Tata Elxsi Coding Questions

April 27, 2023

Tata Elxsi Coding Questions 2023

Tata Elxsi Coding Questions are frequently asked by TATA ELXSI Placement Paper in Written Assessment. It is an elimination round and the student is given a choice to choose any C/C++/JAVA/Python for coding purpose in which he/she is comfortable with.

There is a written assessment of 120 minutes which is necessary for a student to qualify for the further proceedings. Go through the page to know more details.

About Tata Elxsi

One of the top companies in the world for design and technology services, Tata Elxsi works with clients in the automotive, broadcast, communications, healthcare, and transportation sectors.

Below are the key points that everyone should read to have a complete picture about Tata Elxsi Placement Exam, from registration to selection. The process includes two rounds:-

Written Assessment
Technical and HR Interview

Prime Course Trailer

Tata Elxsi Eligibility Criteria

Academic Qualifications :

Class 10th Standard : 60% or Above
Class 12th Standard : 60% or Above
College Graduation : 60% or Above 6.5 CGPA

Eligible Passing Year:

2023

College Qualification Required:

B.E.
B.Tech.

Eligible Branches:

CS/IT/Circuital Branches

Other Important Criteria:

There should be No Backlogs at the time of Selection Process.
The education gap should be of maximum 1 years,if any, is allowed between 10th and graduation.
Should be from a Full-time Degree course recognized by the Central/State Government of India.
The candidates must not have any pending attendance requirement with the college.
Should be Indian Citizen or should carry a PIO or OCI card, in case holding a passport of any other country.

Related Banners

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

Tata Elxsi Written Assessment

Tata Elxsi	Number of Questions	Time	Diifficulty
Verbal Ability	30	48 mins	Medium
Reasoning Ability	30	30 mins	Medium
Coding Round 1	1	Untimed	Hard
Coding Round 2	1	Untimed	Hard

Analysis of Tata Elxsi Written Exam

*Untimed means these sections are without any specific time limit. You can answer these within the total assessment time limit, i.e., Total Time of Untimed Sections = Total Assessment Time Limit – Total Time of Timed Sections. Hence,

Total Time of Untimed Sections = 120 – 78 = 42 minutes

Tata Elxsi Coding Questions

Question 1 : Move Hash to Front

Problem Statement :

You have write a function that accepts, a string which length is “len”, the string has some “#”, in it you have to move all the hashes to the front of the string and return the whole string back and print it.

char* moveHash(char str[],int n);

Example :
Sample Test Case
Input :
Move#Hash#to#Front
Output :
###MoveHashtoFront

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>

using namespace std;
string moveHash(string s) {
    string ans = "";
    for (auto i: s)
        if (i == '#') ans = '#' + ans;
        else ans += i;
    return ans;
}

int main() {
    string s;
    getline(cin, s);
    cout << moveHash(s);
}

JAVA

Run

import java.util.*;
public class Main {
    public static void moveHash(String str, int n) {
        String str1 = new String("");
        String str2 = new String("");
        int i = 0;
        for (i = 0; i < n; i++) {
            if (str.charAt(i) == '#')
                str1 = str1 + str.charAt(i);
            else
                str2 = str2 + str.charAt(i);
        }
        String result = str1.concat(str2);
        System.out.println(result);
    }
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        String a = sc.nextLine();
        int len = a.length();
        moveHash(a, len);
    }
}

Python

Run

def moveHash(s):
    x = s.count("#")
    s = s.replace("#", "")
    return "#" * x + s
s = input()
print(moveHash(s))

Question 2 :Borrow Number

Problem Statement :

You have two numbers number1 and number2, your job is to check the number of borrow operations needed for subtraction of number1 from number2. If the subtraction is not possible then return the string not possible.

Example :

754
658

Answer :
2

654
666

Answer :
Not possible

C++

JAVA

Python

C++

Run

#include &bits/stdc++.h>

using namespace std;

int main() {
    string s1, s2;
    int c = 0, f = 0;
    cin >> s1 >> s2;
    if (stoi(s1) < stoi(s2)) {
        cout << "Impossible";
    }
    reverse(s1.begin(), s1.end());
    reverse(s2.begin(), s2.end());
    for (int i = 0; i < s1.length(); i++)
        if (s1[i] < s2[i]) {
            f = 1;
            c++;
        }
    else if (s1[i] == s2[i]) {
        if (f == 1) {
            c++;
        }
        f = 0;
    } else f = 0;
    cout << c;
}

JAVA

Run

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int number1 = sc.nextInt();
        int number2 = sc.nextInt();
        int count = 0;
        if (number1 < number2) {
            System.out.println("Not possible");
        } else {
            boolean flag = false;
            while (number1 != 0 && number2 != 0) {
                int temp1 = 0;
                int temp2 = number2 % 10;
                if (flag)
                    temp1 = number1 % 10 - 1;
                else
                    temp1 = number1 % 10;
                if (temp1 < temp2) {
                    count++;
                    flag = true;
                } else
                    flag = false;
                number1 = number1 / 10;
                number2 = number2 / 10;
            }
            System.out.println(count);
        }

    }

}

Python

Run

number1 = int(input())
number2 = int(input())
count = 0
if number1 < number2:
    print("Not possible")
else:
    flag = 0
    while number1 != 0 and number2 != 0:
        temp1 = 0
        temp2 = number2 % 10
        if flag:
            temp1 = number1 % 10 - 1
        else:
            temp1 = number1 % 10
        if temp1 < temp2:
            count += 1
            flag = 1
        else:
            flag = 0
        number1 = number1 // 10
        number2 = number2 // 10
    print(count)

Question 3 : Capitalize/Decapitalize

Problem Statement :

You’re given a function that accepts the following, a string1, its length and a character c. Your job is to replace all the occurrences of character c in string1 and capitalize it or decapitalize it based on the character c.

Input :
hello world
l
Output :
heLLo worLd

Input :
prepinsta
p
Output :
PrePinsta

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>

using namespace std;

int main() {
    string s;
    getline(cin, s);
    char k;
    cin >> k;
    for (auto i: s)
        if (i == k) {
            if (i > 95) cout << char(i - 32);
            else cout << char(i + 32);
        }
    else cout << i;
}

JAVA

Run

import java.util.Scanner;
public class Main {
    public static void change(String str, char c, int len) {
        char[] ch = str.toCharArray();
        for (int i = 0; i < ch.length; i++) {
            if (c == ch[i]) {
                if (Character.isUpperCase(ch[i])) {
                    ch[i] = Character.toLowerCase(ch[i]);

                } else if (Character.isLowerCase(ch[i])) {
                    ch[i] = Character.toUpperCase(ch[i]);

                }
            }
        }
        System.out.print(new String(ch));
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        char c = sc.next().charAt(0);
        int len = str.length();
        change(str, c, len);
    }
}

Python

Run

ss = input()
k = input()
if k.isupper():
    s = s.replace(k, chr(ord(k) + 32))
else:
    s = s.replace(k, chr(ord(k) - 32))
print(s)

Question 4 : Individual Character Count

Problem Statement :

You’re given a string where multiple characters are repeated consecutively. You’re supposed to reduce the size of this string using mathematical logic given as in the example below :

Input :
aabbbbeeeeffggg

Output:
a2b4e4f2g3

Input :
abbccccc

Output:
ab2c5

C++

JAVA

Python

C++

Run

#include 

using namespace std;

int main() {
    string s;
    getline(cin, s);
    map < char, int > m;
    for (auto i: s) m[i]++;
    for (auto i: m) cout << i.first << i.second;
}

JAVA

Run

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        int i, j, k = 0, count = 0;
        String uni = new String("");
        for (i = 0; i < str.length(); i++) {
            count = 0;
            for (j = 0; j <= i; j++) {
                if (str.charAt(i) == str.charAt(j)) {
                    count++;
                }
            }
            if (count == 1) {
                uni = uni + str.charAt(i);
            }
        }
        for (i = 0; i < uni.length(); i++) {
            count = 0;
            for (j = 0; j < str.length(); j++) {
                if (uni.charAt(i) == str.charAt(j)) {
                    count++;
                }
            }
            if (count == 1) {
                System.out.printf("%c", uni.charAt(i));
            } else {
                System.out.printf("%c%d", uni.charAt(i), count);
            }
        }
    }
}

Python

Run

s = input()
i = 1
c = 1
while i < len(s):
    if s[i] == s[i - 1]:
        c += 1
    else:
        print(s[i - 1], end="")
        print(c, end="")
        c = 1
    i += 1
print(s[i - 1], end="")
print(c)

Question 5 :Make It Palindrome

Problem Statement :

You’re given a string, you’ve to print additional characters needed to make that string a palindrome.

A Palindrome is a sequence of characters that has the property of reading the same in either direction.

Input :
abede
Output :
ba

Sample Input :
abcfe

Sample output :
fcba

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>

using namespace std;

bool Pal(string s) {
    string s1 = s;
    reverse(s1.begin(), s1.end());
    return s1 == s;
}

int main() {
    string s;
    getline(cin, s);
    int i;
    for (i = 0; i < s.length() - 1; i++)
        if (Pal(s.substr(i, s.length() - i))) break;
    s = s.substr(0, i);
    reverse(s.begin(), s.end());
    cout << s;
}

JAVA

Run

import java.util.*;
public class Main {
  public static boolean isPalindrome(String str) {
    char arr[] = str.toCharArray();
    for (int i = 0, j = arr.length - 1; i < j; i++, j--)
      if (arr[i] != arr[j])
        return false;
    return true;
  }
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String str = sc.next();
    String res = "";
    for (int i = 0; i < str.length(); i++) {
      if (isPalindrome(str.substring(i, str.length()))) {
        res = str.substring(0, i);
        break;
      }
    }
    System.out.println(new StringBuilder(res).reverse());
  }
}

Python

Run

def ispalindrome(s):
    return s == s[::-1]

def solve(s):
    if ispalindrome(s):
        return None
    for i in range(len(s)):
        x = s[:i][::-1]
        if ispalindrome(s + x):
            return x

s = input()
print(solve(s))

FAQs on Tata Elxsi Coding Questions with Solutions

Question 1: Is the Tata Elxsi Assessment hard?

Tata Elxsi Assessment consists of multiple topics including verbal, logical and coding questions. It is not hard, if you practice the questions before exam.

Question 2: Is Coding Questions asked in the Tata Elxsi Assessment?

Yes, two coding questions are asked in the Tata Elxsi Assessment. Both are of hard difficulty level.

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

Checkout list of all the video courses in PrepInsta Prime Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

2 comments on “Tata Elxsi Coding Questions”

Arya
what about the cut off in tata elxsi 1st test

Log in to Reply

PrepInsta Support
Kindly stay updated to your mails and spam folder, company will be mailing you regarding the further updates soon, cut offs depends on the performance and number of students attempting.

Log in to Reply

×

30+ Companies are Hiring

Get Hiring Updates right in your inbox from PrepInsta