How to return answer of three fourths of Booleans in java

Question

Hey guys I was wondering if there was any way to return a certain fraction of a bunch of booleans in java. Simply put I would like to find out if there is a way to create a method that when given four booleans if three of them are true it returns true but if less than three are true it returns false. I know this might be hard to understand and if you don't understand just post a comment saying so.

Answer 1

Count the true's in the array...

public boolean checkBools(boolean[] bools){
    int cnt=0;
    for(boolean b : bools)
        if(b) cnt++;
    return cnt < 3 ? false : true;
}

Answer 2

Weird question ... anyway, here's a possible solution for just 4 booleans:

public boolean booleanFraction(boolean a, boolean b, boolean c, boolean d) {
    int ia = a ? 1 : 0;
    int ib = b ? 1 : 0;
    int ic = c ? 1 : 0;
    int id = d ? 1 : 0;
    return ia + ib + ic + id == 3;
}

For a more general solution, here's a method that receives as parameters the number of booleans that need to be true for considering true the whole expression, and a greater than zero variable number of boolean values:

public static boolean booleanFraction(int number, boolean... bools) {
    int acc = 0;
    for (boolean b : bools)
        acc += b ? 1 : 0;
    return acc == number;
}

Call it like this, for the example in the question:

booleanFraction(3, true, true, true, false);
> true

booleanFraction(3, false, false, true, false);
> false

Answer 3

Just create an int that is incremented by 1 for every true. If it is greater than or equal to 3, return true. Having said this, I'm not sure where your stuck as this seems too rudimentary to even mention.

Answer 4

public boolean checkBooleans(boolean b1, boolean b2, boolean b3, boolean b4) {
    boolean[] array = new boolean[4];
    boolean[0] = b1;
    boolean[1] = b2;
    boolean[2] = b3;
    boolean[3] = b4;
    int j = 0;
    for(int i = 0; i < array.length; i++) {
        if(array[i]) {
            j++;
        }
        if(j == 3) {
            return true;
        }
    }
return false;
}

Answer 5

If you want your answer as a boolean expression, you can try,

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

Counting the number of trues is a little more elegant and easy to understand,

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

Answer 6

 boolean nOfM(int n, boolean... m) {
    for (boolean mi : m) if (mi) n--;

    return n < 1;
 }

Answer 7

Using streams:

static boolean threeOrMore(Boolean... bools) {
    return Stream.of(bools).filter(x -> x).count() >= 3;
}

Answer 8

A variation of my answer for the 2-out-of-3 problem:

boolean atLeastThree(boolean a, boolean b, boolean c, boolean d)
{
    int  n = -3;

    if (a) n++;
    if (b) n++;
    if (c) n++;
    if (d) n++;

    return (n >= 0);
}