给定排序列表如何创建在 O(log(N)) 时间内不会落在某个范围内的整数列表？

Question

I have a sorted arrayList of integers (no duplicates):

{1,2,4,5,8,12,15, 21,23,26,30,32,35,37,40,42,45,48,51,54}

Lets say given range is [21-38] (inclusive) which has to be excluded from the output list of integers.

Then the output should contain the list of integers that does not contain the above range.

{1,2,4,5,8,12,15,40,42,45,48,51,54}

I need to do this in O(log(N) ) time.

The approaches that i could think of at this moment is:

Approach1:
1) Find lower bound index using binary search in O(logN)
2) Find higher bound index using binary search in O(logN)
3) Then loop over the input list and add the integers to new list by 
considering above lower and higher bound index. But this third step takes O(n).

Approach2:

I also feel that the above approach of using binary search is not 
 great as we can directly iterate over original list to remove the 
  integers falling under the range without using binary search.

Do we have any better approach?

Answer 1

Removing of elements may be optimized for better cases by estimating a better option by the size of the remainder and removal:

static List<Integer> removeFrom(List<Integer> sorted, int sv, int ev) {
    int i1 = Collections.binarySearch(sorted, sv);
    int i2 = Collections.binarySearch(sorted, ev);
    int from, to;
    if (i1 < i2) {
        from = i1;
        to = i2;
    } else {
        from = i2;
        to = i1;
    }
    System.out.printf("Removing values %d..%d%n", sv, ev);
    int size = sorted.size();
    int removeLength = to - from + 1;
    int remainLength = size - removeLength;
    if (removeLength < remainLength) {
        System.out.printf("Removing range: [%d, %d]%n", from, to);
        sorted.removeAll(sorted.subList(from, to + 1));
        return sorted;
    } else {
        List<Integer> result = new ArrayList<>();
        if (from > 0) {
            System.out.printf("Keeping head: [%d, %d]%n", 0, from);
            result.addAll(sorted.subList(0, from));
        }
        if (to < size - 1) {
            System.out.printf("Keeping tail: [%d, %d]%n", to, size);
            result.addAll(sorted.subList(to + 1, size));
        }
        return result;
    }
}

Tests:

int[][] tests = {
    {1, 3},
    {7, 10},
    {3, 6},
    {2, 10},
    {1, 9},
    {2, 9}
};
for (int[] test : tests) {
    List<Integer> sorted = IntStream.rangeClosed(1, 10).boxed().collect(Collectors.toList());
    System.out.println("result: " + removeFrom(sorted, test[0], test[1]) + "\n====\n");
}

Output

Removing values 1..3
Removing range: [0, 2]
result: [4, 5, 6, 7, 8, 9, 10]
====

Removing values 7..10
Removing range: [6, 9]
result: [1, 2, 3, 4, 5, 6]
====

Removing values 3..6
Removing range: [2, 5]
result: [1, 2, 7, 8, 9, 10]
====

Removing values 2..10
Keeping head: [0, 1]
result: [1]
====

Removing values 1..9
Keeping tail: [8, 10]
result: [10]
====

Removing values 2..9
Keeping head: [0, 1]
Keeping tail: [8, 10]
result: [1, 10]
====

So, in the best case the complexity is O(M), where M is the size of remaining part.