HashedIn By Deloitte Coding Questions with Answers

April 25, 2024

HashedIn By Deloitte Coding Questions with Answers 2024

HashedIn By Deloitte Coding Questions with Solutions are asked at the time of Online Test/Technical Assessment of HashedIn By Deloitte. The level of questions asked is quiet high.

The page contains a to-do list in order to prepare for Technical Assessment. You’ll also get information about roles and responsibilities, eligibility criteria and salary for the position of Software Engineering Associate (Internship) and Software Engineer – I (Based on the performance) in HashedIn.

About HashedIn By Deloitte

They promote unconventional thinking at HashedIn by Deloitte. But, this does not imply that the only technological skills available to their employees are those. They work to create a workplace of the future that simultaneously values teamwork and individual contributions.

HashedIn always encourage the employees to develop future capabilities and construct what’s next. And they are honoured to work for a company that is known as a leader in attracting great personnel.

Prime Course Trailer

HashedIn By Deloitte Recruitment Process

Here we have mentioned detailed recruitment process of HashedIn by Deloitte to recruit candistes for the designation of Software Engineering Associate (Internship) and later on Software Engineer – I (Based on the performance).

PPT Presentation
Online Test
Technical Interview 1/ 2/3 (Based on the interview performance)
HR (Human Resources) Interview
Offer Confirmation

Eligibility Criteria - HashedIn By Deloitte

Simple average aggregate of 65% or above throughout in academics or 6.5 CGPA overall
B.E/ B.Tech/ M.E/ M.Tech/ M.Sc/ MCA (CSE and aligned branches only
No active backlogs

Technical Assessment/ Online Test

Round 1: Online Coding Exam.

There are three coding questions on it.
The two questions are often classified as easy or medium.
And one question qualifies as difficult.

Platform Used The Codility platform is used to administer the Technical Assessment.

HashedIn By Deloitte	Number of Questions	Time	Diifficulty
Coding	3	90 mins	Medium – Hard

After Online Assessment

Technical Interview Round 1 – 1-1.5 hours
Technical Interview Round 2 – 1-1.5 hours
Technical Interview Round 3 (Optional) – 1-1.5 hours
HR Interview

Related Banners

Get PrepInsta Prime & get Access to all 200+ courses offered by PrepInsta in One Subscription

Salary at HashedIn By Deloitte

Stipend during Internship: INR 25,000 per month
CTC Post Conversion: INR 8,10,000 LPA

HashedIn By Deloitte Coding Questions with Answers

QUESTION 01: Copycat in exam

Problem Statement: Rahul copies in the exam from his adjacent students. But he doesn’t want to be caught, so he changes words keeping the letter constant. That means he interchanges the positions of letters in words. You are the examiner and you have to find if he has copied a certain word from the one adjacent student who is giving the same exam, and give Rahul the markings he deserves.
Note that: Uppercase and lowercase are the same.

Input Format:

First line with the adjacent student’s word
Second line with Rahul’s word

Output Format:

0 if not copied
1 if copied
Constraints:
1<=Length of string<=10^6

Sample Input:

Sample Output:

C++

Java

Python

C++

Run

#include<bits/stdc++.h>
using namespace std;

int main(){
    string s2,s1;
    cin>>s1>>s2;
    transform(s1.begin(),s1.end(),s1.begin(),::tolower);
    transform(s2.begin(),s2.end(),s2.begin(),::tolower);
    sort(s1.begin(),s1.end());
    sort(s2.begin(),s2.end());
    cout<< (s2==s1);
}

Java

Run

import java.util.*;
public class Main{
  public static void main(String[] args){
      Scanner sc=new Scanner(System.in);
      String s1 = sc.next();
      String s2 = sc.next();
      s1 = s1.toUpperCase();
      s2 = s2.toUpperCase(); 
      char arr1[] = s1.toCharArray();
      Arrays.sort(arr1);
      char arr2[] = s2.toCharArray();
      Arrays.sort(arr2);
      String res1 = new String(arr1);
      String res2 = new String(arr2);
      if(res1.equals(res2))
          System.out.println("1");
      else 
          System.out.println("0");  
    }
}

Python

Run

s=input()
s1=input()
s=s.lower()
s=sorted(s)
s1=s1.lower()
s1=sorted(s1)
if s1==s:
    print(1)
else:
    print(0)

Question 02: Array Subarray

Problem Statement: You are given an array, You have to choose a contiguous subarray of length ‘k’, and find the minimum of that segment, return the maximum of those minimums.

Sample input :

1 → Length of segment x =1
5 → size of space n = 5
1 → space = [ 1,2,3,1,2]
2
3
1
2

Sample output :

Explanation :

The subarrays of size x = 1 are [1],[2],[3],[1], and [2],Because each subarray only contains 1 element, each value is minimal with respect to the subarray it is in. The maximum of these values is 3. Therefore, the answer is 3

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>
using namespace std;
vector < int > arr;
int prevmin=-1;
int flag=0;
int x,n,q;

int sorting(int start,int end)
{
    if(start+1==n) {start=0;end=end-n;}
    if(start==end) return arr[start];
    return min(arr[start],sorting(start+1,end));
}

int func(int start,int end)
{
    if(flag==0) {flag++;return prevmin=sorting(start,end);}
    if(arr[start-1]==prevmin) return prevmin;
    return prevmin=(arr[end] <= prevmin)?prevmin:sorting(start,end); } int main() { cin >> x >> n;
    int ans=0;
    for(int i=0;i < n;i++) {cin >> q;arr.push_back(q);}
    for(int i=0;i < n;i++)
    {
       ans=max(ans,func(i,i+x-1));
    }
    cout << ans;
}

JAVA

Run

import java.util.*;
public class DiskSpace
{
  public static void main(String[] args)
{
  Scanner sc=new Scanner(System.in);
  int x=sc.nextInt();
  int n=sc.nextInt();
  int arr[]=new int[n];
  for(int i=0;i < n;i++)
   arr[i]=sc.nextInt();

  int min=Integer.MAX_VALUE;
  int max=Integer.MIN_VALUE;

  for(int i=0;i <= n-x;i++)
 {
    min=Integer.MAX_VALUE;
    for(int j=i;j < (i+x);j++)
     min=Math.min(min,arr[j]);
   max=Math.max(min,max);     
 }
 System.out.println(max);
}
}

Python

Run

n=int(input())
arr=[]
for i in range(n):
    arr.append(int(input()))
s,a=0,0
for i in arr:
    s=s+i   
    if(s < 1):
        a=a+(-1*s) 
        s=0 
print(a)

Question 03: Trapezium pattern

Anirudh is attending an astronomy lecture. His professor who is very strict asks students to
Write a program to print the trapezium pattern using stars and dots as shown below . Since Anirudh is not good at astronomy can you help him?

Sample Input:

N = 3

Output:

.

…

…..

…

.

Java

Python

Run

#include<stdio.h> 
int main(){
    int i,j,n;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            if(j<n-i-1)
                printf("*");
            else
               printf(".");
        }
        for(j=0;j<n-1;j++){
            if(j<i)
                printf(".");
            else
                printf("*");
        }
        printf("\n");
    }

 for(i=2;i<=n;i++){
     for(j=0;j<n;j++){
            if(j<i-1)
                printf("*");
            else
                printf(".");
        }
      for(j=0;j<n-1;j++){
                if(j<n-i)
                   printf(".");
                else
                    printf("*");
            }
            printf("\n");
    }
    return 0;
}

Java

Run

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

      for(i=2;i<=n;i++){
          for(j=0;j<n;j++){
              if(j<i-1)
                  System.out.print("*");
              else
                  System.out.print(".");
      }
      for(j=0;j<n-1;j++){
          if(j<n-i)
              System.out.print(".");
          else
              System.out.print("*");
      }
      System.out.println();
   }
 }
}

Python

Run

n=int(input())
for i in range(n):
    print("*"*(n-1-i)+"."*(2*i+1)+"*"*(n-1-i))
for i in range(n-1):
    print("*"*(i+1)+"."*(2*(n-2-i)+1)+"*"*(i+1))

QUESTION 04: Formatting large Products

Problem Statement: Rohan is weak in mathematics.He is giving mathematics Olympiad , but he got stuck in one of the question .Help rohan to solve the question.In Question there are two positive integer A and B. You have to find the product of all integer between A and B which is represented in the form C=D*10^E , where C is the product of numbers , D and E are non-negative integers and the last digit of D is non-zero.

Function Description

Complete the function formatProducts in the editor below, formatProduct must return a string that represents C in the above described form.
Function has the following parameters
A: an integer
B: an integer

Constraints

A will between 1 and 1,000,000 . Inclusive.
B will be between A and 1,000,000. Inclusive.

Sample Input 0

Sample Output 0

12 * 10^1

Explanation

1*2*3*4*5=120 = 12 * 10^1

Sample Input 1

Sample Output 1

18144 * 10^2

Explanation

3*4*….*10=1814400 =18144 * 10^2

Java

Python

Java

Run

import java.util.*;
public class Main
{
  public static String formatProducts (int a, int b)
  {
    int result = 1;
    for (int i = a; i <= b; i++)
        result = result * i;
    int temp = result;
    int power = 0;
    while ((result % 10) == 0)
      {
	power = power + 1;
	result = result / 10;
      }
    return result + " * 10^" + power;
  }
  public static void main (String[]args)
  {
    Scanner sc = new Scanner (System.in);
    int a = sc.nextInt ();
    int b = sc.nextInt ();
    System.out.println (formatProducts (a, b));
  }
}

Python

Run

def formatProducts(L,H):
    a=1
    for i in range(L,H+1):
        a=a*i 
    c=0
    while(True):
        if a%10!=0:
            break
        else:
            c+=1 
        a=a//10
    return (str(a)+" * 10^"+str(c))
L=int(input())
H=int(input())
print(formatProducts(L,H))

Question 05: Password Creation

Problem Statement: A password manager wants to create new passwords using two strings given by the user, then combined to create a harder-to- guess combination. Given two strings,interleave the characters of the strings to create a new string. Beginning with an empty string, alternately append a character from string a and from string b. If one of the strings is exhausted before the other, append the remaining letters from the other
string all at once. The result is the new password.

Example :

If a = ‘hackerrank’ and b = ‘mountain’,
The result is hmaocuknetrariannk.

Function Description :

Complete the function newPassword in the editor below.

Parameter(s):

Str : string a
Str : string b
Returns:
Str: new password using two strings

Sample input:

abc → a=”abc”
def → b=”def”

Sample output 0:

Adbecf

JAVA

Python

JAVA

Run

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str1 = sc.next();
        String str2 = sc.next();
        String password = "";
        int i, j;
        for (i = 0, j = 0; i < str1.length() && j < str2.length(); i++, j++) {
            password = password + str1.charAt(i) + str2.charAt(j);
        }
        if (i < str1.length())
            password = password + str1.substring(i, str1.length());
        if (j < str2.length())
            password = password + str2.substring(j, str2.length());
        System.out.println(password);
    }
}

Python

Run

def newPassword(a, b):
    ans = ""
    if len(a) > len(b):
        m = len(b)
        x = a[m:]
    if len(a) <= len(b):
        m = len(a)
        x = b[m:]
    for i in range(m):
        ans += a[i]
        ans += b[i]
    return ans + x

a = input()
b = input()
print(newPassword(a, b))

FAQ's Related to HashedIn By Deloitte Coding Questions

Question 1: What does HashedIn really do?

HashedIn specializes in building custom software solutions for clients in various industries, including healthcare, finance, retail, and hospitality. Their services include product ideation, development, and deployment, as well as cloud migration, DevOps consulting, and user experience design.

Question 2: How many technical rounds are there in HashedIn Recruitment process?

A minimum of 2 rounds of Technical interviews is conducted by HashedIn for recruiting the freshers after conducting a Coding Round of 90 Minutes on the Codility Coding Platform.

Get over 200+ course One Subscription

Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others

HashedIn By Deloitte Coding Questions with Answers

HashedIn By Deloitte Coding Questions with Answers 2024

About HashedIn By Deloitte

Prime Course Trailer

HashedIn By Deloitte Recruitment Process

Eligibility Criteria - HashedIn By Deloitte

Technical Assessment/ Online Test

Round 1: Online Coding Exam.

After Online Assessment

Related Banners

Salary at HashedIn By Deloitte

HashedIn By Deloitte Coding Questions with Answers

QUESTION 01: Copycat in exam

Input Format:

Output Format:

Sample Input:

Sample Output:

Question 02: Array Subarray

Sample input :

Sample output :

Explanation :

Question 03: Trapezium pattern

Sample Input:

Output:

**.**

*…*

…..

*…*

**.**

QUESTION 04: Formatting large Products

Function Description

Constraints

Sample Input 0

Sample Output 0

Explanation

Sample Input 1

Sample Output 1

Explanation

Question 05: Password Creation

Example :

Function Description :

Parameter(s):

Sample input:

Sample output 0:

FAQ's Related to HashedIn By Deloitte Coding Questions

Question 1: What does HashedIn really do?

Question 2: How many technical rounds are there in HashedIn Recruitment process?

Get over 200+ course One Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

.

…

…

.