DotPe Coding Questions and Answers

April 28, 2023

DotPe Coding Questions with Solutions

In this page, you will find out DotPe Coding Questions and Answers asked in Online Assessments and Technical Interviews involved in the Recruitment Process of the Company. Further, on this page, you will get more information like Company’s Job Profile, Job Location, CTC Offered, Steps involved in the recruitment process, etc.

About DotPe

DotPe was founded by Anurag Gupta, Gyanesh Sharma, and Shailaz Nag, in 2019. This company provides payments and commerce-related services. They usually help Offline Business Merchants and Organizations to Virtualize or Digitalize their business with Digital Catalogue. With the help of these services, Business Enterprises can easily reach out to their customers, and sell their Goods and Services directly, efficiently, and with faster payment methods.

DotPe also owns an App called Digital Showroom, which help Business Enterprises to create their own Business Domain in less to Digitalize their Business.

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About DotPe Recruitment Process

DotPe Recruitment Process consists of the following steps :

Online Assessment [ MCQ’s Based ]
Technical Interview
HR Interview

We have mentioned further details of the DotPe Recruitment Process in the following Tabular Form

DotPe	Related Information
Position :	Engineering – Intern
Course :	B.E / B.Tech – CS & IT Eligible Batch – 2023
Eligibility Criteria / Academic Qualification Required :	Minimum 80 % or equivalent CGPA required in 10th / 12th/ Graduation. No Current Backlogs.
Offered CTC :	During Internship = ₹ 20K per month Post Completion = ₹ 10 – 15 L.P.A ( Performance Based )
Selection Process :	Online Assessment Technical Interview HR Interview
Joining Location :	Remote

DETAILS OF DOTPE RECRUITMENT EXAM:

Exam Duration: ~ 1 Hour [ MCQ's + Coding ]

Topics: Data Structures & Algorithms, DBMS, Operating Systems & other Core Subjects.

DotPe Job Description

Important Note:

Steps in the selection procedure of the company can be altered according to different colleges.

Please check out DotPe Official Career Page for the Latest updates.

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DotPe Coding Questions and Answers

Question 1 : Coin Distribution Problem

Problem Statement

Find the minimum number of coins required to form any value between 1 to N,both inclusive.Cumulative value of coins should not exceed N. Coin denominations are 1 Rupee, 2 Rupee and 5 Rupee.Let’s Understand the problem using the following example. Consider the value of N is 13, then the minimum number of coins required to formulate any value between 1 and 13, is 6. One 5 Rupee, three 2 Rupee and two 1 Rupee coins are required to realize any value between 1 and 13. Hence this is the answer.However, if one takes two 5 Rupee coins, one 2 rupee coin and two 1 rupee coin, then too all values between 1 and 13 are achieved. But since the cumulative value of all coins equals 14, i.e., exceeds 13, this is not the answer.

Input Format:
- A single integer value.

Output Format:
- Four space separated integer values.
  - 1st – Total number of coins.
  - 2nd – number of 5 Rupee coins.
  - 3rd – number of 2 Rupee coins.
  - 4th – number of 1 Rupee coins.

Constraints:
- 0 < n < 1000

Refer the sample output for formatting

Sample Input

Sample Output

6 1 3 2

Explanation

The minimum number of coins required is 6 with in it:
- minimum number of 5 Rupee coins = 1
- minimum number of 2 Rupee coins = 3
- minimum number of 1 Rupee coins = 2

Using these coins, we can form any value with in the given value and itself, like below:

Here the given value is 13

For 1 = one 1 Rupee coin
For 2 = one 2 Rupee coin
For 3 = one 1 Rupee coin and one 2 Rupee coins
For 4 = two 2 Rupee coins
For 5 = one 5 Rupee coin
For 6 = one 5 Rupee and one 1 Rupee coins
For 7 = one 5 Rupee and one 2 Rupee coins
For 8 = one 5 Rupee, one 2 Rupee and one 1 Rupee coins
For 9 = one 5 Rupee and two 2 Rupee coins
For 10 = one 5 Rupee, two 2 Rupee and one 1 Rupee coins
For 11 = one 5 Rupee, two 2 Rupee and two 1 Rupee coins
For 12 = one 5 Rupee, three 2 Rupee and one 1 Rupee coins
For 13 = one 5 Rupee, three 2 Rupee and two 1 Rupee coins

Python

JAVA

Python

Run

number = int (input ())
five = int ((number - 4) / 5) 
if ((number - 5 * five) % 2)== 0:
  one = 2
else:
  one = 1 
two = (number - 5 * five - one)	//2
print (one + two + five, five, two, one)

JAVA

Run

import java.util.*;
public class Main
{
  public static void main (String[]args)
  {
    System.out.println ("Enter the Number");
    Scanner s = new Scanner (System.in);
    int number = s.nextInt ();
    int one = 0, two = 0;
    int five = (number - 4) / 5;
    if (((number - 5 * five) % 2) == 0)
      {
	one = 2;
      }
    else
      {
	one = 1;
      }
    two = (number - 5 * five - one) / 2;
    System.out.println (one + two + five);
    System.out.println (five);
    System.out.println (two);
    System.out.println (one);

  }
}

Question 2 : Find the homeless

Problem Statement -: There are N Homeless people in the community and N houses in the community. It will be given in the array (people) , height of the person and in the array house capacity of the house is given.

Government decided to give homes for people on the basis of following conditions:

Priority is given for the people from left to right of the array
Each person is allotted to a house if and only if the capacity of house is greater than or equal to persons height
Nearby empty Houses are alloted to the person( starting from extreme left)

You need to find the number of homeless people who have not allotted any home if the government follows the above conditions.So that government will have an idea for how many people they need to allot home for next time.

Constraints:

1 <= N <= 10^3
1 <= people[i] <= 10^5
1 <= house[i] <= 10^5

Input Format for Custom Testing:

The first line contains an integer, N, denoting the number of people and number of houses.
Each line i of the N subsequent lines (where 0 <= i <= N) contains an integer describing peoplei.
Each line i of the N subsequent lines (where 0 <= i <= N) contains an integer describing housei.

Sample Test Cases

Sample Input 1
3
4
2
7
3
5
10
Sample Output 1
0
Explanation
people=[4,2,7]
house=[3,5,10]
People[0] has more priority , from left to right order in houses 5 is the nearest one which fits for people[0]
people[1]=2 will fit in 3 which is nearer from left
people[2]=7 will fit in remaining house of capacity of 10
So no homeless people left so return 0

Sample Input 2
3
3
8
5
1
9
4
Sample Output 2
2
Explanation
people=[3,8,5]
house=[1,9,4]
people[0]=3 can fit in 9 which is nearest from left in array house
people[1]=8 cannot fit in any home which is left (i.e, 1 and 4)
people[2]=5 cannot fit in any home which is left (i.e, 1 and 4)
So return 2,which is number of homeless people

C++

Java

Python

C++

Run

#include<bits/stdc++.h>
using namespace std;
int main(){  
    int n;
    cin>>n;    
    int people[n], house[n];
    for(int i=0; i<n; i++) cin>>people[i];       
    for(int i=0; i<n; i++) cin>>house[i];      
    int count = 0;
    for(int i=0; i<n; i++){      
        for(int j=0; j<n; j++){
    if(people[i]<house[j]){
            count+=1 ;
            house[j]=-1;
            break ;
            }
        }
    } 
    cout<<n-count;
  }

Java

Run

import java.util.*;
class Main
{ 
  public static void main(String args[])
  {
    Scanner sc = new Scanner (System.in);
    int n=sc.nextInt();
    int people[]= new int[n];
    int house[]= new int[n];
    for(int i=0; i<n; i++)
       people[i]=sc.nextInt();      
    for(int i=0; i<n; i++)
       house[i]=sc.nextInt();       
    int count = 0;
    for(int i=0; i<n; i++){       
        for(int j=0; j<n; j++){
         if(people[i]<house[j]){
            count+=1 ;
            house[j]=-1;
            break ;
            }
        }
    }
    System.out.println(n-count);
  }
 }

Python

Run

N=int(input())
people=[]
house=[]
#to read data of people and house arrays
for i in range(N): people.append(int(input()))
for i in range(N): house.append(int(input()))
count=0
for i in range(N):
    for j in range(N):
        if(people[i] < house[j]):
            count+=1 
            house[j]=-1
            break 
print(N-count)

Question 3 :

Ajay has a flight to catch in an hour. So he has to reach the airport as fast at possible. He hires a taxi and promises the taxi driver that if he reaches the airport within k minutes he would pay the taxi driver double the amount.

The city description is as follows –

The taxi is at point 0,0 & airport is at (n-1,m-1) in a 2 – D grid of n rows and m columns. The grid has some blocked (represented as’#’) and some unblocked (represented as’.’) cells.

The starting position of the taxi is in the top – left corner of the grid. It is guaranteed that the starting position & ending positions are not blocked. Each cell of the grid is connected with its right ,left,top,bottom cells (if those cells exist ). It takes 1 second for a taxi to move from a cell to its adjacent cell.

If the taxi can reach the bottom-right (airport) corner of the grid within k seconds, return the string ‘Yes’. Otherwise , return the string ‘No’.

Example

rows =3

grid =[‘..##’,’#.##’,’#…’]

maxTime =5

..##

#.##

#…

It will take the taxi 5 seconds to reach the bottom right corner. As long as k>_5,return

‘Yes’.

Returns:

String;the final string; either ‘yes’ or ‘ No’

Constraints

1<_rows<_500
0<_maxTime<_10^6

Input Format For Custom Testing

The first line contains an integer,rows that denotes the number of rows of the 2-D grid
In each of the next rows lines, the i^th line contains a string denoting the configuration of the i^th row of the grid.
The last line contains an integer, maxTime ,that represents the maximum time in seconds the taxi has to reach the bottom right cell.

Example

Sample Input 0
2 -> size of grid[] rows =2
.. -> grid = [‘..’,’..’]
..
3 -> maxTime = 3

Sample Output 0
Yes

Explanation

The grid has 2 rows and 2 columns and the time within which the taxi needs to reach the bottom-right cell in 3 seconds. Starting from the top-left cell, the taxi can either move to the top-right unblocked

Python

Java

Python

Run

from collections import deque
def shortestPath(grid, k):
    #print(grid)
    n, m = len(grid), len(grid[0])
    queue = deque([(0, 0,0)])
    visited = {(0, 0)}

    while queue:
        x, y,dist = queue.popleft()
        if (x, y) == (n-1, m-1):
            return dist
        else:
            for dx, dy in ((-1, 0), (+1, 0), (0, -1), (0, +1)):
                nx, ny = x + dx, y + dy
                key = (nx, ny)
                if not (0 <= nx < n and 0 <= ny < m) or key in visited:
                    continue      
                if grid[nx][ny] == '.':
                    visited.add(key)
                    queue.append((nx, ny,dist + 1))
    return -1 

n=int(input())
ar=[]
for i in range(n):
    s=list(input())
    ar.append(s) 
k=int(input())    
#a=(shortestPath(ar,k))
print(a)
if(a<=k):
    print('YES')
else:
    print('NO')

Java

Run

import java.util.*;
public class Main
{
  public static void main(String[] args)
  {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    char matrix[][] = new char[n][n];

for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) matrix[i][j] = sc.next().charAt(0); int dp[][] = new int[n][n]; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) dp[i][j] = Integer.MAX_VALUE; long maxTime = sc.nextInt(); dp[0][0] = 0; for (int i = 1; i < n; i++) { if (matrix[0][i] == ‘#’) break; dp[0][i] = dp[0][i – 1] + 1; } for (int i = 1; i < n; i++) { if (matrix[i][0] == ‘#’) break; dp[i][0] = dp[i – 1][0] + 1; } for (int i = 1; i < n; i++) { for (int j = 1; j < n; j++) { if (matrix[i][j] == ‘#’) continue; dp[i][j] = Math.min(dp[i – 1][j], dp[i][j – 1]) + 1; } } if (dp[n – 1][n – 1] < maxTime) System.out.println(“Yes”); else System.out.println(“No”); } }

Question 4: Houses Problem

Problem Description

Question:- There are n houses build in a line, each of which contains some value in it.

A thief is going to steal the maximal value of these houses, but he can’t steal in two adjacent houses because the owner of the stolen houses will tell his two neighbours left and right side.

What is the maximum stolen value?

Sample Input: val[] = {6, 7, 1, 3, 8, 2, 4,12}

Sample Output: 27

C++

JAVA

Python

C++

Run

#include<bits/stdc++.h>
using namespace std;
int maxLoot(int *hval, int n) ; 
int main()
{
    int hval[]= {6, 7, 1, 3, 8, 2, 4, 12};
    int n = sizeof(hval)/sizeof(hval[0]);
    cout << "Maximum loot possible : "
        << maxLoot(hval, n);
    return 0;
}
// calculate the maximum stolen value
int maxLoot(int *hval, int n)
{
    if (n == 0)
        return 0;
    if (n == 1)
        return hval[0];
    if (n == 2)
        return max(hval[0], hval[1]);
 
    // dp[i] represent the maximum value stolen
    // so far after reaching house i.
    int dp[n];
 
    // Initialize the dp[0] and dp[1]
    dp[0] = hval[0];
    dp[1] = max(hval[0], hval[1]);
 
    // Fill remaining positions
    for (int i = 2; i<n; i++)
        dp[i] = max(hval[i]+dp[i-2], dp[i-1]);
 
    return dp[n-1];
}

Output
27

JAVA

Run

import java.util.Scanner;
public class Main {

    static int prep(int hval[],int n) {
        if(n==0)
            return 0;
        if(n==1)
            return hval[0];
        if(n==2)
            return Math.max(hval[0],hval[1]);

        int[] dp = new int[n];
        dp[0]=hval[0];
        dp[1]=Math.max(hval[0],hval[1]);

        for (int i = 2; i < n; i++) {
            dp[i]=Math.max(hval[i]+dp[i-2],dp[i-1]);
        }
        return dp[n-1];
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in); 

        int hval[] = {3,2,4,6,7,1};
        int n=hval.length;
        System.out.println("Max loot value :"+prep(hval,n));
    }

}

Output
6 7 1 3 8 2 5
20

Python

Run

num = list(map(int, input().split()))
sum = num[0]
for i in range(0, len(num)//2):
    sum = sum + int(num[i])
print(sum)

Output
6 7 1 3 8 2 5
20

Question 5 :

Sahil watches TV all day and gets bored. He started playing this dumb game of identifying minimum number of inputs needed to reach a channel. As his cousin, you have to help him, but you live far from his house. So you decide to write a code that will ask Sahil for some inputs and give outputs respectively.

Here are the problems you need to keep in mind,

There are 13 buttons on his remote: 10 buttons for the numbers (0-9) to form integers denoting respective channel index, “Up channel” button and “ Down channel” button for going i +1th channel and i-1th channel from i respectively, and a “Last viewed” button to see what’s the last channel before it.
The number buttons allow you to jump directly to a specific channel (Ex: to go to channel 172 by typing 1,7,2).
If the channel which you are in is ith and that is the max channel index possible, by Up channel, you will reach the first channel possible. Same goes for the down channel button. You can go to the highest channel possible if you go down from the lowest channel possible.

Sahil can get from one channel to the next in one of the two ways.

Sahil’s parents have set some parental control on some channels on Aniruth’s television. The “Up Channel “ and “Down buttons” buttons skip these channels as these channels are not viewable.

Given a list of channels to view, the lowest channel, the highest channel, and a list of blocked channels, your program should return the minimum number of clicks necessary to get through all the shows that Anirudh would like to match.

Input Format

First line is the lowest Channel
Second-line is the highest Channel
Followed by a number of blocked channels B,
and the next B lines contain the actual blocked channels.
Followed by the number of Channels to view V, and the next V lines contain the actual channels to view.

Constraints

The lowest channel on the television will be greater than 0. and less than or equal to 10,000.
The highest channel on the television will be greater than or equal to the lowest channel. and less than or equal to 10.000.
The list of channels that are blocked on Anirudh’s television. All the channels in this list will be valid channels (greater than or equal to lowest channel, less than or equal 1 to highest channel). Duplicates may be Ignored. The blocked list can be a maximum of 40 channels.
The sequence that Sahil must view contains between 1 and 50 elements. inclusive. All channels in this sequence are not in the blocked list and are between lowest channel and highest channel. Inclusive.

Examples

Sample Input 0:
1
20
2
18
19
5
15
14
17
1
17

Sample output 0:
7

C++

Java

C++

Run

#include<bits/stdc++.h>

using namespace std;

unordered_map < int, int > m;
int l, u;
util(int a, int b) {
  if (a == b) return 0;
  if (m[a]) return util(a + 1, b);
  return 1 + util(a + 1, b);
}

int func(int b, int prev) {
  if (b < prev) return min(util(prev, u) + util(l, b) + 1, util(b, prev)); else return min(util(prev, b), util(l, b) + util(prev, u) + 1); } int main() { int flag = 0, ans = 0, prev, prev2; cin >> l >> u;
  int bn, b;
  cin >> bn;
  while (bn--) {
    cin >> b;
    m[b]++;
  }
  cin >> bn;
  while (bn--) {
    cin >> b;
    if (b > 9 && flag == 0) {
      ans += 2;
      flag++;
      prev = b;
    } else if (flag == 0) {
      ans += 1;
      flag++;
      prev = b;
    } else if (prev2 == b) {
      prev2 = prev;
      prev = b;
      ans++;
    } else {
      ans += min(b > 9 ? 2 : 1, func(prev, b));
      prev2 = prev;
      prev = b;
    }
  }

  cout << ans;
}

Java

Run

import java.util.*;
public class Solution {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int l = sc.nextInt();
    int h = sc.nextInt();
    int b = sc.nextInt();
    List < Integer > bl = new ArrayList < > ();
    for (int i = 0; i < b; i++) {
      bl.add(sc.nextInt());
    }
    int v = sc.nextInt();

    List < Integer > vl = new ArrayList < > ();

    for (int i = 0; i < v; i++) {
      vl.add(sc.nextInt());
    }
    Set < Integer > sl = new HashSet < > ();
    int res = 0;

    for (Integer i: vl) {
      if (bl.contains(i))
        continue;
      sl.add(i);
    }

    for (Integer i: sl) {
      String s = i + "";
      res += s.length();
    }
    System.out.println(res);
  }
}

FAQs on DotPe Coding Questions with Solutions

Question 1: Are there any coding-related questions in the DotPe recruitment process?

Yes, DotPe always asks Coding Questions in their Online Assessment and Technical Interview. In this page, you will get all the DotPe Coding Questions.

Question 2: What is the Salary Offered and Eligibility Criteria for the freshers?

DotPe usually asked for 80% or Equivalent C.G.P.A in 10th, 12th & Graduation without current backlog. They offer Rs 20K per month during the Internship, post completion Rs 10 – 15 L.P.A will be offered.

Question 3: What is the difficulty level of Online Assessment and Technical Interview of the DotPe Recruitment Process ?

The difficulty level of Online Assessment and the Coding questions asked in the Technical Interview is Moderate to High.

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DotPe Coding Questions and Answers

DotPe Coding Questions with Solutions

About DotPe

Prime Course Trailer

About DotPe Recruitment Process

DotPe Job Description

Engineering - Intern

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DotPe Coding Questions and Answers

Question 1 : Coin Distribution Problem

Question 2 : Find the homeless

Question 3 :

Question 4: Houses Problem

Question 5 :

FAQs on DotPe Coding Questions with Solutions

Question 1: Are there any coding-related questions in the DotPe recruitment process?

Question 2: What is the Salary Offered and Eligibility Criteria for the freshers?

Question 3: What is the difficulty level of Online Assessment and Technical Interview of the DotPe Recruitment Process ?

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