CareerCup

Hmm... its basically how a multiplexer works . suppose 2 numbers are A and B .
Perform the operation A-B and extract the MSB .let it be k. Now if k =1 then A>B else B>A (2's complement)..

return (1-k)*A+k*B. This will alyways return the maximum of A&B

max_of_two_int(int a , int b)
{

int max[2];
int pos;
max[0] = a ;
max[1] = b;

pos = ((a-b) >> ((sizeof(int)*8 - 1))) & 1;


printf("\nmax = %d \n",max[pos] );

}

It's only one line code if we can use ternary operator.

maxint(int a, int b){
return (a-b >> (sizeof(int)*8 - 1)) ? b:a
}

int max(int x,int y)
{
return(x-((x-y)&(-(x<y))));
}

However the comparison operator is used here.

return (A+B+abs(A-B))/2

#include <stdio.h>

#define MAX(a,b) \
{\
int m = (a - b) & 0x80000000;\
int s = 0;\
s = (m>>31) & 0x1;\
printf("max is %d", (1-s)*a + s*b);\
}

int main(void)
{
MAX(15, 11);
return 0;
}

@technoviking ur solution is just a copy of the first solution given by marley....plz dont repeat the answer just by wrapping it in a different way....avoid syntactic sugar coatingsss, kick ur ass

int foo(int x, int y)
{
return ( x-(((x-y)>>(WORDBITS-1))&(x-y)) );
}

#include<stdio.h>
#include<conio.h>
main()
{
int i=12,j=11;
int x=(i-j+abs(i-j))?i:j;
printf("%d",x);
getch();
}

Here is the answer to this question:
interviewcodesnippets.com/?p=186

There is something called shortcircuiting in C.
If we can use that:
((x>y)&&k=x);
((y>x)&&k=y);

Very Simple :)

int a=x=100;
int b=y=500;
int max;

for (i=0; i<x || i<y; i++ )
{
x++;
y++;
}

max = (x+y-(a+b))/2;
cout<<max;

ex:
100 , 500

max = (x+y-(a+b))/2;
max=(600+1000-(100+500))/2
max=(1600-(600))/2
max=(1000)/2
max=500

my fault.. i used '<' in for loop:
correcting it:

int a=x=100;
int b=y=500;
int sum=x+y=600;
int max;
int flag1=0,flag2=0,flag3=0;

while(!flag3)
{
a--;
b--;

x++;
y++;

while(!a)
{
flag1=1;
break;
}

while(!b)
{
flag2=1;
break;
}

while(flag1 && flag2)
{
flag3=1;
break;
}

}

max = (x+y-(sum))/2;
cout<<max;

ex:
100 , 500

max = (x+y-(sum))/2;
max=(600+1000-(100+500))/2
max=(1600-(600))/2
max=(1000)/2
max=500

1 int getMax(int a, int b) {
2 int c = a - b;
3 int k = (c >> 31) & 0x1;
4 int max = a - k * c;
5 return max;
6 }

int max(int x,int y)
{
return y+((x-y)& ~(x-y)>>31);
}

double mymax (int a, int b)
{ 
double result;
result = (Math.Sqrt(a * a + b * b - 2 * a * b) + a + b) / 2;
return result;

Another solution can be :

int a,b;
while(a/b)
	printf("a is greater b\n");
while(b/a)
	printf("b is greater a");

int max(int a,int b)
{return a*b/b+b/a*a;}

int max = a * (Math.abs(b-a) + a-b)/(2*(a-b)) + b * (Math.abs(a-b) + b-a)/(2*(b-a));

(a+b + abs(a-b))/2

a/b?a:b