InfyTQ is the ultimate technology certification for aspiring professionals looking to verify their industrial preparedness and pursue their dream careers. This annual program, offered by Infosys, puts your programming and database skills to the test. By successfully completing InfyTQ, you will have the opportunity to participate in a pre-placement interview for a specialist technical role at Infosys or to attend a technical interview for the position of systems engineer and become an Infosys Certified Software Programmer.
1. Obtaining the globally recognized InfyTQ Certification will boost your resume and your will stand out in the job market.
2. With InfyTQ Certification you get a chance to become a certified developer.
3. This prestigious certification is open to a wide range of students from various educational backgrounds, including B.Tech, B.E, M.W, M.Tech, M.Sc, and M.C.A.
4. Not only will the InfyTQ Certification give your resume a competitive edge, but it also gives you the chance to participate in off campus interviews at Infosys and potentially secure a position at the company.
5. You can even attempt the InfyTQ Certification up to three times.
1. If you are a student graduating in particular next year with a B.Tech, M.Tech, B.E, M.E, M.Sc, M.C.A., or MCM degree? Then you are eligible for the InfyTQ Certification Program of that year. For example, if you are graduating in 2023 then you are eligible for the InfyTQ Certification Program 2022-2023.
2. The InfyTQ Certification Program is a great opportunity for aspiring software programmers to showcase their skills. If you score at least 65% on the program, you may be contacted for an on-the-spot employment interview at Infosys.
3. It's worth noting that Infosys will not consider your college grades when calculating your InfyTQ Certification Program Score. This is your chance to demonstrate your abilities and potentially secure a position as a software programmer at Infosys.
Don't miss out on this opportunity to launch your career with InfyTQ. Make sure to apply for the InfyTQ Certification Program if you meet the eligibility requirements.
The InfyTQ Certification exam is designed to test your knowledge and skills in a variety of areas, including Python and Java programming, database management systems, and general aptitude. The exam consists of two rounds: the Qualifier Round and the Final Round.
The Qualifier Round is divided into three parts:
The Final Round is divided into two parts:
Make sure to thoroughly prepare for both rounds of the InfyTQ Certification exam in order to increase your chances of success. This prestigious certification is a great way to enhance your resume and stand out in the job market.
InfyTQ Certification covers a wide range of topics related to software development, including algorithms and data structures, requirement engineering, databases, Python programming, and agile development practices such as Scrum. The syllabus also mentions quality assurance and the waterfall development process, which is a traditional, linear approach to software development.
Here is the complete list:
InfyTQ is a comprehensive certification program that consists of two main rounds: the Certification Round and the Advantage Round.
The Certification Round is divided into three sub-sections:
You will have the choice to select your preferred programming language (Java or Python) at the time of booking your slot. This round will last for a total of 3 hours. If you pass the Certification Round with a score of 65% or higher, you will be eligible to move on to the Advantage Round.
The Advantage Round is a higher level coding round that consists of three sections:
This round will also last for 3 hours and is worth a total of 225 marks. To pass the InfyTQ Certification Process, you will need to excel in both the Certification Round and the Advantage Round.
Post the advantage round you have,
InfyTQ Behavioral Interview-1
This is just an Infosys, technical interview to see whether it can offer you a position as a "System Engineer."
InfyTQ Interview-2
If you pass the InfyTQ Upgrade Test cut-off, you will be allowed to take the behavioral interview-2.
There is scope of getting a position at Infosys such as "Power Programming Role," "Systems Engineer Specialist," or "System Engineer," depending on how well you performed in the interview.
You might still be offered the position of system engineer. It entirely relies on how well you perform in the interview.
# Question 1
For the given list of numbers separated with a comma, where the numbers 5 and 8 are present and 8 always comes after 5, how do you calculate the sum of `num1` and `num2`, where `num1` is the sum of all numbers that do not lie between 5 and 8 in the input list and `num2` is the numbers formed by concatenating all numbers from 5 to 8 in the list?
# Example
```sh
Input: 3,2,6,5,1,4,8,9
Output: 5168
```
# Explanation
```
Num1: 3+2+6+9 = 20
Num2: 5148
Output: 5148+20 = 5168
```
# Question 1
Given a 2D array of integers, print the maximum sum of any hourglass shape in the array.
An hourglass shape is a subset of values with indices forming an hourglass shape:
```sh
Input: a b c d e f g
```
# For example,
```sh
Input: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
```
The hourglass with the maximum sum is:
# Solution 1
We can solve this problem by iterating through each element in the 2D array and computing the sum of its hourglass shape. We can keep track of the maximum sum found so far and return it at the end.
To compute the sum of an hourglass shape, we can use the following formula:
# sum = arr[i][j] + arr[i][j+1] + arr[i][j+2] + arr[i+1][j+1] + arr[i+2][j] + arr[i+2][j+1] + arr[i+2][j+2]
# where i and j are the indices of the top-left element of the hourglass shape.
Here is the complete solution:
```sh
from typing import List
def max_hourglass_sum(arr: List[List[int]]) -> int:
max_sum = -float('inf') # Initialize max sum to lowest possible value
# Iterate through each element in the 2D array
for i in range(len(arr) - 2):
for j in range(len(arr[i]) - 2):
# Compute the sum of the hourglass shape
sum = arr[i][j] + arr[i][j+1] + arr[i][j+2] + arr[i+1][j+1] + arr[i+2][j] + arr[i+2][j+1] + arr[i+2][j+2]
# Update the max sum if necessary
max_sum = max(max_sum, sum)
return max_sum
```
# Question 2
What is the largest even number that can be produced from a string that contains a mixture of letters, numbers, and special characters, after removing duplicates, or -1 if no even number can be produced?
# For example,
```sh
Input: 3,2,6,5,1,4,8,9
Output: -1
```
```sh
Input: infosys@337
Output: Hello#81@21349
```
# Question 2
You are given a sequence of integers, and you are allowed to delete any number of elements from the sequence. The goal is to maximize the sum of the remaining elements. However, you must adhere to the following rule: for any two elements a and b in the final sequence, abs(a - b) -> 1.
For example, given the sequence [1, 3, 2, 4], you can delete the 3 and 4 to obtain the maximum sum of 1 + 2 = 3.
Write a function
```sh
def max_sum(nums: List[int]) -> int
```
to solve the problem.
# Solution 2
We can solve this problem using dynamic programming. We will use a 2D array dp where dp[i][j] represents the maximum sum of a subarray of nums ending at index i such that the difference between the minimum and maximum elements in the subarray is j.
# To compute dp[i][j], we can consider the following cases:
* If j > i, then the maximum sum is 0 because it is not possible to have a subarray of length j ending at index i.
* If j = 0, then the maximum sum is the sum of all elements in the subarray nums[0:i+1].
* If j > 0, then we can compute the maximum sum by considering all possible subarrays ending at index i such that the difference between the minimum and maximum elements is j. We can do this by iterating through all possible values of k from 0 to i and taking the maximum of dp[k][j-1] + sum(nums[k+1:i+1]).
We can initialize the array dp with all zeros and compute the values in a bottom-up manner. The final result is the maximum value in dp[n-1], where n is the length of nums.
# Here is the complete solution:
```sh
from typing import List
def max_sum(nums: List[int]) -> int:
n = len(nums)
# Initialize dp array with all zeros
dp = [[0] * n for _ in range(n)]
# Fill in the base cases
for i in range(n):
dp[i][0] = sum(nums[:i+1])
# Fill in the rest of the array in a bottom-up manner
for j in range(1, n):
for i in range(j, n):
for k in range(i):
dp[i][j] = max(dp[i][j], dp[k][j-1] + sum(nums[k+1:i+1]))
# Return the maximum value in dp[n-1]
return max(dp[n-1])
```
# Question 3
What is the least number of unique elements in the array after deleting N elements from the array, given an input number N and an array of elements?
# For example,
```sh
Input: N=2 and the array [1, 2, 3, 3, 4, 4]
Output: 2
```
# Question 3
Given an array of integers, find the length of the longest subarray such that the sum of the elements in the subarray is less than or equal to a given target value.
For example, given the array [1, 2, 3, 4, 5] and the target value 7, the longest subarray with a sum -> 7 is [1, 2, 3], which has a length of 3.
Write a function
```sh
def longest_subarray(arr: List[int], target: int) -> int
```
to solve the problem.
# Solution 3
We can solve this problem using the 2 pointers technique.
We will maintain two pointers start and end pointing to the beginning and end of the current subarray, respectively. We will initialize start to 0 and end to 1, and move end to the right until the sum of the elements in the subarray is greater than the target value.
Then, we will move start to the right until the sum of the elements in the subarray is -> the target value. We will keep track of the maximum length found so far and return it at the end.
# Here is the complete solution:
```sh
from typing import List
def longestsubarray(arr: List[int], target: int) -> int:
start = 0
end = 1
maxlen = 0
currsum = arr[start]
while end < len(arr):
# Move end to the right until the sum is greater than the target
while currsum + arr[end] <= target:
currsum += arr[end]
end += 1
# Update the maximum length found so far
maxlen = max(maxlen, end - start)
# Move start to the right until the sum is less than or equal to the target
while currsum > target:
currsum -= arr[start]
start += 1
# Update the maximum length found so far
maxlen = max(maxlen, end - start)
return maxlen
```
# Question 4
What is the longest prefix in a non-empty string containing only alphabets that is also a suffix, without overlap, and is case-sensitive, or -1 if no such prefix exists?
# For example,
```sh
Input: xxAbcxxAbcxx
Output: xx
```
```sh
Input: Racecar
Output: -1
```
# Question 5
How many distinct subarrays in an array of integers have an odd sum, where two subarrays are considered different if they either start or end at different indexes?
# For example,
```sh
Input: [1, 2, 3]
Output: 4
```
Edyst's training style completely resonated with me. I approached programming as more than a subject. Thanks to Edyst team for the guidance!
I started practising on Edyst platform since my 3rd year of college focused on placements & always liked the way they helped us when we were stuck at a particular problem.
Thank you, Edyst for all the assistance and amazing support!
When I joined the Edyst courses I received personalized mentoring on how to crack coding rounds of different companies. Through a combination of coding skills and great projects, I received multiple offers above 6+ lakhs per annum. Finally I joined for 8+ Lakhs package. Thanks for all the support, from Edyst Team.
Being a mechanical student and getting into an IT company is very tough. One of the main reason I could able to crack TCS CodeVita is because of Edyst.
Aneeq sir, your doubt clearing sessions, daily assignments & incredible mentors support really brushed up my skills.
# Question 1
For the given list of numbers separated with a comma, where the numbers 5 and 8 are present and 8 always comes after 5, how do you calculate the sum of `num1` and `num2`, where `num1` is the sum of all numbers that do not lie between 5 and 8 in the input list and `num2` is the numbers formed by concatenating all numbers from 5 to 8 in the list?
# Example
```sh
Input: 3,2,6,5,1,4,8,9
Output: 5168
```
# Explanation
```
Num1: 3+2+6+9 = 20
Num2: 5148
Output: 5148+20 = 5168
```
# Question 1
Given a 2D array of integers, print the maximum sum of any hourglass shape in the array.
An hourglass shape is a subset of values with indices forming an hourglass shape:
```sh
Input: a b c d e f g
```
# For example,
```sh
Input: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
```
The hourglass with the maximum sum is:
# Solution 1
We can solve this problem by iterating through each element in the 2D array and computing the sum of its hourglass shape. We can keep track of the maximum sum found so far and return it at the end.
To compute the sum of an hourglass shape, we can use the following formula:
# sum = arr[i][j] + arr[i][j+1] + arr[i][j+2] + arr[i+1][j+1] + arr[i+2][j] + arr[i+2][j+1] + arr[i+2][j+2]
# where i and j are the indices of the top-left element of the hourglass shape.
Here is the complete solution:
```sh
from typing import List
def max_hourglass_sum(arr: List[List[int]]) -> int:
max_sum = -float('inf') # Initialize max sum to lowest possible value
# Iterate through each element in the 2D array
for i in range(len(arr) - 2):
for j in range(len(arr[i]) - 2):
# Compute the sum of the hourglass shape
sum = arr[i][j] + arr[i][j+1] + arr[i][j+2] + arr[i+1][j+1] + arr[i+2][j] + arr[i+2][j+1] + arr[i+2][j+2]
# Update the max sum if necessary
max_sum = max(max_sum, sum)
return max_sum
```
# Question 2
What is the largest even number that can be produced from a string that contains a mixture of letters, numbers, and special characters, after removing duplicates, or -1 if no even number can be produced?
# For example,
```sh
Input: 3,2,6,5,1,4,8,9
Output: -1
```
```sh
Input: infosys@337
Output: Hello#81@21349
```
# Question 2
You are given a sequence of integers, and you are allowed to delete any number of elements from the sequence. The goal is to maximize the sum of the remaining elements. However, you must adhere to the following rule: for any two elements a and b in the final sequence, abs(a - b) -> 1.
For example, given the sequence [1, 3, 2, 4], you can delete the 3 and 4 to obtain the maximum sum of 1 + 2 = 3.
Write a function
```sh
def max_sum(nums: List[int]) -> int
```
to solve the problem.
# Solution 2
We can solve this problem using dynamic programming. We will use a 2D array dp where dp[i][j] represents the maximum sum of a subarray of nums ending at index i such that the difference between the minimum and maximum elements in the subarray is j.
# To compute dp[i][j], we can consider the following cases:
* If j > i, then the maximum sum is 0 because it is not possible to have a subarray of length j ending at index i.
* If j = 0, then the maximum sum is the sum of all elements in the subarray nums[0:i+1].
* If j > 0, then we can compute the maximum sum by considering all possible subarrays ending at index i such that the difference between the minimum and maximum elements is j. We can do this by iterating through all possible values of k from 0 to i and taking the maximum of dp[k][j-1] + sum(nums[k+1:i+1]).
We can initialize the array dp with all zeros and compute the values in a bottom-up manner. The final result is the maximum value in dp[n-1], where n is the length of nums.
# Here is the complete solution:
```sh
from typing import List
def max_sum(nums: List[int]) -> int:
n = len(nums)
# Initialize dp array with all zeros
dp = [[0] * n for _ in range(n)]
# Fill in the base cases
for i in range(n):
dp[i][0] = sum(nums[:i+1])
# Fill in the rest of the array in a bottom-up manner
for j in range(1, n):
for i in range(j, n):
for k in range(i):
dp[i][j] = max(dp[i][j], dp[k][j-1] + sum(nums[k+1:i+1]))
# Return the maximum value in dp[n-1]
return max(dp[n-1])
```
# Question 3
What is the least number of unique elements in the array after deleting N elements from the array, given an input number N and an array of elements?
# For example,
```sh
Input: N=2 and the array [1, 2, 3, 3, 4, 4]
Output: 2
```
# Question 3
Given an array of integers, find the length of the longest subarray such that the sum of the elements in the subarray is less than or equal to a given target value.
For example, given the array [1, 2, 3, 4, 5] and the target value 7, the longest subarray with a sum -> 7 is [1, 2, 3], which has a length of 3.
Write a function
```sh
def longest_subarray(arr: List[int], target: int) -> int
```
to solve the problem.
# Solution 3
We can solve this problem using the 2 pointers technique.
We will maintain two pointers start and end pointing to the beginning and end of the current subarray, respectively. We will initialize start to 0 and end to 1, and move end to the right until the sum of the elements in the subarray is greater than the target value.
Then, we will move start to the right until the sum of the elements in the subarray is -> the target value. We will keep track of the maximum length found so far and return it at the end.
# Here is the complete solution:
```sh
from typing import List
def longestsubarray(arr: List[int], target: int) -> int:
start = 0
end = 1
maxlen = 0
currsum = arr[start]
while end < len(arr):
# Move end to the right until the sum is greater than the target
while currsum + arr[end] <= target:
currsum += arr[end]
end += 1
# Update the maximum length found so far
maxlen = max(maxlen, end - start)
# Move start to the right until the sum is less than or equal to the target
while currsum > target:
currsum -= arr[start]
start += 1
# Update the maximum length found so far
maxlen = max(maxlen, end - start)
return maxlen
```
# Question 4
What is the longest prefix in a non-empty string containing only alphabets that is also a suffix, without overlap, and is case-sensitive, or -1 if no such prefix exists?
# For example,
```sh
Input: xxAbcxxAbcxx
Output: xx
```
```sh
Input: Racecar
Output: -1
```
# Question 5
How many distinct subarrays in an array of integers have an odd sum, where two subarrays are considered different if they either start or end at different indexes?
# For example,
```sh
Input: [1, 2, 3]
Output: 4
```
All InfyTQ Questions
All InfyTQ Questions