HashedIn By Deloitte Coding Questions with Answers
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.
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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.
| 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
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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:
- CAR
- Acr
Sample Output:
- 1
#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);
}
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");
}
}
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 :
- 3
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
#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;
}
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);
}
}
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:
**.**
*…*
…..
*…*
**.**
#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;
}
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();
}
}
}
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
- 1
- 5
Sample Output 0
- 12 * 10^1
Explanation
- 1*2*3*4*5=120 = 12 * 10^1
Sample Input 1
- 3
- 10
Sample Output 1
- 18144 * 10^2
Explanation
- 3*4*….*10=1814400 =18144 * 10^2
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));
}
}
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
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);
}
}
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.
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