The other simpler approach

#1

Can’t we approach that way like find the number of bit permutations of n digit number having no consecutive ones
for 2 digit – 00 01 10
for 3 digit – 000 001 010 100 101
for 4 digit – 0000 0001 0010 0100 1000 1001 1010

for this question also

#2

hi @arjun_121 yes you can do that as well. However you should also know the DP approach.

#3

#include
using namespace std;

int main() {
int t;
int A[26];
A[0]=0;A[1] =2;A[2] = 3;
for(int i=3;i<26;i++){
A[i] = A[i-1]+A[i-2];
}
cin>>t;
while(t–){
int n;
cin>>n;
cout<<"#"<<n<<" : “<<A[n]<<”\n";
}
return 0;
}

I have used that approach but that is giving me a wrong answer

#4

@arjun_121 this is just bottom up DP, you have done nothing different. Also please share your code using CB IDE and explain the logic of this part

#5

https://ide.geeksforgeeks.org/hJupcr8Xga

i have just tried to precompute the number of computations a N digit number formed from { 0 , 1} can have without consecutive ones.

#6

and it follows a recurrence as f(n) = f(n-1) + f(n-2)

#7

hi @arjun_121
your wrong answer is basically because of the output format. See the output format given in question, you have to print the answer only, nothing else.

#8

The sample output shows as such
#1 : 2
#2 : 3
#3 : 5

and I have printed it the same way

#9

@arjun_121 instead of “n” you have to print the case number.
like #1 for first test case, #2 for second and so on.

#10

Thanks mam , I solved that

1 Like
#11

@arjun_121 dont forget to mark your doubt as resolved!