Find second min element from array

Sandeep Tiwari :

Anyone can convert this in the Java functional style (lambda):

public int findSecondMin(int arr[]) {

    int min = Integer.MAX_VALUE, secondMin = Integer.MAX_VALUE;
    for (int i = 0; i < arr.length; i++) {
        if (min > arr[i]) {
            secondMin = min;
            min = arr[i];
        } else if (secondMin > arr[i]) {
            secondMin = arr[i];
        }
    }
    return secondMin;
}

I tried by applying the filter over this but it is not working.

Malte Hartwig :

Using an IntStream, you can easily sort it and skip the first element:

public int findSecondMin(int[] arr)
{
    return IntStream.of(arr).sorted().skip(1).findFirst().orElse(Integer.MAX_VALUE);
}

But of course, you don't have to use streams. java.util.Arrays has a nice sort method, and then you can just take the second element:

public int findSecondMin(int[] arr)
{
    Arrays.sort(arr);
    return arr.length < 2 ? Integer.MAX_VALUE : arr[1];
}

To avoid sorting the whole array, we can take your approach and adapt it into a custom reduction on the stream:

public int findSecondMin(int[] arr)
{
    return IntStream.of(arr).boxed().reduce(
        new int[] {Integer.MAX_VALUE, Integer.MAX_VALUE},
        (mins, i) -> {
            return new int[] {Math.min(i, mins[0]), Math.min(Math.max(i, mins[0]), mins[1])};
        }, (mins1, mins2) -> {
            int[] lesser = mins1[0] < mins2[0] ? mins1 : mins2;
            int[] larger = mins1[0] < mins2[0] ? mins2 : mins1;
            return new int[] {lesser[0], Math.min(lesser[1], larger[0])};
        }
    )[1];
}

Compared to a for loop based implementation, it might be harder to read, but can work in parallel.