Infosys Exam Solution SOLUTIONS


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Репост из: Off campus jobs/Intership Uodates Daily
Microsoft Hiring Software Engineer | 2022 Batch | BE MTech

careers.microsoft.com/us/en/job/1072686/Software-Engineer-Full-Time-Opportunity

✅ Note - Share the Post Link with your college 2022 Batches
Telegram -
https://t.me/offcampusjobs_0


Репост из: Off Campus Job Update
TCS Verbal
-------------------

1) take to his books blow steam
2)bend over. Backwards
3)cut her some slak
4)powdery snow
5)effected
6) Modulate
7)Swindle
8)Full
9)commend
10)confiscate
11)Withdrawn
12)involved
13) obliterated
14)Obviates
15)BDAC
16)DACB
17)detained
18)consicous about
19)RP
20)RQ
21)The history of
22)Paris
23)Boulangers
24)Each member must sent a

http://t.me/Coding_solution_0




Репост из: Off Campus Job Update
N =int(input())
L = [int(i) for i in input().split()]
S=sum(L)
moves=0
If S%N!=0:
Print( -1)
else:
x=S//N
for i in range(len(L)):
L[i] = L[i] - x
Print(max(L))

Python
1st Qsn
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0






Репост из: Off Campus Job Update
#include
using namespace std;

int main(){
int n ;
cin>>n;
vectora(n);
vectorb(n);
for(int i=0;i>a[i];
for(int i=0;i>b[i];
int ans=1;
for(int i=0;i






Репост из: Off campus jobs/Intership Uodates Daily
DELL Off Campus Drive

Freshers = 💰7 - 8.5LPA
Experienced = 💰11 - 13LPA

firstnaukri.com/careers/customised/landingpage/dell/index.html

Telegram - https://t.me/offcampusjobs_0


Репост из: Off Campus Job Update
Join this till i am doing your Solution
🚀🚀 Join join join

https://t.me/offcampusjobs_0
https://t.me/offcampusjobs_0
https://t.me/offcampusjobs_0


Репост из: Off Campus Job Update
Add your 5 friends ❤️

Will provide all the solution here


Репост из: Off Campus Job Update
def minOps(A, B):
m = len(A)
n = len(B)

# This part checks whether conversion is possible or not
if n != m:
return -1

count = [0] * 256

for i in range(n): # count characters in A
count[ord(B[i])] += 1
for i in range(n): # subtract count for every char in B
count[ord(A[i])] -= 1
for i in range(256): # Check if all counts become 0
if count[i]:
return -1

# This part calculates the number of operations required
res = 0
i = n-1
j = n-1
while i >= 0:

# if there is a mismatch, then keep incrementing
# result 'res' until B[j] is not found in A[0..i]
while i>= 0 and A[i] != B[j]:
i -= 1
res += 1

# if A[i] and B[j] match
if i >= 0:
i -= 1
j -= 1

return res

# Driver program
A = "EACBD"
B = "EABCD"
print ("Minimum number of operations required is " + str(minOps(A,B)))


Python
Minimum number of operations required Code

http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0


Репост из: Off Campus Job Update
Revature Exam pattern

✅All solution Available at
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Репост из: Off campus jobs/Intership Uodates Daily
Your_Name-Resume.docx
29.3Кб
📝 Resume Template

✅ Kindly ask your friends or collegemates to join

🚀🚀🚀
https://t.me/offcampusjobs_0


Репост из: Off campus jobs/Intership Uodates Daily
Its a fantastic News for Final Year Students📖, Graduates👨🏻‍🎓 and people who have just started their career .


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Only few seats remain!!
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Last date for the registration- 30th Jan


Репост из: Off campus jobs/Intership Uodates Daily
MINDTREE Off Campus Drive | 🎖2020 & 2021 Batch | 💰3LPA

⚠️ Deadline - 30th
Jan
mindtree.com/careers/campus-careers

Telegram -https://t.me/offcampusjobs_0




Репост из: Off campus jobs/Intership Uodates Daily
Cognizant Off Campus Drive | 🎖2020 & 2021 Batch | 💰4LPA

⚠️ Deadline - 22 June 2022
app.joinsuperset.com/join/#/signup/student/jobprofiles/86721847-7ecb-4dbf-b02f-3155e594e760

Telegram -
https://t.me/offcampusjobs_0

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