Found a really interesting problem on Codewars today ! Gonna discuss the solution
You got two numbers and you have to find either the Min or the Max of that number depending on the symbol < or > you got as input BUT you have to do this without if / ?: / switch / while / for which is basically everything conditional
In the end you have a function that goes like this
solve(2, 1, '>') => 2
solve(2, 1, '<') => 1
Now we have two conditions that we have to resolve without using conditional statements or Irony
- Is a > b or vice versa
- Should I calculate Min or Max depending on
<or>
We can solve the first problem using Bitshift and the fact that integers are of 32 bits
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// if a > b 0 else -1
int value = (a-b) >> 31;
If a > b then a-b would be positive which would make the value 0 whereas the diff being negative would
make the value -1 while telling us that actually b > a
We the solve the second we gonna to look at how the symbols in ASCII
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60 - <
61 - =
62 - >
There is an = symbol between < & > so to mimic a - b we just gotta the passed ascii symbol with = and do
some math mumbo jumbo to make sure that > gives 0 & < gives -1.
The math basically makes sure that 1 returned for > gets converted to 0 and -1 remains the same
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// if `>` 0 else -1
int operation = (symbol - '=' - 1) / 2
Now all that is left is to combine these conditions into a mega condition
- If a > b and Symbol = > then return A (0, 0) Or Vice versa (-1, -1)
- If a > b and Symbol = < then return B (0, -1) Or vice versa (-1, 0)
Which is actually just an XOR gate where we just produce a 0 when both inputs match else it’s 1, now to convert this boolean into our actual MinMax result we convert our inputs to a List and treat our bool as the Index
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public class Puzzle {
public static int solve(int a, int b, char symbol) {
/*
Since integers are 32 bits
Doing an Right Bit Shift by 31
- Negative values -> -1
- Positive values -> 0
*/
// if a > b 0 else -1
int value = (a-b) >> 31;
// if `>` 0 else -1
int operation = (c - '=' - 1) / 2 ;
// if a > b and ` >` return a else b
int index = (value ^ operation) * -1;
// Expression that gives us 0 or 1 to pick a or b.
return new int[]{a, b}[index];
}
}
Neat !
There’s a whole lot more on https://www.codewars.com/kata/61211f9d27e72200567dfd3d/solutions/java
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public class Puzzle {
public static int solve(int a, int b, char symbol) {
return a + (b - a & (b - a ^ symbol << 30) >> 31);
}
}