Check if at least 2 out of 3 booleans is true

Question

An interviewer recently asked me this question: given 3 boolean variables a, b, c, return true if at least 2 out of the 3 are true.

My solution follows:

boolean atLeastTwo(boolean a, boolean b, boolean c) {
  if ((a && b) || (b && c) || (a && c)) {
    return true;
  } else {
    return false;
  }
}

He said that this can be improved further, but I'm not sure how?

Answer 1

Rather than writing:

    if (someExpression) {
        return true;
    } else {
        return false;
    }

Write:

    return someExpression;

---

As for the expression itself, something like this:

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    return a ? (b || c) : (b && c);
}

or this (whichever you find easier to grasp):

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    return a && (b || c) 
             || (b && c);
}

It tests a and b exactly once, and c at most once.

References

• JLS 15.25 Conditional Operator ? :

Answer 2

Why not implement it literally? :)

(a?1:0)+(b?1:0)+(c?1:0) >= 2

In C you could just write a+b+c >= 2 (or !!a+!!b+!!c >= 2 to be very safe).

In response to TofuBeer's comparison of java bytecode, here is a simple performance test:

class Main
{
    static boolean majorityDEAD(boolean a,boolean b,boolean c)
    {
        return a;
    }

    static boolean majority1(boolean a,boolean b,boolean c)
    {
        return a&&b || b&&c || a&&c;
    }

    static boolean majority2(boolean a,boolean b,boolean c)
    {
        return a ? b||c : b&&c;
    }

    static boolean majority3(boolean a,boolean b,boolean c)
    {
        return a&b | b&c | c&a;
    }

    static boolean majority4(boolean a,boolean b,boolean c)
    {
        return (a?1:0)+(b?1:0)+(c?1:0) >= 2;
    }

    static int loop1(boolean[] data, int i, int sz1, int sz2)
    {
        int sum = 0;
        for(int j=i;j<i+sz1;j++)
        {
            for(int k=j;k<j+sz2;k++)
            {
                sum += majority1(data[i], data[j], data[k])?1:0; 
                sum += majority1(data[i], data[k], data[j])?1:0; 
                sum += majority1(data[j], data[k], data[i])?1:0; 
                sum += majority1(data[j], data[i], data[k])?1:0; 
                sum += majority1(data[k], data[i], data[j])?1:0; 
                sum += majority1(data[k], data[j], data[i])?1:0; 
            }
        }
        return sum;
    }

    static int loop2(boolean[] data, int i, int sz1, int sz2)
    {
        int sum = 0;
        for(int j=i;j<i+sz1;j++)
        {
            for(int k=j;k<j+sz2;k++)
            {
                sum += majority2(data[i], data[j], data[k])?1:0; 
                sum += majority2(data[i], data[k], data[j])?1:0; 
                sum += majority2(data[j], data[k], data[i])?1:0; 
                sum += majority2(data[j], data[i], data[k])?1:0; 
                sum += majority2(data[k], data[i], data[j])?1:0; 
                sum += majority2(data[k], data[j], data[i])?1:0; 
            }
        }
        return sum;
    }

    static int loop3(boolean[] data, int i, int sz1, int sz2)
    {
        int sum = 0;
        for(int j=i;j<i+sz1;j++)
        {
            for(int k=j;k<j+sz2;k++)
            {
                sum += majority3(data[i], data[j], data[k])?1:0; 
                sum += majority3(data[i], data[k], data[j])?1:0; 
                sum += majority3(data[j], data[k], data[i])?1:0; 
                sum += majority3(data[j], data[i], data[k])?1:0; 
                sum += majority3(data[k], data[i], data[j])?1:0; 
                sum += majority3(data[k], data[j], data[i])?1:0; 
            }
        }
        return sum;
    }

    static int loop4(boolean[] data, int i, int sz1, int sz2)
    {
        int sum = 0;
        for(int j=i;j<i+sz1;j++)
        {
            for(int k=j;k<j+sz2;k++)
            {
                sum += majority4(data[i], data[j], data[k])?1:0; 
                sum += majority4(data[i], data[k], data[j])?1:0; 
                sum += majority4(data[j], data[k], data[i])?1:0; 
                sum += majority4(data[j], data[i], data[k])?1:0; 
                sum += majority4(data[k], data[i], data[j])?1:0; 
                sum += majority4(data[k], data[j], data[i])?1:0; 
            }
        }
        return sum;
    }

    static int loopDEAD(boolean[] data, int i, int sz1, int sz2)
    {
        int sum = 0;
        for(int j=i;j<i+sz1;j++)
        {
            for(int k=j;k<j+sz2;k++)
            {
                sum += majorityDEAD(data[i], data[j], data[k])?1:0; 
                sum += majorityDEAD(data[i], data[k], data[j])?1:0; 
                sum += majorityDEAD(data[j], data[k], data[i])?1:0; 
                sum += majorityDEAD(data[j], data[i], data[k])?1:0; 
                sum += majorityDEAD(data[k], data[i], data[j])?1:0; 
                sum += majorityDEAD(data[k], data[j], data[i])?1:0; 
            }
        }
        return sum;
    }

    static void work()
    {
        boolean [] data = new boolean [10000];
        java.util.Random r = new java.util.Random(0);
        for(int i=0;i<data.length;i++)
            data[i] = r.nextInt(2) > 0;
        long t0,t1,t2,t3,t4,tDEAD;
        int sz1 = 100;
        int sz2 = 100;
        int sum = 0;

        t0 = System.currentTimeMillis();

        for(int i=0;i<data.length-sz1-sz2;i++)
            sum += loop1(data, i, sz1, sz2);

        t1 = System.currentTimeMillis();

        for(int i=0;i<data.length-sz1-sz2;i++)
            sum += loop2(data, i, sz1, sz2);

        t2 = System.currentTimeMillis();

        for(int i=0;i<data.length-sz1-sz2;i++)
            sum += loop3(data, i, sz1, sz2);

        t3 = System.currentTimeMillis();

        for(int i=0;i<data.length-sz1-sz2;i++)
            sum += loop4(data, i, sz1, sz2);

        t4 = System.currentTimeMillis();

        for(int i=0;i<data.length-sz1-sz2;i++)
            sum += loopDEAD(data, i, sz1, sz2);

        tDEAD = System.currentTimeMillis();

        System.out.println("a&&b || b&&c || a&&c : " + (t1-t0) + " ms");
        System.out.println("   a ? b||c : b&&c   : " + (t2-t1) + " ms");
        System.out.println("   a&b | b&c | c&a   : " + (t3-t2) + " ms");
        System.out.println("   a + b + c <= 2    : " + (t4-t3) + " ms");
        System.out.println("       DEAD          : " + (tDEAD-t4) + " ms");
        System.out.println("sum: "+sum);
    }

    public static void main(String[] args) throws InterruptedException
    {
        while(true)
        {
            work();
            Thread.sleep(1000);
        }
    }
}

This prints the following on my machine (running Ubuntu on Intel Core 2 + sun java 1.6.0_15-b03 with HotSpot Server VM (14.1-b02, mixed mode)):

First and second iterations:

a&&b || b&&c || a&&c : 1740 ms
   a ? b||c : b&&c   : 1690 ms
   a&b | b&c | c&a   : 835 ms
   a + b + c <= 2    : 348 ms
       DEAD          : 169 ms
sum: 1472612418

Later iterations:

a&&b || b&&c || a&&c : 1638 ms
   a ? b||c : b&&c   : 1612 ms
   a&b | b&c | c&a   : 779 ms
   a + b + c <= 2    : 905 ms
       DEAD          : 221 ms

I wonder, what could java VM do that degrades performance over time for (a + b + c <= 2) case.

And here is what happens if I run java with a -client VM switch:

a&&b || b&&c || a&&c : 4034 ms
   a ? b||c : b&&c   : 2215 ms
   a&b | b&c | c&a   : 1347 ms
   a + b + c <= 2    : 6589 ms
       DEAD          : 1016 ms

Mystery...

And if I run it in GNU Java Interpreter, it gets almost 100 times slower, but the a&&b || b&&c || a&&c version wins then.

Results from Tofubeer with the latest code running OS X:

a&&b || b&&c || a&&c : 1358 ms
   a ? b||c : b&&c   : 1187 ms
   a&b | b&c | c&a   : 410 ms
   a + b + c <= 2    : 602 ms
       DEAD          : 161 ms

Answer 3

The most obvious set of improvements are:

// no point in an else if you already returned.
boolean atLeastTwo(boolean a, boolean b, boolean c) {
  if ((a && b) || (b && c) || (a && c)) {
    return true;
  } 

  return false;
}

and then

// no point in an if(true) return true otherwise return false.
boolean atLeastTwo(boolean a, boolean b, boolean c) {
  return ((a && b) || (b && c) || (a && c));
}

But those improvements are minor.

Answer 4

You don't need to use the short circuiting forms of the operators.

return (a & b) | (b & c) | (c & a);

This performs the same number of logic operations as your version, however is completely branchless.

Answer 5

Readability should be the goal. Someone who reads the code must understand your intent immediatelly. So here is my solution.

int howManyBooleansAreTrue = 
    (a ? 1 : 0) 
  + (b ? 1 : 0) 
  + (c ? 1 : 0);

return howManyBooleansAreTrue >= 2;

Answer 6

This kind of questions can be solved with a Karnaugh Map:

      | C | !C
------|---|----
 A  B | 1 | 1 
 A !B | 1 | 0
!A !B | 0 | 0
!A  B | 1 | 0

from which you infer that you need a group for first row and two groups for first column, obtaining the optimal solution of polygenelubricants:

(C && (A || B)) || (A && B)  <---- first row
       ^
       |
   first column without third case

Answer 7

That's a trick question. It's obviously not a question about programming ability, because the answer is trivial. Whenever you see a question that simple, then the interviewer is really trying to assess your coding style.

Do you code for readability, with sensible data types and sensible structure? Or do you optimize the heck out of two lines of C code using bitwise operations and numerical conversions?

You know the correct answer, and you know the other answer. Now you have to asses which answer the interviewer wants. Does he think bit-hacks are neat tricks and demonstrate coding prowess? Or does he seem concerned about maintaining a large code base for an extended period of time?

Answer 8

Taking the answers (so far) here:

public class X
{
    static boolean a(final boolean a, final boolean b, final boolean c)
    {
    return ((a && b) || (b && c) || (a && c));
    }

    static boolean b(final boolean a, final boolean b, final boolean c)
    {
    return a ? (b || c) : (b && c);
    }

    static boolean c(final boolean a, final boolean b, final boolean c)
    {
    return ((a & b) | (b & c) | (c & a));
    }

    static boolean d(final boolean a, final boolean b, final boolean c)
    {
    return ((a?1:0)+(b?1:0)+(c?1:0) >= 2);
    }
}

and running them through the decompiler (javap -c X > results.txt):

Compiled from "X.java"
public class X extends java.lang.Object{
public X();
  Code:
   0:   aload_0
   1:   invokespecial   #1; //Method java/lang/Object."<init>":()V
   4:   return

static boolean a(boolean, boolean, boolean);
  Code:
   0:   iload_0
   1:   ifeq    8
   4:   iload_1
   5:   ifne    24
   8:   iload_1
   9:   ifeq    16
   12:  iload_2
   13:  ifne    24
   16:  iload_0
   17:  ifeq    28
   20:  iload_2
   21:  ifeq    28
   24:  iconst_1
   25:  goto    29
   28:  iconst_0
   29:  ireturn

static boolean b(boolean, boolean, boolean);
  Code:
   0:   iload_0
   1:   ifeq    20
   4:   iload_1
   5:   ifne    12
   8:   iload_2
   9:   ifeq    16
   12:  iconst_1
   13:  goto    33
   16:  iconst_0
   17:  goto    33
   20:  iload_1
   21:  ifeq    32
   24:  iload_2
   25:  ifeq    32
   28:  iconst_1
   29:  goto    33
   32:  iconst_0
   33:  ireturn

static boolean c(boolean, boolean, boolean);
  Code:
   0:   iload_0
   1:   iload_1
   2:   iand
   3:   iload_1
   4:   iload_2
   5:   iand
   6:   ior
   7:   iload_2
   8:   iload_0
   9:   iand
   10:  ior
   11:  ireturn

static boolean d(boolean, boolean, boolean);
  Code:
   0:   iload_0
   1:   ifeq    8
   4:   iconst_1
   5:   goto    9
   8:   iconst_0
   9:   iload_1
   10:  ifeq    17
   13:  iconst_1
   14:  goto    18
   17:  iconst_0
   18:  iadd
   19:  iload_2
   20:  ifeq    27
   23:  iconst_1
   24:  goto    28
   27:  iconst_0
   28:  iadd
   29:  iconst_2
   30:  if_icmplt   37
   33:  iconst_1
   34:  goto    38
   37:  iconst_0
   38:  ireturn
}

You can see that the ?: ones are slightly better then the fixed up version of your original. The one that is the best is the one that avoids branching altogether. That is good from the point of view of fewer instructions (in most cases) and better for branch prediction parts of the CPU, since a wrong guess in the branch prediction can cause CPU stalling.

I'd say the most efficient one is the one from moonshadow overall. It uses the fewest instructions on average and reduces the chance for pipeline stalls in the CPU.

To be 100% sure you would need to find out the cost (in CPU cycles) for each instruction, which, unfortunately isn't readily available (you would have to look at the source for hotspot and then the CPU vendors specs for the time taken for each generated instruction).

See the updated answer by Rotsor for a runtime analysis of the code.

Answer 9

boolean atLeastTwo(boolean a, boolean b, boolean c) 
{
  return ((a && b) || (b && c) || (a && c));
}

Answer 10

Another example of direct code:

int  n = 0;
if (a) n++;
if (b) n++;
if (c) n++;
return (n >= 2);

It's not the most succinct code, obviously.

Answer 11

Here's a test-driven, general approach. Not as "efficient" as most of the solutions so far offered, but clear, tested, working, and generalized.

public class CountBooleansTest extends TestCase {
    public void testThreeFalse() throws Exception {
        assertFalse(atLeastTwoOutOfThree(false, false, false));
    }

    public void testThreeTrue() throws Exception {
        assertTrue(atLeastTwoOutOfThree(true, true, true));
    }

    public void testOnes() throws Exception {
        assertFalse(atLeastTwoOutOfThree(true, false, false));
        assertFalse(atLeastTwoOutOfThree(false, true, false));
        assertFalse(atLeastTwoOutOfThree(false, false, true));
    }

    public void testTwos() throws Exception {
        assertTrue(atLeastTwoOutOfThree(false, true, true));
        assertTrue(atLeastTwoOutOfThree(true, false, true));
        assertTrue(atLeastTwoOutOfThree(true, true, false));
    }

    private static boolean atLeastTwoOutOfThree(boolean b, boolean c, boolean d) {
        return countBooleans(b, c, d) >= 2;
    }

    private static int countBooleans(boolean... bs) {
        int count = 0;
        for (boolean b : bs)
            if (b)
                count++;
        return count;
    }
}

Answer 12

Here's another implementation using map/reduce. This scales well to billions of booleans^© in a distributed environment. Using MongoDB:

Creating a database values of booleans:

db.values.insert({value: true});
db.values.insert({value: false});
db.values.insert({value: true});

Creating the map, reduce functions:

Edit: I like CurtainDog's answer about having map/reduce apply to generic lists, so here goes a map function which takes a callback that determines whether a value should be counted or not.

var mapper = function(shouldInclude) {
    return function() {
        emit(null, shouldInclude(this) ? 1 : 0);
    };
}

var reducer = function(key, values) {
    var sum = 0;
    for(var i = 0; i < values.length; i++) {
        sum += values[i];
    }
    return sum;
}

Running map/reduce:

var result = db.values.mapReduce(mapper(isTrue), reducer).result;

containsMinimum(2, result); // true
containsMinimum(1, result); // false


function isTrue(object) {
    return object.value == true;
}

function containsMinimum(count, resultDoc) {
    var record = db[resultDoc].find().next();
    return record.value >= count;
}

Answer 13

Yet another way to do this but not a very good one:

return (Boolean.valueOf(a).hashCode() + Boolean.valueOf(b).hashCode() + Boolean.valueOf(c).hashCode()) < 3705);

The Boolean hashcode values are fixed at 1231 for true and 1237 for false so could equally have used <= 3699

Answer 14

The simplest way (imo) that is not confusing and easy to read:

// three booleans, check if 2+ are true

return ( a && ( b || c ) ) || ( b && c );

Answer 15

Ternary operators get the nerd juices flowing, but they can be confusing (making code less maintainable, thus increasing the potential for bug injection). Jeff Attwood said it well here:

It's a perfect example of trading off an utterly meaningless one time write-time savings against dozens of read-time comprehension penalties-- It Makes Me Think.

Avoiding ternary operators, I've created the following function:

function atLeastTwoTrue($a, $b, $c) {
        $count = 0;

        if ($a) { $count++; }
        if ($b) { $count++; }
        if ($c) { $count++; }

        if ($count >= 2) {
                return true;
        } else {
                return false;
        }
}

Is this as cool as some of the other solutions? No. Is it easier to understand? Yes. Will that lead to more maintainable, less buggy code? Yes.

Answer 16

Function ReturnTrueIfTwoIsTrue(bool val1, val2, val3))
{
     return (System.Convert.ToInt16(val1) +
             System.Convert.ToInt16(val2) +
             System.Convert.ToInt16(val3)) > 1;
}

Too many ways to do this...

Answer 17

I don't like ternary (return a ? (b || c) : (b && c); from the top answer) and I don't think I've seen anyone mention it written like this so

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    if (a) {
        return b||c;
    } else {
        return b&&C;
    }

Answer 18

Just for the sake of using XOR to answer a relatively straight-forward problem...

return a ^ b ? c : a

Answer 19

function atLeastTwoTrue($a, $b, $c) {

  int count = 0;
  count = (a ? count + 1 : count);
  count = (b ? count + 1 : count);
  count = (c ? count + 1 : count);
  return (count >= 2);
}

Answer 20

A C solution.

int two(int a, int b, int c) {
  return !a + !b + !c < 2;
}

or you may prefer:

int two(int a, int b, int c) {
  return !!a + !!b + !!c >= 2;
}

Answer 21

Why restrict it to 3 booleans?

(define (_true? args #!optional (flag #f))
  (if (null? args)
      flag
      (if (car args)
          (if flag
              #t
              (true? (cdr args) (or flag (car args))))
          (_true? (cdr args) flag))))

(define (true? . args)
  (_true? args))

(true? #t #t #f)

I know this isn't Java. Just playing around...

Answer 22

If the non-functional requirements are more readability than performance, one way is:

public int Count(params bool[] conditions)
{
    return conditions.Count(a => a);
}

public bool AtLeast(int n, params bool[] conditions)
{
    return Count(conditions) >= n;
}

public bool AtLeastTwo(bool a, bool b, bool c)
{
    return AtLeast(2, a, b, c);
}

Answer 23

return 1 << $a << $b << $c >= 1 << 2;

Answer 24

Sum it up. It's called boolean algebra for a reason:

  0 x 0 = 0
  1 x 0 = 0
  1 x 1 = 1

  0 + 0 = 0
  1 + 0 = 1
  1 + 1 = 0 (+ carry)

If you look at the truth tables there, you can see that multiplication is boolean and, and simply addition is xor.

To answer your question:

return (a + b + c) >= 2

Answer 25

My first thought was

return (a||b)&&(b||c)

but for ease of reading I liked the proposed a+b+c>=2 solution that you guys suggested better

Answer 26

Definitely a question that's more about how you solve problems/think than your actual coding ability.

A slightly terser version could be

return ((a ^ b) && (b ^ c)) ^ b

But like a previous poster said, if I saw this in any code I was working on, someone would be getting an earful. :)

Answer 27

literal interpretation will work in all major languages

return (a ? 1:0) + (b ? 1:0) + (c ? 1:0) >= 2;

but I would probably make it easier for people to read, and expandable to more than 3 - something that seems to be forgotten by many programmers

boolean testBooleans(Array bools)
{
     int minTrue = ceil(bools.length * .5);
     int trueCount = 0;

     for(int i = 0; i < bools.length; i++)
     {
          if(bools[i])
          {
               trueCount++;
          }
     }

     return trueCount >= minTrue;
}

Answer 28

I don't think I've seen this solution yet:

boolean atLeast(int howMany, boolean[] boolValues) {
  // check params for valid values

  int counter = 0;
  for (boolean b : boolValues) {
    if (b) {
      counter++;

      if (counter == howMany) {
        return true;
      }
    }
  }
  return false;
}

Its advantage is that once it reaches the number that you're looking for, it breaks. So if this was "at least 2 out of this 1,000,000 values are true" where the first two are actually true, then it should go faster than some of the more "normal" solutions.

Answer 29

clojure:

(defn at-least [n & bools]
  (>= (count (filter true? bools)) n)

usage:

(at-least 2 true false true)

Answer 30

X = OR(a+b,c)

a b c X

1 1 0 1

0 0 1 1

0 1 1 1

Answer 31

If the goal is to return a bitwise two-out-of-three value for three operands, arithmetic and iterative approaches are apt to be relatively ineffective. On many CPU architectures, a good form would be "return ((a | b) & c) | (a & b);". That takes four boolean operations. On single-accumulator machines (common in small embedded systems) that's apt to take a total of seven instructions per byte. The form "return (a & b) | (a & c) | (b & c);" is perhaps nicer looking, but it would require five boolean operations, or nine instructions per byte on a single-accumulator machine.

Incidentally, in CMOS logic, computing "not two out of three" requires twelve transistors (for comparison, an inverter requires two, a two-input NAND or NOR requires four, and a three-input NAND or NOR requires six).

Answer 32

Objective Caml version:

let at_least_two = function
  | (true, true, _)
  | (true, _, true)
  | (_, true, true) -> true
  | _ -> false
;;

For people who are curious about Objective Caml, these pipe operators are a pattern match; they're like switch statements on steroids. The underscores mean "match anything", and that final clause is like a default: case.

It's functionally identical to:

let at_least_two (a, b, c) =
  if a && b then
    true
  else if a && c then
    true
  else if b && c then
    true
  else
    false
;;

With some practice, I find the OCaml pattern match syntax both more succinct and clearer at representing the problem being solved than any of the other solutions. Of course, this isn't immediately true for other programmers since most programmers are much more familiar with Java than some obscure academic language like OCaml. :)

Examples of the function being called at a REPL:

# at_least_two (true, false, false) ;;
- : bool = false
# at_least_two (false, false, false) ;;
- : bool = false
# at_least_two (true, false, true) ;;
- : bool = true

Answer 33

I believe using plain boolean operators (a || b) && (b || c) is fine and is simpler.

You can swap any of the 3 letters with any of the other two and it's still the same expresion.

Answer 34

One thing I haven't seen others point out is that a standard thing to do in the "please write me some code" section of the job interview is to say "Could you improve that?" or "Are you completely happy with that" or "is that as optimized as possible?" when you say you are done. It's possible you heard "how would you improve that" as "this might be improved; how?". In this case changing the if(x) return true; else return false; idiom to just return x is an improvement - but be aware that there are times they just want to see how you react to the question. I have heard that some interviewers will insist there is a flaw in perfect code just to see how you cope with it.

Answer 35

Clearly, the right way is to use C++ templates. That way, you can compute the answer in compile time - this is surely what your employer had in mind.

template<bool A, bool B>
struct TwoEqual
{
    static const bool value = A == B;
};

template<bool A, bool B, bool C>
struct AtLeastTwo
{
    static const bool value = TwoEqual<A, B>::value || TwoEqual<B, C>::value || TwoEqual<A, C>::value;
};

You can then do something like

    cout << AtLeastTwo<true, false, false>::value << endl;

Answer 36

The CORRECT answer, and the only one that I accept from the programmers I interview is:

improved by what metric?

I do not want to see a "process" or the variety of answers available. I want to know that he or she can ask for more details when problems are ambiguous and can take direction. THAT is the test.

Answer 37

Sorta like reading this better:

if (a) {
  return b || c;
} else {
  return b && c;
}

Answer 38

Here's a list comprehension solution in Erlang:

at_least_two(BoolList) ->
   length( [X || X <- BoolList, X] ) >= 2;

Or more generically:

at_least(N, BoolList) ->
   length( [X || X <- BoolList, X] ) >= N;

Answer 39

The best answer to the question should be "As an employee, it's important I write it so that my meaning is clear while maintaining efficiency where necessary for performance." Here's how I'd write it:

function atLeastTwoAreTrue(a, b, c) {
    return (a && b) || (b && c) || (a && c);
}

In reality, the test is so contrived that writing a method that's the fastest, most cryptic possible is perfect acceptable if you accomodate it with a simple comment. But, in general, in this one-liner world, we need more readable code in this world. :-)

Answer 40

Calculated via truth table.

return (A and B) or (c and (a ^ b))

The ^ symbol is the xor operator.

Answer 41

(a==b) ? a : c;

Answer 42

FYI, this is just the carry-out bit of a full adder. In hardware, you could use the logical effort to determine the optimal circuit based on the different boolean expressions. I would guess that the traditional XOR solution would take more effort than the less succinct expression that the poster presented.

Answer 43

I know you tagged this as Java, but I'll give you a nudge towards Scala anyway, you can make this much nicer :)

First of all add an automatic way to transform a boolean into a number:

implicit def bool2num(a:Boolean) = if (a) 1 else 0

Now the solution becomes trivial:

def atLeast2(a:Boolean,b:Boolean, c:Boolean) = a + b + c >= 2

But why stop at testing 2 out of 3? It's even easier to make something generic:

def atLeastN(l:List[Boolean], n:Int) = l.foldLeft(0)(_+_)>=n

But for that you don't even need the conversion to a number:

def atLeastN(l:List[Boolean], n:Int) = l.count(a=>a)>=n

Typesafe and elegant ;)

Answer 44

Python:

def at_least(things, count):
     return len(filter(lambda x: bool(x), things)) >= count

at_least([True, False, True], 2)

Has the nice benefit to work on iterables of arbitrary length.

Answer 45

Not Java, but here are a couple different approaches in scheme.

Not the most efficient thing in the world, but it's functionally correct and far more flexible.

(define (count-true . vals)
   (apply + (map #L(if _ 1 0) vals)))

(define (n-majority? n . vals)
   (>= (apply count-true vals) n))

(define (2-majority? . vals)
   (apply n-majority? 2 vals))

Another stricter implementation, using a sum.

(define (2-majority? v0 v1 v2)
  (define (->0/1 bool) (if bool 1 0))
  (>= (+ (->0/1 v0)
         (->0/1 v1)
         (->0/1 v2))
       2))

And a transliteration of the ternary expression in the highest rated answer (which is my favorite Java solution, particularly if it had a nice comment or was wrapped in a function with a descriptive name):

(define (2-majority? v0 v1 v2)
   (if v0
      (or v1 v2)
      (and v1 v2)))

And here's a macro that should produce majority expressions at will. (n is the number of trues you need.)

(defmacro (n-majority? n . vs)
  (cond ((> n (length vs)) #f)
        ((= n 0)           #t)
        (#t
          `(if ,(car vs)
             `(n-majority? ,(- n 1) ,@(cdr vs))
             `(n-majority? ,n ,@(cdr vs))))))

This last solution gives you both compile-time flexiblity and speed. I'm also pretty sure that it guarantees that no expression in vs is evaluated more than once (which is quite useful). The aspect of it that still needs improvement (besides much more testing) is that it needs better error handling.

Answer 46

c#

off of the top of my head:

 public bool lol( int minTrue, params bool[] bools )
 {
  return bools.Count( ( b ) => b ) >= minTrue;
 }

should be pretty quick.

A call would look like this:

lol( 2, true, true, false );

This way you are leaving the rules (two must be true) up to the caller, instead of embedding them in the method.

Answer 47

As an addition to TofuBeer's excellent post consider pdox's answer:

static boolean five(final boolean a, final boolean b, final boolean c)
{
    return a == b ? a : c;
}

Consider also it's disassembled version as given by "javap -c"

static boolean five(boolean, boolean, boolean);
  Code:
   0:   iload_0
   1:   iload_1
   2:   if_icmpne   9
   5:   iload_0
   6:   goto    10
   9:   iload_2
   10:  ireturn

pdox's answer compiles to less byte code than any of the previous answers. How does its execution time compare to the others?

one                5242 ms
two                6318 ms
three (moonshadow) 3806 ms
four               7192 ms
five  (pdox)       3650 ms

At least on my computer pdox's answer is just slightly faster than moonshadow's answer, making pdox's the fastest overall (at least on my HP/intel laptop).

Answer 48

It really depends what you mean by "improved":

Clearer?

boolean twoOrMoreAreTrue(boolean a, boolean b, boolean c)
{
    return (a && b) || (a && c) || (b && c);
}

Terser?

boolean moreThanTwo(boolean a, boolean b, boolean c)
{
    return a == b ? a : c;
}

More general?

boolean moreThanXTrue(int x, boolean[] bs)
{
    int count = 0;

    for(boolean b : bs)
    {
        count += b ? 1 : 0;

        if(count > x) return true;
    }

    return false;
}

More scalable?

boolean moreThanXTrue(int x, boolean[] bs)
{
    int count = 0;

    for(int i < 0; i < bs.length; i++)
    {
        count += bs[i] ? 1 : 0;

        if(count > x) return true;

        int needed = x - count;
        int remaining = bs.length - i;

        if(needed >= remaining) return false;
    }

    return false;
}

Faster?

// Only profiling can answer this.

Which one is "improved" depends heavily on the situation.

Answer 49

int count=0;

boolean atLeastTwo(boolean a, boolean b, boolean c) {


if (a)
count++;
if (b)
count++;
if (c)
count++;

if (count>1)
return true;
else
return false;
}

Answer 50

In Ruby:

[a, b, c].count { |x| x } >= 2

Which could be run in JRuby on the JavaVM. ;-)

Answer 51

The 2 and 3 in the question posed are decidedly magic-numberish. The 'correct' answer will depend on whether the interviewer was trying to get at your grasp of boolean logic (and I don't think pdox's answer could be bested in this respect) or your understanding of architectural issues.

I'd be inclined to go with a map-reduce solution that will accept any sort of list with any arbitrary condition.

Answer 52

I found this solution.

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    bool result = !(a ^ b ^ c) && !(!a & !b & !c) || (a & b & c);
    return result;
}

Answer 53

My first thought when I saw the question was:

int count=0;
if (a) ++count;
if (b) ++count;
if (c) ++count;
return count>=2;

After seeing other posts, I admit that

return (a?1:0)+(b?1:0)+(c?1:0)>=2;

is much more elegant. I wonder what the relative runtimes are.

In any case, though, I think this sort of solution is much better than a solution of the

return a&b | b&c | a&c;

variety because is is more easily extensible. What if later we add a fourth variable that must be tested? What if the number of variables is determined at runtime, and we are passed an array of booleans of unknown size? A solution that depends on counting is much easier to extend than a solution that depends on listing every possible combination. Also, when listing all possible combinations, I suspect that it is much easier to make a mistake. Like try writing the code for "any 3 of 4" and make sure you neither miss any nor duplicate any. Now try it with "any 5 of 7".

Answer 54

He's probably not looking for anything convoluted like bitwise comparison operators (not normally convoluted but with booleans, it's extremely odd to use bitwise operators) or something that is very roundabout like converting to int and summing them up.

The most direct and natural way to solve this is with an expression like this:

a ? (b || c): (b && c)

Put it in a function if you prefer, but it's not very complicated. The solution is logically concise and efficient.

Answer 55

In C# the following LINQ query could be used:

bool atLeastTwo(bool a, bool b, bool c)
{
    return (from v in new[] {a, b, c} where v select v).Count() >= 2;
}

It clearly expresses the intention and is scalable to any number of booleans.

Answer 56

#include<stdio.h>

int two(int a, int b, int c){
    return  (a+b+c)>>1;
}

int main(){
    printf("%d %d %d %d\n", 
           two(0,0,0), two(0,0,1), two(0,1,1), two(1,1,1));
    return 0;
}

Output: 0 0 1 1

Answer 57

Let the three boolean values be A,B and C....

You can use a k-MAP and come with a boolean expression ...

In this case boolean expression will be A(B+C) + C

or if((A && (B || C )) || C ) { return true; } else return false;

Answer 58

C:

if (!!a + !!b + !!c >= 2)

Answer 59

C'mon guys... Why nobody gave this:

(a || b && c) && (b || c && a)

Also, if true is automatically converted to 1 and false to 0:

(a + b*c) * (b + c*a) > 0