CODEWITHSUNDEEP
javaarraysmin
MArio
MArio

Reputation: 122

How to find closest number in array above target?

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;

    }

Upvotes: 1

Views: 1759

Answers (3)

Nowhere Man
Nowhere Man

Reputation: 19545

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

Upvotes: 4

Aniketh Malyala
Aniketh Malyala

Reputation: 2660

One alternative is the method below:

  public static int findClosest(Integer[] arr, int target) {
        int temp = Integer.MAX_VALUE;
        for(int i: arr){
          if(i > target && i < temp){
            temp = i;
          }
        }
        return temp;
    }

The way this code works is that it starts by initializing an integer, temp to the largest possible integer. Then, we start looping through arr and whenever we find a value greater than the target but less than the current value of temp, we set temp equal to that value. Finally, after we iterate through, we return that value.

Upvotes: 1

Vinod
Vinod

Reputation: 446

Here's one way to do it (requires importing java.util package) - sort the array in ascending order, then find the first higher element, if it exists. A higher value may not exist in the array, so that error has to be handled.

public static int findClosestHigher(Integer[] arr, int target) throws NoSuchElementException {
       
    List<Integer> arrList = Arrays.asList(arr);
    Collections.sort(arrList);
    
    for (int element : arrList) {
        if (element > target) {
            System.out.println("FOUND!!! CLOSEST HIGHER IS --->" + element);                
            return element;
        }
    }
    
    throw new NoSuchElementException();
}

Upvotes: 1

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