Compute the minimum or maximum of two integers without branching | GeeksforGeeks

Method 2(Use subtraction and shift)

Minimum of x and y will be

y + ((x - y) & ((x - y) >>(sizeof(int) * CHAR_BIT - 1)))

This method shifts the subtraction of x and y by 31 (if size of integer is 32). If (x-y) is smaller than 0, then (x -y)>>31 will be -1. If (x-y) is greater than or equal to 0, then (x -y)>>31 will be 0.
So if x >= y, we get minimum as y + (x-y)&0 which is y.
If x < y, we get minimum as y + (x-y)&1 which is x.

Similarly, to find the maximum use

x - ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1)))

int min(int x, int y)

{

  return  y + ((x - y) & ((x - y) >> 

            (sizeof(int) * CHAR_BIT - 1))); 

}

/*Function to find maximum of x and y*/

int max(int x, int y)

{

  return x - ((x - y) & ((x - y) >>

            (sizeof(int) * CHAR_BIT - 1)));

}

Java Version: This will not work if the minus operation overflows.
public static int min(int x, int y) {
int diff = (x - y);
return y + (diff & diff >> 31);
}

public static int max(int x, int y) {
int diff = (x - y);
return x - (diff & diff >> 31);
}

Method 1(Use XOR and comparison operator)

int min(int x, int y)

{

  return y ^ ((x ^ y) & -(x < y));

}

 

/*Function to find maximum of x and y*/

int max(int x, int y)

{

  return x ^ ((x ^ y) & -(x < y));

}

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