Off Campus Job Update
前往频道在 Telegram
500+ Placement & Competitive Exam Materials 75+ Companies Materials 100+ JOB Updates in a Month 🚀DM for Promotion @Wren_0 🚀Placement Updates: @offcampusjobs_0 🌟 Placment Material: @placement_materials0
显示更多8 266
订阅者
无数据24 小时
无数据7 天
-12430 天
帖子存档
#include<bits/stdc++.h>
using namespace std;
int main(){
int n ;
cin>>n;
vector<int>a(n);
vector<int>b(n);
for(int i=0;i<n;i++)
cin>>a[i];
for(int i=0;i<n;i++)
cin>>b[i];
int ans=1;
for(int i=0;i<n;i++)
{
int temp=1;
for(int j=i+1;j<n;j++)
{
if((a[i]>=a[j] && a[i]<=b[j]) || (a[j]>=a[i] && a[j]<=b[i] ))
temp++;
}
ans= max(ans,temp);
}
cout<<ans;
return 0;
}
Python
Courth house Code✅✅
Telegram
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
Python
Courth house Code✅✅
Telegram
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
Join Us For Daily Job Updates 📣
https://t.me/offcampusjobs_0
https://t.me/offcampusjobs_0
https://t.me/offcampusjobs_0
Reasoning 10 am slot
1 197
2 341
3 304
4.73
5.858
6.1192
7.I
8.A
9.C
Telegram-
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
Join Us For Daily Job Updates 📣
https://t.me/offcampusjobs_0
https://t.me/offcampusjobs_0
https://t.me/offcampusjobs_0
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
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
def minOps(A, B):
m = len(A)
n = len(B)
# This part checks whether conversion is possible or not
if n != m:in
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)))
✅Telegram- Send the Question here ❤️🚀
https://t.me/good_coders
https://t.me/good_coders
https://t.me/good_coders
Send the Question here ❤️🚀
https://t.me/good_coders
https://t.me/good_coders
https://t.me/good_coders
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
def check(a,n)
if n==1:
return 1
a.sort()
c=0
i=0
while(i<n-1):
if a[i+1]-a[i]==1:
i=i+2
c=c+1
else:
i=i+1
if c:
return c
else:
return 1
n=int(input())
a=[]
for i in range(n):
a.append(int(input()))
print(check(a,n))
Python
Whales code
Telegram -
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0.
http://t.me/Coding_solution_0
Revature Exam pattern
✅All solution Available at
http://t.me/Coding_solution_0
Its a fantastic News for Final Year Students📖, Graduates👨🏻🎓 and people who have just started their career .
🤯• Pay Only When you land a JOB. Yes, You are right!! Why pay hefty amount on courses Initially ?
💻• A 9 Months coding bootcamp on Web Development- Learn Full Stack development, DS and Algo, System Design and work on various course projects.
💵. An Average Package of 7.8 Lakhs offered among all the batches of career camp.
🤝. A dedicated Placement support for the students with over 50 opportunities a week
🚀• Get Your career Started in product based organisations like RazorPAy, Paytm, Snaphunt etc.
Only few seats remain!!
Register Now- https://bit.ly/3H2IUL4
Last date for the registration- 30th Jan
