Help with the algorithm

Question

Algorithms

There is an ordered array of integers and two more integers that define some interval. It is necessary to determine how many numbers from the array belong to a given interval using the binary search algorithm.

Answer 1

    static int binarySearch(int[] A, int n) {
        int left = 0;
        int right = A.length;
        if (right == 0 || n > A[right - 1]) {
            return right;
        }

        while (left < right - 1) {
            int mid = (left + right) / 2;
            if (A[mid] <= n) {
                left=mid;
            } else {
                right=mid;
            }
        }
        return A[left] >= n ? left : right;
    }

    int leftIndex = binarySearch(A, left);
    int rightIndex = binarySearch(A, right + 1);
    int countBetweenLeftAndRight = rightIndex - leftIndex;

Answer 2

Everything is very simple. Using the binary search algorithm, first find the first number, then the second.
The difference between their indices will be the required amount. Depending on the exact setting, it may be necessary to remove the unit.
I suspect that the language is Pascal. Then the search code will be similar to this.
{foo is the desired value. a and b are search boundaries}
procedure Find(foo: integer; a: integer; b: integer);
var c: integer;
begin
if (ba) > 1 then
begin
c:= (a+b) div 2;
find(foo,a,c);
find(foo,c,b);
end else
begin
if (array_[a] = foo) then Result := a;
if (array_[b] = foo) then Result := b;
end;
end;

Answer 3

If the language allows (C, for example), you can somewhat speed up finding the second number after the first is found by setting the search to a smaller interval.
Although on modern very smart processors it will not necessarily be faster.