Check if at least two out of three booleans are true

Question

An interviewer recently asked me this question: given three boolean variables, a, b, and c, return true if at least two out of the three 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 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

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

return a ^ b ? c : a

Answer 3

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

Results from Paul Wagland with a Mac Java 1.6.0_26-b03-383-11A511

a&&b || b&&c || a&&c : 394 ms 
   a ? b||c : b&&c   : 435 ms
   a&b | b&c | c&a   : 420 ms
   a + b + c >= 2    : 640 ms
   a ^ b ? c : a     : 571 ms
   a != b ? c : a    : 487 ms
       DEAD          : 170 ms

Answer 4

return (a==b) ? a : c;

Explanation:

If a==b, then both are true or both are false. If both are true, we have found our two true booleans, and can return true (by returning a). If both are false there cannot be two true booleans even if c is true, so we return false (by returning a). That's the (a==b) ? a part. What about : c ? Well if a==b is false, then exactly one of a or b must be true, so we have found the first true boolean, and the only thing left that matters is if c is also true, so we return c as the answer.

Answer 5

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 6

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

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

return howManyBooleansAreTrue >= 2;

Answer 7

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 8

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 9

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 10

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

Answer 11

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 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

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 14

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.

Addendum

Another (slightly optimized) version of this:

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

This might run slightly faster, assuming that the comparison against 0 will use faster (or perhaps less) code than the comparison against 2.

Answer 15

The most obvious set of improvements are:

// There is 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

// There is 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 16

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 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. It is written like this:

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

Answer 18

In Clojure:

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

Usage:

(at-least 2 true false true)

Answer 19

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 20

We can convert the bools to integers and perform this easy check:

(int(a) + int(b) + int(c)) >= 2

Answer 21

Since it wasn't specified how the code should be improved, I shall endeavour to improve the code by making it more amusing. Here's my solution:

boolean atLeastTwo(boolean t, boolean f, boolean True) {
    boolean False = True;
    if ((t || f) && (True || False)) 
        return "answer" != "42";
    if (t && f) 
        return !"France".contains("Paris");
    if (False == True) 
        return true == false;
    return Math.random() > 0.5;
}

In case anyone's wondering if this code works, here's a simplification using the same logic:

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    if ((a || b) && (c)) 
        return true;
    if (a && b) 
        return true;
    if (true) 
        return false;
    // The last line is a red herring, as it will never be reached:
    return Math.random() > 0.5; 

}

This can be boiled down further to the following:

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

But now it's not funny any more.

Answer 22

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 23

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 24

A 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 three - 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 25

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

// Three booleans, check if two or more are true

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

Answer 26

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

Answer 27

Calculated via a truth table:

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

Answer 28

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

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

Consider also its 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 moonshadow's answer, making pdox's the fastest overall (on my HP/Intel laptop).

Answer 29

Currenty with Java 8, I really prefer something like this:

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    return Stream.of(a, b, c).filter(active -> active).count() >= 2;
}

Answer 30

In C:

return !!a + !!b + !!c >= 2;

Answer 31

This sort of is reading better:

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

Answer 32

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 33

In Ruby:

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

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

Answer 34

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 35

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 36

It should be:

(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

Answer 37

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 38

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 39

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 40

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 41

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 42

In 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 43

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 44

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 45

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 46

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 47

The non-reduced solution to this problem is:

a'bc + abc' + abc + ab'c

Reducing using K-Maps, one gets:

bc + ab + ac

One could reduce this further by using exclusive or on the a'bc and abc' minterms and combining the abc and ab'c minterms:

b(a ^ c) + ac

Answer 48

How about (a||b) && (a||c) - Java, uses three comparisons instead of the OP's six.

Wrong, I should have checked earlier.

Answer 49

Not in context of performance but good code(extensible and readable code that can be reused)

     static boolean trueBooleans (int howMany,boolean ... bools)
     {
      int total = 0;

      for (boolean b:bools)
        if (b && (++total == howMany)) return true;


      return false;
    }

In my humble opinion when writing Java, easy handling unexpected changes and no duplicated code are more important than concise (domain of script languages) or fast program.

Answer 50

How about this one:

(a - b) ? c : a

Answer 51

Taking another approach to this using Java 8's Stream functionality, for any number of booleans with an arbitrary required amount. The Stream short circuits if it hits the limit before processing all of the elements:

public static boolean atLeastTrue(int amount, Boolean ... booleans) {
    return Stream.of(booleans).filter(b -> b).limit(amount).count() == amount;
}

public static void main(String[] args){
    System.out.println("1,2: " + atLeastTrue(1, true, false, true));
    System.out.println("1,1: " + atLeastTrue(1, false, true));
    System.out.println("1,0: " + atLeastTrue(1, false));
    System.out.println("1,1: " + atLeastTrue(1, true, false));
    System.out.println("2,3: " + atLeastTrue(2, true, false, true, true));
    System.out.println("3,2: " + atLeastTrue(3, true, false, true, false));
    System.out.println("3,3: " + atLeastTrue(3, true, true, true, false));
}

Output:

1,2: true
1,1: true
1,0: false
1,1: true
2,3: true
3,2: false
3,3: true

Answer 52

X = OR(a+b,c)

a b c X

1 1 0 1

0 0 1 1

0 1 1 1

Answer 53

C:

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

Answer 54

If I convert the booleans into a number, and if the number is not a power of two, it has at least two trues.

a*4 + b*2 + c*1 = N
return( N != 0 && (N&(N-1)) != 0)

I am just giving an alternative.

Answer 55

public static boolean atLeast(int atLeastToBeTrue, boolean...bools){
    int booleansTrue = 0;
    for(boolean tmp : bools){
        booleansTrue += tmp ? 1 : 0;
    }
    return booleansTrue >= atLeastToBeTrue;
}

You can choose how many at least you want to be true from varargs a.k.a boolean[] :-)

Answer 56

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 57

It seems to me that three out of three are quite arbitrary numbers, and the function should work with an arbitrary number. So in order to answer the question, I'd write a function that would work out if x in an array were true, for example,

bool istrue ( int x, bool[] list)
    y = count true in list
    return y >= x

Answer 58

This is very simple with the help of arithmatic operation.

boolean a = true;
boolean b = false;
boolean c = true;


// Exactly One boolean value true.
if((a?1:0)+(b?1:0)+(c?1:0)==1) 
   return true;
else
   return false;

// Exactly 2 boolean value true.
if((a?1:0)+(b?1:0)+(c?1:0)==2) 
   return true;
else
   return false;

and this is how you can just increase the value of constant to check how many boolean values are true

Answer 59

Its easy with operator overloading if you have a lot of booleans.

operator fun Boolean.unaryPlus() = if (this) 1 else 0
// ...
if(+bool1 + +bool2 + ... + +boolN > 2) {
    // ...
}

Answer 60

The simplest form using ternary operators to solve the problem is:

return a ? (b ? true : c) : (b ? c : false);

You may also want to invest finding a solution by using double negation of the requirement, meaning to say, instead of at least two true values, you need to satisfy the condition at most one false value.

Answer 61

Function ko returns the answer:

static int ho(bool a)
{
    return a ? 1 : 0;
}

static bool ko(bool a, bool b, bool c)
{
    return ho(a) + ho(b) + ho(c) >= 2 ? true : false;
}

Answer 62

Another one:

return a? b||c : b&&c

Answer 63

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 64

I think the simplest solution is:

return (a && b) || c;

Answer 65

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