How to find closest number in array above target?

Question

I use this code to find closest number in array, but this method also return number below target number. So if target is 5 and array have 4 a 7 it will return 4. How can i change this code to show clostest number above target.

    public static int findClosest(Integer[] arr, int target) {
        int idx = 0;
        int dist = Math.abs(arr[0] - target);


        for (int i = 1; i < arr.length; i++) {
            int cdist = Math.abs(arr[i] - target);
//TODO MAKE CLOSEST NUMBER BE ABOVE TARGET
            if (cdist < dist) {
                idx = i;
                dist = cdist;
            }
        }
        Log.e("FIND!!!", "CLOSEST MINUTE IS --->" + arr[idx]);
        int minute_of_day = arr[idx];

        return minute_of_day;

    }

Answer 1

Just fix the code according to your requirement: find the minimal number of those above the target. Also it would make sense to check for null values in the input array.

public static Integer findClosestAbove(Integer[] arr, int target) {
    Integer min = null;

    for (Integer x : arr) {
        if (x != null && x > target && (min == null || min > x)) {
            min = x;
        }
    }
    return min; // may be null
}

Similar solution using Java Stream API is as follows:

public static Integer findClosestAboveStream(Integer[] arr, int target) {
    return Arrays.stream(arr)
                 .filter(x -> x != null && x > target)
                 .min(Integer::compareTo) // Optional<Integer>
                 .orElse(null); // or orElseGet(() -> null)
}

Test:

Integer[] arr = new Integer[]{1, 2, null, 7, 4};
System.out.println("findClosest:\t" + findClosestAbove(arr, 5));
System.out.println("findClosest2:\t" + findClosestAboveStream(arr, 5));

Output:

findClosest:    7
findClosest2:   7