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# Compute the minimum or maximum of two integers without branching

On some rare machines where branching is expensive, the below obvious approach to find minimum can be slow as it uses branching.

## C++

 `/* The obvious approach to find minimum (involves branching) */``int` `min(``int` `x, ``int` `y)``{``  ``return` `(x < y) ? x : y``}` `//This code is contributed by Shubham Singh`

## Java

 `/* The obvious approach to find minimum (involves branching) */``static` `int` `min(``int` `x, ``int` `y)``{``  ``return` `(x < y) ? x : y;``}` `// This code is contributed by rishavmahato348.`

## Python3

 `# The obvious approach to find minimum (involves branching)``def` `min``(x, y):``    ``return` `x ``if` `x < y ``else` `y` `  ``# This code is contributed by subham348.`

## C#

 `/* The obvious approach to find minimum (involves branching) */``static` `int` `min(``int` `x, ``int` `y)``{``  ``return` `(x < y) ? x : y;``}` `// This code is contributed by rishavmahato348.`

## Javascript

 ``

## C

 `/* The obvious approach to find minimum (involves branching) */``int` `min(``int` `x, ``int` `y)``{``  ``return` `(x < y) ? x : y``}`

Below are the methods to get minimum(or maximum) without using branching. Typically, the obvious approach is best, though.

Method 1(Use XOR and comparison operator)
Minimum of x and y will be

`y ^ ((x ^ y) & -(x < y))`

It works because if x < y, then -(x < y) will be -1 which is all ones(11111….), so r = y ^ ((x ^ y) & (111111…)) = y ^ x ^ y = x.

And if x>y, then-(x<y) will be -(0) i.e -(zero) which is zero, so r = y^((x^y) & 0) = y^0 = y.

On some machines, evaluating (x < y) as 0 or 1 requires a branch instruction, so there may be no advantage.
To find the maximum, use

`x ^ ((x ^ y) & -(x < y));`

## C++

 `// C++ program to Compute the minimum``// or maximum of two integers without``// branching``#include``using` `namespace` `std;` `class` `gfg``{``    ` `    ``/*Function to find minimum of x and y*/``    ``public``:``    ``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));``    ``}``    ``};``    ` `    ``/* Driver code */``    ``int` `main()``    ``{``        ``gfg g;``        ``int` `x = 15;``        ``int` `y = 6;``        ``cout << ``"Minimum of "` `<< x <<``             ``" and "` `<< y << ``" is "``;``        ``cout << g. min(x, y);``        ``cout << ``"\nMaximum of "` `<< x <<``                ``" and "` `<< y << ``" is "``;``        ``cout << g.max(x, y);``        ``getchar``();``    ``}` `// This code is contributed by SoM15242`

## Java

 `// Java program to Compute the minimum``// or maximum of two integers without``// branching``public` `class` `AWS {` `    ``/*Function to find minimum of x and y*/``    ``static` `int` `min(``int` `x, ``int` `y)``    ``{``    ``return` `y ^ ((x ^ y) & -(x << y));``    ``}``    ` `    ``/*Function to find maximum of x and y*/``    ``static` `int` `max(``int` `x, ``int` `y)``    ``{``    ``return` `x ^ ((x ^ y) & -(x << y));``    ``}``    ` `    ``/* Driver program to test above functions */``    ``public` `static` `void` `main(String[] args) {``        ` `        ``int` `x = ``15``;``        ``int` `y = ``6``;``        ``System.out.print(``"Minimum of "``+x+``" and "``+y+``" is "``);``        ``System.out.println(min(x, y));``        ``System.out.print(``"Maximum of "``+x+``" and "``+y+``" is "``);``        ``System.out.println( max(x, y));``    ``}` `}`

## Python3

 `# Python3 program to Compute the minimum``# or maximum of two integers without``# branching` `# Function to find minimum of x and y` `def` `min``(x, y):` `    ``return` `y ^ ((x ^ y) & ``-``(x < y))`  `# Function to find maximum of x and y``def` `max``(x, y):` `    ``return` `x ^ ((x ^ y) & ``-``(x < y))`  `# Driver program to test above functions``x ``=` `15``y ``=` `6``print``(``"Minimum of"``, x, ``"and"``, y, ``"is"``, end``=``" "``)``print``(``min``(x, y))``print``(``"Maximum of"``, x, ``"and"``, y, ``"is"``, end``=``" "``)``print``(``max``(x, y))` `# This code is contributed``# by Smitha Dinesh Semwal`

## C#

 `using` `System;` `// C# program to Compute the minimum``// or maximum of two integers without ``// branching``public` `class` `AWS``{` `    ``/*Function to find minimum of x and y*/``    ``public`  `static` `int` `min(``int` `x, ``int` `y)``    ``{``    ``return` `y ^ ((x ^ y) & -(x << y));``    ``}` `    ``/*Function to find maximum of x and y*/``    ``public`  `static` `int` `max(``int` `x, ``int` `y)``    ``{``    ``return` `x ^ ((x ^ y) & -(x << y));``    ``}` `    ``/* Driver program to test above functions */``    ``public` `static` `void` `Main(``string``[] args)``    ``{` `        ``int` `x = 15;``        ``int` `y = 6;``        ``Console.Write(``"Minimum of "` `+ x + ``" and "` `+ y + ``" is "``);``        ``Console.WriteLine(min(x, y));``        ``Console.Write(``"Maximum of "` `+ x + ``" and "` `+ y + ``" is "``);``        ``Console.WriteLine(max(x, y));``    ``}` `}` `  ``// This code is contributed by Shrikant13`

## Javascript

 ``

## C

 `// C program to Compute the minimum``// or maximum of two integers without``// branching``#include` `/*Function to find minimum of x and y*/``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));``}` `/* Driver program to test above functions */``int` `main()``{``int` `x = 15;``int` `y = 6;``printf``(``"Minimum of %d and %d is "``, x, y);``printf``(``"%d"``, min(x, y));``printf``(``"\nMaximum of %d and %d is "``, x, y);``printf``(``"%d"``, max(x, y));``getchar``();``}`

## PHP

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Output

```Minimum of 15 and 6 is 6
Maximum of 15 and 6 is 15```

Time Complexity: O(1)
Auxiliary Space: O(1)

Method 2(Use subtraction and shift)
If we know that

`INT_MIN <= (x - y) <= INT_MAX`

, then we can use the following, which are faster because (x – y) only needs to be evaluated once.
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)))`

## C++

 `#include ``using` `namespace` `std;``#define CHARBIT 8` `/*Function to find minimum of x and y*/``int` `min(``int` `x, ``int` `y)``{``    ``return` `y + ((x - y) & ((x - y) >>``            ``(``sizeof``(``int``) * CHARBIT - 1)));``}` `/*Function to find maximum of x and y*/``int` `max(``int` `x, ``int` `y)``{``    ``return` `x - ((x - y) & ((x - y) >>``            ``(``sizeof``(``int``) * CHARBIT - 1)));``}` `/* Driver code */``int` `main()``{``    ``int` `x = 15;``    ``int` `y = 6;``    ``cout<<``"Minimum of "``<

## Java

 `// JAVA implementation of above approach``class` `GFG``{``    ` `static` `int` `CHAR_BIT = ``4``;``static` `int` `INT_BIT = ``8``;``/*Function to find minimum of x and y*/``static` `int` `min(``int` `x, ``int` `y)``{``    ``return` `y + ((x - y) & ((x - y) >>``                ``(INT_BIT * CHAR_BIT - ``1``)));``}` `/*Function to find maximum of x and y*/``static` `int` `max(``int` `x, ``int` `y)``{``    ``return` `x - ((x - y) & ((x - y) >>``            ``(INT_BIT * CHAR_BIT - ``1``)));``}` `/* Driver code */``public` `static` `void` `main(String[] args)``{``    ``int` `x = ``15``;``    ``int` `y = ``6``;``    ``System.out.println(``"Minimum of "``+x+``" and "``+y+``" is "``+min(x, y));``    ``System.out.println(``"Maximum of "``+x+``" and "``+y+``" is "``+max(x, y));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach``import` `sys;``    ` `CHAR_BIT ``=` `8``;``INT_BIT ``=` `sys.getsizeof(``int``());` `#Function to find minimum of x and y``def` `Min``(x, y):``    ``return` `y ``+` `((x ``-` `y) & ((x ``-` `y) >>``                ``(INT_BIT ``*` `CHAR_BIT ``-` `1``)));` `#Function to find maximum of x and y``def` `Max``(x, y):``    ``return` `x ``-` `((x ``-` `y) & ((x ``-` `y) >>``                ``(INT_BIT ``*` `CHAR_BIT ``-` `1``)));` `# Driver code``x ``=` `15``;``y ``=` `6``;``print``(``"Minimum of"``, x, ``"and"``,``                    ``y, ``"is"``, ``Min``(x, y));``print``(``"Maximum of"``, x, ``"and"``,``                    ``y, ``"is"``, ``Max``(x, y));` `# This code is contributed by PrinciRaj1992`

## C#

 `// C# implementation of above approach``using` `System;` `class` `GFG``{``    ` `static` `int` `CHAR_BIT = 8;` `/*Function to find minimum of x and y*/``static` `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*/``static` `int` `max(``int` `x, ``int` `y)``{``    ``return` `x - ((x - y) & ((x - y) >>``            ``(``sizeof``(``int``) * CHAR_BIT - 1)));``}` `/* Driver code */``static` `void` `Main()``{``    ``int` `x = 15;``    ``int` `y = 6;``    ``Console.WriteLine(``"Minimum of "``+x+``" and "``+y+``" is "``+min(x, y));``    ``Console.WriteLine(``"Maximum of "``+x+``" and "``+y+``" is "``+max(x, y));``}``}` `// This code is contributed by mits`

## Javascript

 ``

## C

 `#include``#define CHAR_BIT 8` `/*Function to find minimum of x and y*/``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)));``}` `/* Driver program to test above functions */``int` `main()``{``  ``int` `x = 15;``  ``int` `y = 6;``  ``printf``(``"Minimum of %d and %d is "``, x, y);``  ``printf``(``"%d"``, min(x, y));``  ``printf``(``"\nMaximum of %d and %d is "``, x, y);``  ``printf``(``"%d"``, max(x, y));``  ``getchar``();``}`

Output

```Minimum of 15 and 6 is 6
Maximum of15 and 6 is 15```

Time Complexity: O(1)
Auxiliary Space: O(1)