Colt Technology Coding Questions and Answers
Colt Technology Coding Questions with Solutions
This page will help you to get Colt Technology Coding Questions and Answers asked in the Recruitment Process of the company. Other than that you will find Insights on Online Assessments, Steps involved in the Recruitment Process, Eligibility Criteria, CTC Offered, Interview Rounds, and multiple Job Profiles offered.
Colt Technology offers different domains for Graduate Trainee Engineer Job Profile like Information Technology, Data Engineer, and many more. For getting details about mentioned profiles, please go through this page.
About Colt Technology
Colt Technology, formally known as Colt Technologies Services is a top-class Product Based Company to provide services regarding Network, Data, and Telecommunications services that include cybersecurity, cloud computing, Data Centers, networking system, and their security, specialized Business Related Networks, and VPNs.
About Colt Technology Recruitment Process
This Recruitment process consists of the following steps :
- Online Aptitude Assessment
- Technical Interview
- HR Interview
Colt Technology Job Descriptions
The Eligibility Criteria are different for all roles coming under Trainee Engineer Job Profile. For your reference and understanding, we have discussed each and every Role below:
Trainee Engineer - Information Security
- Course: BE/B.Tech
- Specialization: Electronic & Communication/ Electrical and Electronics Engineering/ Information Technology/ Computer Science/ Information Security.
- Basic understanding of operating systems, networking and security enforcing technologies.
- Basic understanding of Cybersecurity.
- Basic understanding of privacy requirements (including GDPR).
- Information security or information risk certification desirable (or willingness to achieve this).
Trainee Engineer - Data Engineer
- Course: M.Sc/MBA or BE/B.Tech
- Specialization: MBA/M.Sc – Maths, Statistics, Data Science, BE/B.Tech – CS/IT, ECE.
- Knowledge of Python, Relational Database Management Systems (SQL), and No SQL.
- Basic knowledge of Computer Networks, Operating Systems, Object Oriented Programming, and Software Development Principles.
- Prior experience in development in some of the widely used technologies – .Net, Java, JavaScript, etc. will be preferred.
- Knowledge of data Visualization techniques and working experience with a few visualization tools will be handy.
Trainee Engineer - Project Coordinator
- Course: BE/B.Tech
- Specialization: Electronics and Telecommunication/ Electronics and Communication/ Instrumentation/ Electronics and Instrumentation.
- Builds effective relationships with other team members and key stakeholders such as internal customers and suppliers.
- Aptitude for learning, to build knowledge of the Colt business, products, systems, processes and customers.
- Good with numbers, able to operate in ways that either enhance revenue or contain costs.
- Good presentation & communication skills.
Trainee Engineer - Data Strategy & Innovation
- Course: M.Sc/MBA or BE/B.Tech
- Specialization: MBA/M.Sc – Maths, Statistics, Data Science, BE/B.Tech – CS/IT, ECE
- Business and Functional knowledge and appreciation of use of data for Insights generation.
- Knowledge of statistical models, techniques to handle large volume data, AI/ML, Augmented data etc.
- Prior experience of development in some of the widely used technologies – C, Python, etc. will be preferred.
Trainee Engineer - Quality Assurance Group
- Course: BE/B.Tech
- Specialization: Computer Science/ Information Technology/ Electronics
- Knowledge of software testing.
- Knowledge in Telecommunication Domain will be an added advantage.
- Able to logically analyze problems and identify solutions.
- Should have excellent analytical skills.
Trainee Engineer - Identity & Access Governance
- Course: BE/B.Tech
- Specialization: Computer Science/ Information Technology
- Knowledge of Python, Relational Database Management Systems (SQL), and No SQL.
- Basic knowledge of Computer Networks, Operating Systems, Object Oriented Programming, and Software Development Principles.
- Prior experience in development in some of the widely used technologies – .Net, Java, JavaScript, etc. will be preferred.
- Ability to multitask and prioritize based on business requirements.
- Must be able to collaborate and effectively communicate with colleagues in disparate geographical locations.
We have mentioned further details of the Colt Technology Recruitment Process in the following Tabular Form
Colt Technology | Related Information |
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Position : | Trainee Engineer |
Course : |
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Eligibility Criteria / Academic Qualification Required : |
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Cost to Company (CTC) | Rs. 5.5 L.P.A |
Selection Process : |
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Joining Location : |
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Sample Colt Technology Coding Questions With Solutions
Question 1 : 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.
newPassword has the following parameter(s):
Str : string a
Str : string b
Returns:
Str: new password using two strings
Sample case 0 :
Sample input0:
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))
Question 2 : Mr. Robot’s Password (R)
Mr. Robot is making a website, in which there is a tab to create a password. As other websites, there are rules so that the password gets complex and none can predict the password for another. So he gave some rules like:
– At least one numeric digit
– At Least one Small/Lowercase Letter
– At Least one Capital/Uppercase Letter
– Must not have space
– Must not have slash (/)
– At least 6 characters
If someone inputs an invalid password, the code prints: “Invalid password, try again”.
Otherwise, it prints: “password valid”.
Input Format:
A line with a given string as a password
Output Format:
If someone inputs an invalid password, the code prints: “Invalid password, try again”.
Otherwise, it prints: “password valid”, without the quotation marks.
Constraints:
Number of character in the given string <=10^9
Sample input 1:
abjnlL09
Sample output 1:
password valid
Sample input 2:
jjnaskpk
Sample output 2:
Invalid password, try again
#include<bits/stdc++.h> using namespace std; int CheckPassword (char str[], int n) { if (n < 4) return 0; int a = 0, cap = 0, nu = 0, low = 0; while (a < n) { if (str[a] == ' ' || str[a] == '/') return 0; if (str[a] >= 65 && str[a] <= 90) { cap++; } if (str[a] - 32 >= 65 && str[a] - 32 <= 90) { low++; } else if (str[a] - '0' >= 0 && str[a] - '0' <= 9) nu++; a++; } return cap > 0 && nu > 0 && low > 0; } int main () { string s; getline (cin, s); int len = s.size (); char *c = &s[0]; if (CheckPassword (c, len)) cout << "password valid"; else cout << "Invalid password, try again"; }
def CheckPassword(s, n): if n < 4: return 0 cap = 0 nu = 0 lo = 0 for i in range(n): if s[i] == " " or s[i] == "/": return 0 if s[i] >= "A" and s[i] <= "Z": cap += 1 if s[i] >= "a" and s[i] <= "z": lo += 1 elif s[i].isdigit(): nu += 1 if cap > 0 and nu > 0 and lo > 0: return 1 else: return 0 s = input() a = len(s) if CheckPassword(s, a): print("password valid") else: print("Invalid password, try again")
Question 3 : 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))
Question 4 : Death Note
Problem Statement :
Ryuk, the Shinigami (God of death) had allowed Light Yagami, a school student, to kill as many people as he can by using a death note. But writing the names barely will allow other people to watch them. So he encrypts the names using digits, where a means 1, b means 2 and so on upto z is 26. Now if he gives numbers, there is a communication error because a number string can be decrypted by the death note in various ways and eventually killing them all. If everyone in the world has a unique name, for a given number, how many people can die?
NOTE THAT: There is every possible name in the world with the 26 letters, and capital or small letters is not a problem.
Input format:
A number stream denoting the first name’s encrypted version
Output Format:
Number of people dying by this.
Constraints:
1<=stream length<=10^8
Sample Input: 1267
Sample Output: 3
Output Specification:Two people of the name azg and lfg die.
#include <bits/stdc++.h> using namespace std; string s; int ans; void func(int i, int n) { if (i == n - 1 || i == n) { ans++; return; } if (s[i] == '1') func(i + 2, n); else if (s[i] == '2' && s[i + 1] < '7') func(i + 2, n); func(i + 1, n); } int main() { ans = 0; cin >> s; func(0, s.length()); cout << ans; }
import java.util.*; class Main { static String s; static int ans; public static void func(int i, int n) { if (i == n - 1 || i == n) { ans++; return; } if (s.charAt(i) == '1') func(i + 2, n); else if (s.charAt(i) == '2' && s.charAt(i + 1) < '7') func(i + 2, n); func(i + 1, n); } public static void main(String[] args) { Scanner sc = new Scanner(System.in); s = sc.next(); func(0, s.length()); System.out.println(ans); } }
def solve(s, n): if n == 0 or n == 1: return 1 ans = 0 if s[n - 1] > "0": ans = solve(s, n - 1) if s[n - 2] == "1" or (s[n - 2] == "2" and s[n - 1] < "7"): ans += solve(s, n - 2) return ans s = input() print(solve(s, len(s)))
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
#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; }
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()); } }
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 Colt Technology Coding Questions with Solutions
Question 1: What is the recruitment process for Colt Technology?
Colt Technology Recruitment process consists of the following rounds:-
- Online Assessment
- Technical Interview
- HR Interview
Question 2: Is Colt Technology exam difficult to crack?
Colt Technology exam is of medium to high difficulty level. It gives higher preference to coding questions and to those with an analytical mind.
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