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OR with 10101010101..

Ummmm, flip ALL odd bits of a number, that will just set it to 0.

AND with 0 and call it.... wtf kind of question is this?

nvm misunderstood the question

> count no. of bits

int t=n,c=0;

while(t)
{
c++;
t&=t-1;
}

int k=0;

while(k<c)
{
if(k&1)n^=0<<k;

k++;

}

sorry.....plz ignore the above

> find posn of highest set bit .

int t=n,c=0;
while(t)
{
c++;
t>>=1;
}

int k=0;
while(k<c-1)
{

if(k&1)n^=1<<k;

k++;

}

Above solution assumes the length of bits used.

int ToggleOddBits(int input){
int x=2;
while(x != 0){
input ^= x;
x= x<<2;
}
return input;
}

Please revert back if you dont agree with the above function.

That doesn't work, because you are trying to do xor till x becomes 0, not till the end of the given number

public static int flipAllOddBitsOfNumber( final int num){
// Convert the no into a binary String
int N = num;
String binary="";
while(N!=0){
binary = N%2 + binary ;
N/=2;
}
// Flip bits at odd position in binary String
char [] car = binary.toCharArray();
for(int i =1 ; i<car.length ;i+=2)
if (car[i]=='0' )
car[i] = '1';
else
car[i] = '0';

// Convert from binary to decimal
int result =0;
for(int i = car.length - 1 ; i>=0 ; --i){
int powerTwo=1;

for(int j = 1 ; j < car.length -i ;++j )
powerTwo*= 2;
result+= powerTwo* (car[i] - '0');
}

return result ;
}

The following program should work. Havent tested it though.

int main ()
{
	int num;
	cin >> num;
	
	char num_char[33];
	int i = 0;

	itoa(num, num_char, 10);
	while (num_char[i] != '\0')
	{
		if ( (i%2) != 0)
		{
			if (num_char[i] == '0')
				num_char[i] = '1';
			else
				num_char[1] = '0';
		}

		i++;
	}

	num = atoi(num_char);
	
	cout << num;
	return 0;
}

//Flip odd bits of a number
//Ex : 10001001->11011100
int FlipOddBits(int num)
{
int size = sizeof(num) * 8;
int i = 0;
int tmp=1;
for(i=0;i<size;i=i+2)
{
if(num & tmp)
{
num = num & (~tmp);
}
else
{
num = num | tmp;
}
tmp = tmp<<2;
}
return num;
}

size_t = sizeof(int)*8;
int tmpVal

int flipOddBits(int num)
{
size_t = sizeof(int)*8;
int tmpVal = 1;
while(tmpVal < power(2,size))
{
num = num ^ tmp;
tmpVal = tmpVal <<2;
}
return num;
}

nt FlipOddBits(int num)
{
int size = sizeof(num) * 8;
int i = 0;
int tmp=1;
for(i=0;i<size;i=i+2)
{
if(num & tmp)
{
num = num & (~tmp);
}
else
{
num = num | tmp;
}
tmp = tmp<<2;
}
return num;
}