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Python 3
Alfred McDonald's Code
✅Telegram- 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

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

Send the Question here ❤️🚀
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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: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)))

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Reasoning 10 am slot

1 197
2 341
3 304
4.73
5.858
6.1192
7.I
8.A
9.C
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Python
Courth house Code✅✅
Telegram
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http://t.me/Coding_solution_0
http://t.me/Coding_solution_0

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

Python
Jack Code
Telegram-
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http://t.me/Coding_solution_0
http://t.me/Coding_solution_0

C++
Tcs Homies and pizza code
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
http://t.me/Coding_solution_0

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