Experience
It consisted of 2 section
- He fired somewhere like 5-7 question on scenerio based design , like given scenerio if present flow work or not / what can done to build etc like this.
- DSA question expection was to implement 1 question and explaon approach for the second if time is there implement it it also has some sort of partial/ step marking as well.
Question 1 : was given a word and a list of words we need to find if given word can be formed from same character present in any word of the list.
Word1 : sky
list - <cry,edt,srk,srrfkrey>
-> answer = srrfkrey
string findSimilar(vector<string> words,string note){
unordered_map<char, int>noteFreq;
for(auto ch:note){
noteFreq[ch]++;
}
unordered_map<char, int>wordFreq;
unordered_map<char, int>duplicateNoteFreq;
bool notFound= false;
for(string word:words){
duplicateNoteFreq = noteFreq;
notFound = false;
for(auto ch:word){
duplicateNoteFreq[ch]--;
if(duplicateNoteFreq[ch]<0){
notFound= true;
}
}
if(!notFound){
return word;
}
}
return "-";
}Question 2:
grid1 = [
['b', 'b', 'b', 'a', 'l', 'l', 'o', 'o'],
['b', 'a', 'c', 'c', 'e', 's', 'c', 'n'],
['a', 'l', 't', 'e', 'w', 'c', 'e', 'w'],
['a', 'l', 'o', 's', 's', 'e', 'c', 'c'],
['w', 'o', 'o', 'w', 'a', 'c', 'a', 'w'],
['i', 'b', 'w', 'o', 'w', 'w', 'o', 'w']
]
word1_1 = "access" # [(1, 1), (1, 2), (1, 3), (2, 3), (3, 3), (3, 4)]
word1_2 = "balloon" # [(0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (1, 7)]
Able to do question this impl but need some extra time to make it working
void bfsHelper(vector<vector<char>>grid1,string word,int ind, vector<pair<int,int>>&ans , vector<vector<bool>>visited, int r,int c, int row, int col){
if(ind == word.length())
return;
int dx[2]={0,-1};
int dy[2]={1, 0};
for(int i =0;i<2;i++){
int xx = r+dx[i];
int yy = c +dy[i];
if(xx < row && xx >=0 && yy<col && yy>=0 ){
if(grid1[xx][yy]==word[ind] && visited[r][c]== false){
ans.push_back({xx,yy});
visited[r][c]= true;
bfsHelper(grid1,word,ind+1,ans,visited,xx,yy,row,col);
ans.pop_back();
visited[r][c]= false;
} else {
continue;
}
}
}
return;
}
vector<pair<int,int>> findPath(vector<vector<char>>grid1, string word){
int row = grid1.size();
int col = grid1[0].size();
vector<vector<bool>>visited(row,vector<bool>(col,false));
vector<pair<int,int>>ans;
for(int i=0;i<row;i++){
for(int j=0;j<col;j++){
if(grid1[i][j]==word[0]){
ans.push_back({i,j});
visited[i][j]=true;
bfsHelper(grid1,word,1,ans,visited,i,j,row,col);
ans.pop_back();
visited[i][j]=false;
}
}
}
return ans;
}Interview Questions (2)
was given a word and a list of words we need to find if given word can be formed from same character present in any word of the list.
Word1 : sky
list - <cry,edt,srk,srrfkrey>
-> answer = srrfkrey
grid1 = [
['b', 'b', 'b', 'a', 'l', 'l', 'o', 'o'],
['b', 'a', 'c', 'c', 'e', 's', 'c', 'n'],
['a', 'l', 't', 'e', 'w', 'c', 'e', 'w'],
['a', 'l', 'o', 's', 's', 'e', 'c', 'c'],
['w', 'o', 'o', 'w', 'a', 'c', 'a', 'w'],
['i', 'b', 'w', 'o', 'w', 'w', 'o', 'w']
]
word1_1 = "access" # [(1, 1), (1, 2), (1, 3), (2, 3), (3, 3), (3, 4)]
word1_2 = "balloon" # [(0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (1, 7)]