Amazon Software System Development Engineer Jobs - Apply Now for On-Demand Interviews!"

Position: System Development Engineer
Job type: FULL TIME
Salary: 15 LPA

Apply Now:-

https://www.amazon.jobs/en/jobs/2598111/system-development-engineer-i

SAP Off Campus Hiring Fresher For IT Technology Services Associate

Location: Bangalore
Qualification: Bachelor's Degree
Work Experience: Fresher
Salary: 7 - 8 LPA


Apply Now:-
https://jobs.sap.com/job/Bangalore-IT-Technology-Services-Associate-560066/1090655601/?feedId=244601&utm_campaign=limitedlistings&utm_source=LinkedinJobPostings

Accenture HackDiva Test Pattern - 2 Codes

Round 1-2 Codes (90 min)

Round 2-2 Codes (120 min)

Round 3-2 Codes (120 min)

Note: Each Round is Elimination

Round 1 (DATE: 07 JULY 2024)

Round 2 (DATE: 14 JULY 2024)

Round 3 (DATE: 21 JULY 2024)

📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
📌Linux
📌Networking
📌Design patterns
📌Donet
📌Docker
📌Entity framework
📌Node.js
📌ASP. Net
📌Aps. Net cro
📌java
📌JavaScript
📌full stack developer

Tutorials + Books + Courses + Trainings + Workshops + Educational Resources

🔹Data science
🔹Python
🔹Artificial Intelligence
🔹AWS Certified
🔹Cloud
🔹BIG DATA
🔹Data Analytics
🔹BI
🔹Google Cloud Platform
🔹IT Training
🔹MBA
🔹Machine Learning
🔹Deep Learning
🔹Ethical Hacking
🔹SPSS
🔹Statistics
🔹Data Base
🔹Learning language resources  English , 🇫🇷




𝐂𝐘𝐁𝐄𝐑 𝐒𝐄𝐂𝐔𝐑𝐈𝐓𝐘 𝐀𝐋𝐋  𝐂𝐎𝐔𝐑𝐒𝐄

⚡️ Basics


⚡️ Reconnaissance and Footprinting


⚡️ Network Scanning


⚡️ Enumeration


⚡️ Firewalls HIDs Honeypot


⚡️ Malware and Threats


⚡️ Mobile Platform


⚡️ Pentesting



⚡️ Sql Injection



⚡️ System Hacking



⚡️ Web Application


⚡️ Wireless Network



⚡️ Cloud Computing



⚡️ Web Server



⚡️ Social Engineering



⚡️ Session Hijacking



⚡️ Sniffing



⚡️ BufferOverflow


⚡️ Cryptography



⚡️ Denial Of Service


All courses (100 rupees)

Contact:- @meterials_available

def split_string_cost(S):
# Length of the string S
len_S = len(S)

# To store the cost of the split parts
max_cost = 0

# Set to keep track of distinct characters in the first part
distinct_chars_A = set()

# List to keep track of the cost for the second part from each split position
cost_B = [0] * len_S

# Set to keep track of distinct characters in the second part
distinct_chars_B = set()

# Calculate cost for second part from the end
for i in range(len_S - 1, -1, -1):
distinct_chars_B.add(S[i])
cost_B[i] = len(distinct_chars_B)

# Calculate maximum sum of cost for parts A and B
for i in range(len_S - 1):
distinct_chars_A.add(S[i])
cost_A = len(distinct_chars_A)
cost = cost_A + cost_B[i + 1]
max_cost = max(max_cost, cost)

# Calculate the result as |S| - X
result = len_S - max_cost
return result

# Example usage
S = "aaabbb"
print(split_string_cost(S)) # Output: 3

def minimum_unique_sum(A):
N = len(A)
A.sort()
total = A[0]
for i in range(1, N):
if A[i] <= A[i-1]:
A[i] = A[i-1] + 1
total += A[i]
return total

# Input format
N = int(input())
A = []
for i in range(N):
A.append(int(input()))

# Output
result = minimum_unique_sum(A)
print(result)

def count_distinct_strings(S):
distinct_strings = set()

for i in range(len(S) - 1):
new_string = S[:i] + S[i+2:]
distinct_strings.add(new_string)

return len(distinct_strings)

# Read input string
S = input().strip()

# Get the number of distinct strings that can be generated
result = count_distinct_strings(S)
print(result)

Minimum substring ..

def solve(N, A):
unique_sums = set()

for start in range(N):
current_sum = 0
for end in range(start, N):
current_sum += A[end]
unique_sums.add(current_sum)

print(len(unique_sums))

import sys

det solve(N, A)

for i in range(N):

if A[i]=0;

A[1]

ps=0

M-1

pm={0:-1)

for i in range(N):

ps+=A[i]

if ps in pm:

m=max(m,i-pm[ps])

else:

pm[ps]=i

return m

def main():

Nint(sys.stdin.readline().strip())

A-[]

for_ in range(N):

A.append(int(sys.stdin.readline().strip()))

result = solve(N, A)

print(result)




Largest Subarray with equal number


Infosys

def max_sum_of_distinct_characters(S):

n = len(S)

left_chars = set()

right_chars = set()

left_count = [0] * n

right_count = [0] * n

for i in range(n):

left_chars.add(S[i])

left_count[i] = len(left_chars)

for i in range(n-1, -1, -1):

right_chars.add(S[i])

right_count[i] = len(right_char)

max_sum = 0

for i in range(n-1):

max_sum = max(max_sum, left_count[i] + right_count[i+1])

return n- max_sum



Split String code

Python 3

All passed

def max_cost_split(s):
n = len(s)

max_cost = 0

for i in range(1, n):
a = s[:i]
b = s[i:]
cost_a = len(set(a))
cost_b = len(set(b))
max_cost = max(max_cost, cost_a + cost_b)

return n - max_cost




Infosys max cost split code in python

Minimum unique sum

Minimum difference pairs in python

Sub set with LCM

Python

class TreeNode:
def init(self, value=0, left=None, right=None):
self.value = value
self.left = left
self.right = right

def count_nodes(node, counts):
if node is None:
return
if node.value in counts:
counts[node.value] += 1
else:
counts[node.value] = 1
count_nodes(node.left, counts)
count_nodes(node.right, counts)

def find_double_roots(root):
counts = {}
count_nodes(root, counts)
double_roots = [value for value, count in counts.items() if count > 1]
return double_roots

def main():
# Example tree:
# 1
# / \
# 2 3
# / \
# 2 4
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(2)
root.left.right = TreeNode(4)

result = find_double_roots(root)
print("Nodes with double roots:", result)

if name == "main":
main()