Pattern 110 : Digit DP

#1

I understood this question but unable to proceed with the implementation part. Can you please help me with that.

#2

did you try the naive solution?
the naive solution can be run a loop from L to R, and check in every number if 110 exists in its binary form , if yes , count++.
how to check for 110 in a number?
while(n){
int x = n&1;
if(x==0){
//wrote your code to check for two consecutive 1’s. }
n= n>>1;
}

then think of a better solution by establishing some relationship of binary form between two consecutive numbers.

thanks

#3

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