Checking if values in an array are within a range

Question

I have 2 arrays. One is twice the size of the other. array2 represents pairs of doubles, for example, array2[0] is a lower bound and array2[1] is the upper bound. I want to check if array1[0] is between the values array2[0] and array2[1] , if array1[1] is between values of array2[2] and array2[3] and so on. If this is the case for the entire array I want to return true, otherwise return false.

Here's what I have been trying so far but it does not work.

double[] array1 = new double[10];
double[] array2 = new double[20];

public boolean match(double[] array1, double[] array2) {
    boolean b = false;
    int count = 0;
    for (int i = 0; i < array1.length; i++) {
        if (array1[i] > array2[i * 2] && array1[i] < array2[i * 2 + 1]) {
            count++;  
        }
    }
    if (count == array1.length) {
        b = true; 
    }
    return b;
}

Any help with this would be appreciated

Answer 1

Your loop is incorrect:

for (int i = 0; i < array1.length; i++) {
    if (array1[i] > array2[i * 2] && array1[i] < array2[i * 2 + 1]) {
        count++;  
    }
}

You check the index i*2 , which will yield 8 when i is 4 . However you want to check at the index before and behind i . One way to do this is to keep track of what index you are on and use that to check the elements in array2 :

public static boolean match(double[] array1, double[] array2) {
    boolean b = false;
    int index = 0;
    int count = 0;
    for (int i = 0; i < array1.length; i++) {
        if (array1[i] > array2[index++] && array1[i] < array2[index++]) {
            count++;  
        }
    }
    if (count == array1.length) {
        b = true; 
    }
    return b;
}

Using the post increment operator we can up the counter every time we call the index, so that we will call the correct index of array2 .

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Also we can make this method more efficient by exiting as soon as the condition becomes false:

if (array1[i] <= array2[i * 2] || array1[i] >= array2[i * 2 + 1]) {
    return false;  
}