π―IBM is hiring Data Engineer
Qualifications: Bachelorβs/ Masterβs Degree
Salary: 6.8 - 11 LPA (Expected)
Experience: Entry Level
Location: Bangalore; Hyderabad; Mumbai; Pune, India
Apply Now: https://careers.ibm.com/job/20558139/data-engineer-pune-in/?codes=SN_LinkedIn&Codes=SN_LinkedIn (Pune)
https://careers.ibm.com/job/20256584/data-engineer-navi-mumbai-in/?codes=SN_LinkedIn&Codes=SN_LinkedIn (Mumbai)
https://careers.ibm.com/job/20623494/data-engineer-bangalore-in/?codes=SN_LinkedIn&Codes=SN_LinkedIn (Bengalore)
Qualifications: Bachelorβs/ Masterβs Degree
Salary: 6.8 - 11 LPA (Expected)
Experience: Entry Level
Location: Bangalore; Hyderabad; Mumbai; Pune, India
Apply Now: https://careers.ibm.com/job/20558139/data-engineer-pune-in/?codes=SN_LinkedIn&Codes=SN_LinkedIn (Pune)
https://careers.ibm.com/job/20256584/data-engineer-navi-mumbai-in/?codes=SN_LinkedIn&Codes=SN_LinkedIn (Mumbai)
https://careers.ibm.com/job/20623494/data-engineer-bangalore-in/?codes=SN_LinkedIn&Codes=SN_LinkedIn (Bengalore)
π3
Bottomline Off Campus Hiring Fresher For Associate Software Engineer
Location: Bangalore
Qualification: B.E / B.Tech / MCA / M.E / M.Tech / M.Sc
Work Experience: Fresher
CTC: Approx 5.5 - 9 LPA
Apply Now:- https://boards.greenhouse.io/bottomlinetechnologies/jobs/7498875002?gh_src=e1c2a8322us
Location: Bangalore
Qualification: B.E / B.Tech / MCA / M.E / M.Tech / M.Sc
Work Experience: Fresher
CTC: Approx 5.5 - 9 LPA
Apply Now:- https://boards.greenhouse.io/bottomlinetechnologies/jobs/7498875002?gh_src=e1c2a8322us
π2
Urban Company off campus Hiring Fresher For SDE 1- Growth
Location : Gurgaon
Qualification : Bachelorβs/Masterβs graduates
Work Experience : Fresher - 2 years
CTC : 12 - 13.5 LPA
Apply now :- https://careers.urbancompany.com/jobDetail?id=9af0656f-fad3-421a-802e-caf1591f8a4f
Location : Gurgaon
Qualification : Bachelorβs/Masterβs graduates
Work Experience : Fresher - 2 years
CTC : 12 - 13.5 LPA
Apply now :- https://careers.urbancompany.com/jobDetail?id=9af0656f-fad3-421a-802e-caf1591f8a4f
π2
Cognizant | Voice Call Assistant Job | Freshers and Experienced both can apply
https://forms.office.com/pages/responsepage.aspx?id=B8QI3rkZfUKf6O3yVDAMp5sSAwA1DEhDg5LC-vaHTGlUNTZWRTVUMlFHWDZKTERWNTJOVVI0UVdKMS4u
https://forms.office.com/pages/responsepage.aspx?id=B8QI3rkZfUKf6O3yVDAMp5sSAwA1DEhDg5LC-vaHTGlUNTZWRTVUMlFHWDZKTERWNTJOVVI0UVdKMS4u
π2π1
Apple is hiring for Full Stack Developer role
45 - 70 LPA
Apply Now:-
https://jobs.apple.com/en-us/details/200551354/full-stack-engineer?team=SFTWR
45 - 70 LPA
Apply Now:-
https://jobs.apple.com/en-us/details/200551354/full-stack-engineer?team=SFTWR
Intel Freshers Recruitment | Graduate Intern Bangalore
Qualification:- BTech, MTech
Apply Now:-
https://jobs.intel.com/en/job/-/-/599/66537757808
Qualification:- BTech, MTech
Apply Now:-
https://jobs.intel.com/en/job/-/-/599/66537757808
π3
Razorpay is hiring for SDE l role
1+ year experience required
CTC : 15 - 20 LPA
https://boards.greenhouse.io/razorpaysoftwareprivatelimited/jobs/4438927005
1+ year experience required
CTC : 15 - 20 LPA
https://boards.greenhouse.io/razorpaysoftwareprivatelimited/jobs/4438927005
π2
Revature | Software Engineer | 4 - 6 LPA | 2023/2024 passouts
Apply Now:-
https://revature.com/apply/
Apply Now:-
https://revature.com/apply/
β€1
Infosys
Job ID/Reference Code
INFSYS-EXTERNAL-184946
Work Experience
2-5 Years
Job Title
SQL Database
https://career.infosys.com/jobdesc?sourceId=4003&jobReferenceCode=INFSYS-EXTERNAL-184946
Job ID/Reference Code
INFSYS-EXTERNAL-184946
Work Experience
2-5 Years
Job Title
SQL Database
https://career.infosys.com/jobdesc?sourceId=4003&jobReferenceCode=INFSYS-EXTERNAL-184946
π2
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
Position: System Development Engineer
Job type: FULL TIME
Salary: 15 LPA
Apply Now:-
https://www.amazon.jobs/en/jobs/2598111/system-development-engineer-i
π2
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
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)
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)
π4
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
# 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
π1
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)
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)
π1
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)
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)
π1
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))
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
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
π3
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
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
π3