Binary search to find the range in which the number lies

Question

I have an array

Values array: 12 20 32 40 52
              ^  ^  ^  ^  ^
              0  1  2  3  4

on which I have to perform binary sear

Answer 1

My python implementation:

Time complexity: O(log(n)) Space complexity: O(log(n))

def searchForRange(array, target):
    range = [-1, -1]
    alteredBinarySerach(array, target, 0, len(array) -1, range, True)
    alteredBinarySerach(array, target, 0, len(array) -1, range, False)
    return range

def alteredBinarySerach(array, target, left, right, range, goLeft):
    if left > right:
        return

    middle = (left+ right)//2

    if array[middle] > target:
        alteredBinarySerach(array, target, left, middle -1, range, goLeft)
    elif array[middle] < target:
        alteredBinarySerach(array, target, middle +1, right, range, goLeft)
    else:
        if goLeft:
            if middle == 0 or array[middle -1] != target:
                range[0] = middle
            else:
                alteredBinarySerach(array, target, left, middle -1 , range, goLeft)
        else:
            if middle == len(array) -1 or array[middle+1] != target:
                range[1] = middle
            else:
                alteredBinarySerach(array, target, middle +1, right , range, goLeft)