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
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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
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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
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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
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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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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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Channel name was changed to Β«TCS NQT Infosys Exam SOLUTION / ALL EXAM SOLUTIONΒ»
Reasoning 10 am slot
1 197
2 341
3 304
4.73
5.858
6.1192
7.I
8.A
9.C
Telegram-
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http://t.me/Coding_solution_0
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1 197
2 341
3 304
4.73
5.858
6.1192
7.I
8.A
9.C
Telegram-
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http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
Telegram
Off Campus Job Update
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πDM for Promotion @Wren_0
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75+ Companies Materials
100+ JOB Updates in a Month
πDM for Promotion @Wren_0
πPlacement Updates: @offcampusjobs_0
π Placment Material: @placement_materials0
Forwarded from Off Campus Job Update
Join Us For Daily Job Updates π£
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Python
Courth house Codeβ β
Telegram
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Courth house Codeβ β
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 Codeβ β
Telegram
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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 Codeβ β
Telegram
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Python
Jack Code
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Jack Code
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C++
Tcs Homies and pizza code
http://t.me/Coding_solution_0
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Tcs Homies and pizza code
http://t.me/Coding_solution_0
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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
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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
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β
Share to all 2pm slot students
Group π
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Channel π
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Group π
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Channel π
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Channel name was changed to Β«TCS NQT Revature Infosys Exam SOLUTION / ALL EXAM SOLUTIONΒ»
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
β
http://t.me/Coding_solution_0
-------------------
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
β
http://t.me/Coding_solution_0
TCS REASONING
1)113
2)222
3)14
4)607
5)42
6)143
7)
β Telegram- http://t.me/Coding_solution_0
http://t.me/Coding_solution_0
1)113
2)222
3)14
4)607
5)42
6)143
7)
β Telegram- http://t.me/Coding_solution_0
http://t.me/Coding_solution_0