CODEWITHSUNDEEP
java
Baustein
Baustein

Reputation: 49

Java: Check if a multiple of 100 lies between two doubles

I want to check if something between (including) two doubles is a multiple of one hundred.

With if (double1 % 100 == 0) I can check one double, but I want to check two doubles.

Example:

double double1 = 398, double2 = 399.6; should return false but double double1 = 797.5, double2 = 801.2; should return true.

Upvotes: 0

Views: 1352

Answers (4)

Connor Clark
Connor Clark

Reputation: 706

Given doubles d1 and d2, is there an integer x such that x is a multiple of 100 and in the range [d1, d2]?

1) If the difference between the two numbers is >= 100, return true.

2) If d1 / 100 (integer division) != d2 / 100, return true

3) If d1 * d2 <= 0, return true

4) If either d1 or d2 is a multiple of 100, return true

5) return false

Step 1 handles cases such as (100, 1100). Step 2 handles cases such as (250, 300). Step 3 handles cases such as (-5, 5), or (0, 10). Step 4 handles cases such as (100, 101).

function test(d1, d2) {
  if (d2 - d1 >= 100) return true;
  if (((d1/100)|0) != ((d2/100)|0)) return true;
  if (d1*d2 <= 0) return true;
  if ((d1|0) % 100 == 0 || (d2|0) % 100 == 0) return true;
  return false;
}

var cases = [
             true, 100, 200, 
             true, 0, 1, 
             true, -5, 5, 
             true, 99, 101, 
             true, 10.2, 101.2, 
             true, -0.1, 1.1,
             true, 100, 100,
             true, 0, 0,
             false, 50, 99, 
             false, 101, 102, 
             false, 394.3, 399.9,
             false, 1, 1,
             false, 1001.2, 1001.4
            ];

for (var i = 0; i < cases.length; i += 3) {
  if (test(cases[i + 1], cases[i + 2]) != cases[i]) {
      alert("fail! " + cases[i + 1] + " " + cases[i + 2]);
  }
}

Upvotes: 0

peter.petrov
peter.petrov

Reputation: 39457

I think the other solutions suggested here are a bit too complex.
Here is how you can do it in O(1).

public class Test011 {

    public static void main(String[] args) {
        System.out.println(containsMultipleOf100(99.5, 101.2));
        System.out.println(containsMultipleOf100(101.1, 201.2));
        System.out.println(containsMultipleOf100(100.001, 199.992));
    }

    static boolean containsMultipleOf100(double double1, double double2) {
        long int1 = (long) Math.ceil(double1);
        long int2 = (long) Math.floor(double2);

        if ((int2 - (int1-1)) >= 100){
            return true;
        }else{
            return (int2 / 100) * 100 >= int1;
        }
    }
}

Upvotes: 0

DoubleDouble
DoubleDouble

Reputation: 1493

Consider that there will always be a different digit in the hundreds position (and greater), when the case is true

double1 = 398
          ^ //look at that digit - (and greater)
double2 = 399.6
          ^ //they are the same, return false!



double1 = 797.5
          ^ //look at that digit - (and greater)
double2 = 801.2
          ^ //they are not the same, return true!

double1 = 2300
          ^^ //look at that digit - (and greater)
double2 = 230
          ^ //they are not the same, return true!

You could easily compare them by turning them into an int value (thus ignoring any decimal values), and dividing by 100 (chops off anything below the hundreds position since it is now an int)

In the case where your values could also be negative, -75 and 75 would both divide out to zero and be the same, resulting in returning false.

For this case, one simple way to check is to multiply the doubles and see if it is less than or equal to 0

double1 = -50;
double2 = 1;
//multiply, result is <= 0, return true

double1 = -5;
double2 = -2;
//multiply, result is > 0, return false

double1 = 1;
double2 = 3;
//multiply, result is > 0, return false

double1 = 0;
double2 = 3;
//multiply, result is <= 0, return true

Upvotes: 0

tegaphilip
tegaphilip

Reputation: 21

If I understand your question correctly, then the following steps might just help:

  • Ceil the smaller and floor the larger double number then cast both to integers
  • user a for loop to check if any of the integers from the smallest to the largest is a multiple of 100.

Based on the assumption that the first number is always less than the second, an example code is shown below, this is not very efficient though.

static boolean isDivisibleby100(double double1, double double2) {
    int int1 = (int) Math.ceil(double1);
    int int2 = (int) Math.floor(double2);

    for (int i = int1; i <= int2; i++) {
        if (i % 100 == 0) {
            return true;
        }
    }

    return false;
}

Upvotes: 1

Related Questions