CareerCup

int median(int a, int b, int c){
   int m = min(a,b);
   int n = min(b,c);
   int o = min(c,a);
   return(m^n^o);
}

Sorry a small mistake... First we should calculate the max of 2 numbers and then min of max and third number.
Let x=5,y=8,z=6
int w= x^((x^y)&-(x<y)) /// w=Max(x,y)
int median= w^((z^w)&-(z<w)) /// Min(w,z)

Mn1 = (a<=b)?a:b; //min(a,b)
Mn2 = (b<=c)?b:c; //min(b,c)
median = (Mn1>=Mn2)?Mn1:Mn2; //Max(Mn1,Mn2)

We need to examine all possible pairs in order to handle all possible cases.

static int median(int a, int b, int c) {
		int m = (a<=b)?a:b; // min(a,b)
		int n = (a<=c)?a:c;  // min(a,c)
		m = (m>=n)?m:n;  // max(m,n)
		n = (b<=c)?b:c;  // min(b,c)
		return (m>=n)?m:n;  // max(m,n)
	}

int median (int a,int b,int c)
{
  if (a <= b)
  {
    if (b <= c) return b;  // a <= b <= c ; 2 ops
    if (a <= c) return c;  // a <= c < b  ; 3 ops
    return a; // c < a <= b ; 3 ops
  }
  // b < a
  if (b => c) return b; // c <= b < a ; 2 ops
  if (a <= c) return a; // b < a <= c ; 3 ops
  return c; b < c < a ; 3 ops
}

e.g : median(5, 8, 4) == 5

int median (int a = 5, int b = 8, int c = 4)
{
  if (a <= b) // 5 <= 8 : true
  {
    if (b <= c) return b;  // a <= b <= c ; 2 ops
	//  8 <= 4 : false
	
	
    if (a <= c) return c;  // a <= c < b  ; 3 ops
	// 5 <= 4 : false
	
    return a; // c < a <= b ; 3 ops
	// a==5 ; return 5  ; BINGO!
	
  }
  // b < a
  if (b => c) return b; // c <= b < a ; 2 ops
  if (a <= c) return a; // b < a <= c ; 3 ops
  return c; b < c < a ; 3 ops
}

int diffab=a-b;
if(diffab*(a-c)<=0)
return a;
if(diffab*(b-c)>=0)
return b;

return c;

Can we construct a binary search tree and in-order traverse the tree to get the median? Then we only need to compare at most 4 times.

int median(int a, int b, int c){
   int l = min(a,b);
   int m = min(b,c);
   int n = min(c,a);
   if(l==m){ return n;}
   else if(l!=m){ return max(l,m);}
}

We can construct a BST from these nodes. For 3 elements, it'll have total-order property.So, inorder traversal. qed!
total comparisons: 3
result : sorted order !

public static int med(int i, int j, int k) {
if (i == j)
return i;
else if (i == k)
return i;
else if (j == k)
return j;
else if (i > j) {
if (j > k)
return j;
else
return (i < k)? i : k;
}
else if (i > k)
return i;
else
return (j < k)? j : k;
}

why people are using excessive integer and comparison operator?
when you can do it in just two comparison operator

min(a, b) = a^(((b-a) >> 31) ^ (a^b))

public int median(int a, int b, int c)
        {
                return a ^ (((b - a) >> 31) & (a ^ b)) ^
                        b ^ (((c - b) >> 31) & (b ^ c)) ^
                        a ^ (((c - a) >> 31) & (a ^ c));
        }

int median(int a, int b, int c)
{
return min(a, b) ^ min(b, c) ^ min(c, a);
}

Let x=5,y=8,z=6
int w= y^((x^y)&-(x<y)) /// w=Min(x,y)
int median= z^((z^w)&-(z<w)) /// Max(w,z)

It's not really clear what is allowed from the question. Can you please clarify?