Byjus – whitehat junior : Online Technical Coding Interview Questions
I got called from instahyre for php developer.
Round 1 :
1.) Implement Explode functionality of php
<?php
$string = "I am |~| I am ....|~||~||~||~||~||~||~|[email protected]@#@#|~|@#@# fdsds |~|dfdf~ |~|";
$deli = "|~|";
print_r(explode($deli , $string));
print_r(myExplode($deli , $string ));
function myExplode( $deli , $string ){
$ans = array();
$deliLen = strlen($deli);
$index = 0;
$temp ="";
$subStr = "";
$deliIndex = 0;
for( $i = 0 ; $i < strlen($string) ; $i++){
if( $string[$i] == $deli[0] ){
while($index < $deliLen && $i < strlen($string) && $string[$i] == $deli[$index]){
$temp .= $string[$i];
$index++;
$i++;
}
if($index == $deliLen){
$ans[] = $subStr;
$subStr = "";
}else{
$subStr .= $temp;
}
$temp = "";
$index = 0;
}
else{
$subStr .= $string[$i];
}
}
if(strlen($subStr) > 0){
$ans[] = $subStr;
}
return $ans;
}
?>2.) Mysql Query to find
a.) ranks of given roll number based on marks . Same marks should be at same rank.
b.) kth Largest numner in the column , duplcate & non duplicate case both
select rolln from englis_test where marks = (select distinct marks from englis_test order by marks desc LIMIT 1,1 );
Round 2 : system Desgin
Design a scalable Cab finding system based on user and cab location. Given a list of cabs with their location, and user's current location, return a set of nearby cabs which lie within r km radius of the user.
Input:
A list of cabs with their location.
cabs = [[id1, x1, y1], [id2, x2, y2], [id3, x3, y3]...[idn, xn, yn]]
where xi, yi are x and y coordinates of the ith cab and idi is unique id of ith cab
User's current location
userLocation = [xu, yu]
where xu, yu are x and y coordinates of the user.
Radius r denoting the radius of circle within which cab is considered as nearby
Design following methods:
// takes above defined inputs and returns a set of nearby cabs
Set<cabs> findNearbyCabs ( List<cabs>, Tuple userLocation, int r)
Requirements:
Above method should have minimum latency. The list of cabs is huge. Iterating all of them is not an option.
// Takes in a cab and it's new location and updates the location
void updateCabLocation ( Cab cab, Tuple newCabLocation)
Requirements:
This method will be called a lot of times, hence latency should be minimum.
// optional method
Void preprocessing ()
Test Cases:
Input:
cabs = [1, 0, 0], [2, 5, 10], [3, 5, 1000], [4, 1000, 2], [5, 3, 4], [6, -4, -5], [7, -2, 3]
userLocation = [0,0]
r = 5
Method calls:
findNearbyCabs(cabs, userLocation, r)
Output: [ 1, 5, 7]
updateCabLocation([3,5,1000], [1,1])
updateCabLocation([7,-2, 3], [1,100])
findNearbyCabs(cabs, userLocation, r)
Output: [ 1, 3, 5]
Class Cabs(){
vector<vector<int>> cabX(n , vector<int>(2 , 0));
vector<vector<int>> cabY(n , vector<int>(2 , 0));
unorderd_map<int , vector<int>> m;
Void preprocessing (){
Int index = 0;
foreach(vector<int> cab : cabs){
cabX[index][1][0] = cab[1];
cabY[index][1][0] = cab[2];
vector<int> coordinates{cab[1] , cab[2]};
M[cab[0]] = coordinates;
cabX[index][1][1] = cabY[index][1][1] = cab[0];
index++;
}
sort(CabX);
sort(CabY);
}
vector<int> findNearbyCabs ( vector<int> &cabs, vector<int> userLocation, int r){
Int xUser = userLocation[0];
Int yUser = userLocation[1];
vector<int> ans ;
Int xStartIndex = getStartIndex(cabX ,xUser-r );
Int xEndIndex = getIndex(cabX ,xUser+r);
Int yStartIndex = getStartIndex(cabY ,yUser-r );
Int yEndIndex = getIndex(cabY ,yUser+r);
for(int i =xStartIndex ; i <= xEndIndex ; i++ ){
Int cabId = cabX[i][1];
Int yIndex = m[cabId][1];
if( yIndex <= (yUser+r) && yIndex >= (yUser-r) ){
ans.push_back(cabId);
}
}
Return ans;
}
void updateCabLocation ( int cabId, vector<int> newCabLocation){
m[cabId] = newCabLocation;
preprocessing();
}
// Online C++ compiler to run C++ program online
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
using namespace std;
class Cabs{
public :
vector<vector<int>> cabX;
vector<vector<int>> cabY;
map<int , vector<int>> m;
vector<vector<int>> cabs ;
void setCabs(vector<vector<int>> c){
cabs = c;
preprocessing();
}
void preprocessing(){
int index = 0;
for(vector<int> cab : cabs){
vector<int>x{cab[1] ,cab[0] };
vector<int>y{cab[2] , cab[0] };
cabX.push_back(x);
cabY.push_back(y);
vector<int> coordinates{cab[1] , cab[2] };
m[cab[0]] = coordinates;
cabX[index][1] = cabY[index][1] = cab[0];
index++;
}
sort(cabX.begin() , cabX.end() , [](vector<int> a, vector<int> b ){
return a[1] <= b[1];
});
/*
sort(cabY.begin() , cabY.end() , [](vector<int> a, vector<int> b ){
return a[2] <= b[2];
});
*/
}
vector<int> findNearbyCabs ( vector<int> userLocation, int r){
return findCab(userLocation , r);
}
vector<int> findCab ( vector<int> userLocation, int r){
int xUser = userLocation[0];
int yUser = userLocation[1];
vector<int> ans ;
int xStartIndex = 0 ;//getStartIndex(cabX ,xUser-r );
int xEndIndex = cabs.size();//getIndex(cabX ,xUser+r);
for(int i =xStartIndex ; i < xEndIndex ; i++ ){
// cout<<"1.."<<endl;
int cabId = cabX[i][1];
// cout<<"2.."<<cabId<<endl;
int yIndex = m[cabId][1];
int xIndex = m[cabId][0];
//cout<<"3.."<<yIndex<<endl;
if( insideCircle(xUser , yUser , xIndex ,yIndex , r) ){
ans.push_back(cabId);
}
}
for(int i : ans){
cout<<i<<"--";
}
return ans;
}
bool insideCircle(int a , int b , int c , int d , int r){
return r*r >= ( (a -c)*(a -c) + (b-d)*(b-d));
}
void updateCabLocation ( int cabId, vector<int> newCabLocation){
m[cabId] = newCabLocation;
preprocessing();
}
};
int main() {
vector<vector<int>> cabs{
vector<int>{1, 0, 0},
vector<int>{2, 5, 10},
vector<int>{3, 5, 1000},
vector<int>{4, 1000, 2},
vector<int>{5, 3, 4},
vector<int>{6, -4, -5},
vector<int>{7, -2, 3}
};
Cabs c ;//
c.setCabs(cabs);
vector<int> userLocation{0,0};
int r = 5;
c.findCab(userLocation , r);
return 0;
}
