Java – finds the second min element from the array

Anyone can convert it in the Java function 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 applying a filter, but it didn't work

Solution

With 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 good sorting method, and then you can 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 entire array, we can adopt your method and adjust it to a custom reduction in the stream:

public int findSecondMin(int[] arr)
{
    return IntStream.of(arr).@R_398_2419@ed().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[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 with the implementation based on the for loop, it may be more difficult to read, but it can work in parallel