count the no of bits lets say n
then answer becomes 2^n-1
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I hope the code is self explainatory.
int main()
{
//num is the number
int prev =1, next=1,count=0;
while(num)
{
if(count>1)
{
print false
break;
}
prev=next;
next=num%10;
num=num/10;
if(next!=prev)
count++;
}
if(count<=1)
print true
}
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I want to see if all the ones in a number appear on the right side of
the number and all zeros appear on the left, how can I do this most
efficiently? (i.e. 0111 is true but 100010 is false)
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