Status: Final Year Student, B.Tech. IT MMMUT
Position: SDE Intern at Probo
Location: Gurugram, Haryana
Date: July 2024
Here is the interview experience with Probo:
Round 1
Taken By: Akash Utreja Sir
First Question
Link: Set Matrix Zeroes
Problem: Given an m x n integer matrix, if an element is 0, set its entire row and column to 0's. You must do it in place.
Follow-up: Devise a constant space solution.
Answer:
class Solution {
public:
void setZeroes(vector<vector<int>>& matrix) {
int n = matrix.size();
int m = matrix[0].size();
bool flag1 = false, flag2 = false;
for(int i=0; i<n; i++){
if(matrix[i][0] == 0){
flag1 = true;
}
}
for(int j=0; j<m; j++){
if(matrix[0][j] == 0){
flag2 = true;
}
}
for(int i=1; i<n; i++){
for(int j=1; j<m; j++){
if(matrix[i][j] == 0){
matrix[i][0] = matrix[0][j] = 0;
}
}
}
for(int i=1; i<n; i++){
for(int j=1; j<m; j++){
if(matrix[i][0] == 0 || matrix[0][j] == 0){
matrix[i][j] = 0;
}
}
}
if(flag1 == true){
for(int i=0; i<n; i++){
matrix[i][0] = 0;
}
}
if(flag2 == true){
for(int j=0; j<m; j++){
matrix[0][j] = 0;
}
}
}
};Second Question
Link: Best Time to Buy and Sell Stock II
Problem: Given an integer array prices where prices[i] is the price of a given stock on the ith day, find and return the maximum profit you can achieve.
Answer:
class Solution {
public:
int maxProfit(vector<int>& prices) {
int sum = 0;
int val = prices[0];
int n = prices.size();
for(int i=1;i<n;i++){
if(prices[i]>val){
sum += (prices[i]-val);
val = prices[i];
}
else {
val = prices[i];
}
}
return sum;
}
};Third Question
Problem: Design a data structure that supports the following operations in O(1) time.
insert(x): Inserts an item x to the data structure if not already present.remove(x): Removes item x from the data structure if present.search(x): Searches an item x in the data structure.getRandom(): Returns a random element from the current set of elements.
Fourth Question
Link: Best Time to Buy and Sell Stock III
Problem: Given an array prices where prices[i] is the price of a given stock on the ith day, find the maximum profit you can achieve. You may complete at most two transactions.
Answer:
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n = prices.size();
vector<int> suf(n+1,0);
int lr = prices[n-1];
for(int i=n-1;i>=0;i--){
lr = max(lr,prices[i]);
suf[i] = max(suf[i+1],lr-prices[i]);
}
int val = prices[0];
int ans = 0;
for(int i=1;i<n;i++){
if(prices[i]>val){
ans = max(ans,prices[i]-val+suf[i+1]);
}
val = min(val,prices[i]);
}
return ans;
}
};Round 2
Taken By: Karan Varshney Sir
First Question
Problem: Get the data from a given list of APIs, then combine all the data and return that data.
Answer:
const list = [{},{}];
const app = async () => {
const combinedData = [];
list.map((listItem)=>{
const data = await fetch(listItem);
const response = await data.json();
combinedData = [...combinedData,response];
});
return combinedData;
}Follow-up Question: Suppose there are as many as 100 lists of APIs, now what will be your approach?
Answer: By using Promise.all() we can do this easily and it is also an optimized approach.
Second Question
Problem: Given multiple numbers, you have to find the numbers between 1 and n (inclusive) which are divisible by all the given numbers.
Answer:
vector<int> fun(int n, int x, int y){
vector<int> vec;
int gc = __gcd(x, y);
if(gc == 0) return {};
int new_x = ((x*y) / gc);
if(new_x == 0) return {};
new_x = abs(new_x);
for(int i = new_x; i <= n; i += new_x){
vec.push_back(i);
}
return vec;
}Third Question
Problem: Explain Inheritance using a Real-Life Example.
Answer:
class FourWheeler{
int wheels = 4;
}
class Car extends FourWheeler{
int define(){
// Car specific functionality
}
}Fourth Question
Problem: Explain Polymorphism.
Answer: Polymorphism means many forms. I explained it with a simple example of method overriding and overloading in object-oriented programming.
Fifth Question
Problem: Design a system for the tic-tac-toe game.
Answer:
vector<vector<int>> vec;
void assign(){
// For assigning the matrix to the default value
}
void play(int player, pair<int,int> ind){
// For assigning the value in the matrix in every turn
}
void switch_turn(int &player){
// For switching the turns between players
}
void check(){
// For checking if any player wins the game or not in each and every turn
}Sixth Question
Problem: Design a data structure that can store key-value pairs in cache memory for size 100 and can remove the last inserted key-value pair if the limit is reached or the size of the data structure is reached to 100 and then insert the new value.
Answer:
set<pair<string, int>> st2; // For storing key with their priority
set<pair<int, pair<string, string>>> st1; // For storing key-value with their priorityRound 3
Taken By: Prashant Sir
Description: Had a general HR interview and some managerial questions. We also talked about my previous projects and internships.
