Tower Research Interview Experience
Role: Core Software Engineer Intern
College: Tier 2
Experience: 6-month SDE Intern at MakeMyTrip
Rounds: 3 (2 Technical + 1 HR)
Round 1 (DSA + Problem Solving)
Question:
Given a binary grid with exactly two islands (groups of 1s), find the minimum number of 0s that must be flipped to 1s to connect the two islands.
I have provided the code I wrote in the interview below
Show Code
#include <climits>
#include <cmath>
#include <cstdio>
#include <iterator>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
#include <limits>
#include <bits/stdc++.h>
using namespace std;
vector<pair<int,int>>direction={{-1,0},{0,-1},{0,1},{1,0}};
bool insideBoundary(int x,int y,int height,int width){
return (x>=0 and y>=0 and x<height and y<width);
}
int minDistance(int height,int width,vector<vector<int>>&grid){
queue<pair<int,int>> multiSourceBfs;
vector<vector<int>>dist(height,vector<int>(width,INT_MAX));
pair<int,int>source={-1,-1};
for(int row=0;row<height;row++){
for(int col=0;col<width;col++){
if(grid[row][col]==1){
source={row,col};
break;
}
}
}
queue<pair<int,int>> bfs;
bfs.push(source);
while(!bfs.empty()){
pair<int,int> node= bfs.front();
bfs.pop();
if(dist[node.first][node.second]<INT_MAX) continue;
dist[node.first][node.second]=0;
multiSourceBfs.push({node.first,node.second});
for(auto dir:direction){
int nx=node.first+dir.first,ny=node.second+dir.second;
if(insideBoundary(nx,ny,height,width) and grid[nx][ny]==1){
bfs.push({nx,ny});
}
}
}
while(!multiSourceBfs.empty()){
pair<int,int> node= multiSourceBfs.front();
multiSourceBfs.pop();
if(grid[node.first][node.second]==1 and dist[node.first][node.second]>0) return dist[node.first][node.second]-1;
for(auto dir:direction){
int nx=node.first+dir.first,ny=node.second+dir.second;
if(insideBoundary(nx,ny,height,width) and dist[nx][ny]>dist[node.first][node.second]+1){
dist[nx][ny]=dist[node.first][node.second]+1;
multiSourceBfs.push({nx,ny});
}
}
}
return -1;
}
int main() {
int height, width;
cin>>height>>width;
vector<vector<int>>grid(height,vector<int>(width,0));
for(int row=0;row<height;row++){
for(int col=0;col<width;col++){
cin>>grid[row][col];
}
}
int ans=minDistance(height,width,grid);
cout<<"Ans: "<<ans<<endl;
return 0;
}Q2: Is vector memory contiguous? Why? How?
Q3: Difference between pass-by-value and pass-by-reference.
Q4: Threads, cache types (L1, L2, L3), why L1 cache is fast, what is its size, and why don’t we increase it?
Q5: Indexes in databases — why not use multiple indexes, what is partitioning, and what is sharding?
Round 2 (DSA + OOP)
Given multiple orders with statuses, identify completed and cancelled orders.
Completed orders have gone through all stages (placed, processed, shipped, delivered).
Cancelled orders can have cancelled after any stage.
I have provided the code I wrote in the interview below
Show Code
#include <cmath>
#include <cstdio>
#include <unordered_map>
#include <vector>
#include <iostream>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
struct order{
int order_id;
int customer_id;
string status;
order(int order_id,int customer_id,string status): order_id(order_id),customer_id(customer_id),status(status){}
};
class markCompletedOrder{
unordered_map<string,int> statusToStatusCode={
{"cancelled",0},
{"placed",1},
{"processed",2},
{"shipped",3},
{"delivered",4}
};
vector<order>orders;
unordered_map<int,set<int>>orderStatusmap; // key=>order_id, value: set<status_code>
vector<int>cancelled,completed;
void CountCompletedOrder(){
for(auto ord:orders){
if(statusToStatusCode.find(ord.status)==statusToStatusCode.end()){
cout<<"[Error] : invalid status code"<<endl;
return;
}
orderStatusmap[ord.order_id].insert(statusToStatusCode[ord.status]);
}
for(auto val:orderStatusmap){
set<int>allStatus=val.second;
if(*allStatus.begin()==0){
cancelled.push_back(val.first);
continue;
}
if(allStatus.size()==statusToStatusCode.size()-1){
completed.push_back(val.first);
}
}
}
public:
markCompletedOrder(vector<order>&orders){
this->orders=orders;
CountCompletedOrder();
}
int getCompletedOrder(){
return completed.size();
}
int getCancelledOrder(){
return cancelled.size();
}
vector<int>completedOrder(){
return completed;
}
};
int main() {
int countOrders;
cin>>countOrders;
vector<order>orders;
for(int i=0;i<countOrders;i++){
int order_id,customer_id;
string status;
cin>>order_id>>customer_id>>status;
orders.push_back(order(order_id,customer_id,status));
}
markCompletedOrder *completed= new markCompletedOrder(orders);
cout<<completed->getCompletedOrder()<<endl;
cout<<"completed order ids are: ";
for(auto it:completed->completedOrder()){
cout<<it<<" ";
}
cout<<endl;
return 0;
}Round 3 (HR)
Mostly standard HR questions:
- Why Tower Research?
- Walk me through your projects
- Internship experience at MakeMyTrip
- Any offers in hand?
Summary
It was a rigorous yet enjoyable interview experience.
The technical rounds focused heavily on problem-solving, memory management, and understanding of computer architecture.
System design concepts like indexing, sharding, and caching were explored in detail.
Verdict: Selected
