CareerCup

bool patternCheck(int a)
{
int b=a>>1;
/*if a=000000101010 then b=0=00000010101,,,,rightshift of a by 1*/
//now xor of a and b
int n=(a^b)+1;//n is in 2^n format if alternate bits are there
if(n & (n-1)) == 0) return true;
return false;
}

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You can just write a for loop and right shift by one bit each time.
And use flag to check for alternate 1's and 0's.

I wasn't sure of the question until I start reading answers. But what if numbers such as BB are consider "alternating"?

hey, i am confused about the first example.
since the first 6 bits are not alternating pattern, why would it return true?

can anybody pls tell me wat actually test cases mean... i.e. wat interviewer wants us to write when he asks for the test cases... it wud be gr8 if u can give a small eg... thanks ...

//num represents the equivalent decimal
//logic is simple implementation using finite automata
// i am assuming tat strings like 0001010,00000101 are considered as
//alternate.
//the code is simple but i'd like to get a more compact code



int alternate(int num)
{
int a=0;

if(num&1)
{
a=4;
num>>=1;

while(num)
{
if((a==4)&&(num&1))
return 0;

else if((a==4)&&(!(num&1)))
a=5;

else if((a==5)&&(num&1))
{
a=4;

if(num==1)
a=5;
}
num>>=1;
}
}

else
{
a=1;
num>>=1;

while(num)
{
if((a==1)&&(!(num&1)))
return 0;

else if((a==1)&&(num&1))
a=2;

else if((a==2)&&(!(num&1)))
a=1;

num>>=1;
}
}


if((a==5)||(a==2))
return 1;
return 0;
}

bool alternate(int x)
{

int y = (x*3);
y>>=1;
y+=1;
if((y & (y-1))==0)
return true;
else return false;
}

am i missing something.. this sounded really simple.

int findAlternate(int num)
{
int n=1;
while(n <= num)
n<<=1;
n-=1; n = n^num; n<<=1;
return( n==num?1:0);

}

I think shortest & effective way-

isAlternate(int num)
{
  return (num & 0x5555)==0 || (num & 0xaaaa)==0);
}

bool isAlternate(int x){
If((x & 0xAAAAAAAA)==x)
return true;
}
Assuming int is of 4bytes.

bool isAlternate(int x){
If((x & 0xAAAAAAAA)==x)
return true;
}
Assuming int is of 4bytes.

public boolean isAlternate(int num)
{
  return (num ^ 0x55555555)==0 || (num ^ 0xaaaaaaaa)==0);
}