For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.
If the target number does not exist in the array, return -1.
Example 1:
Input: [1,4,4,5,7,7,8,9,9,10],1
Output: 0
Explanation:
the first index of 1 is 0.
Example 2:
Input: [1, 2, 3, 3, 4, 5, 10],3
Output: 2
Explanation:
the first index of 3 is 2.
Example 3:
Input: [1, 2, 3, 3, 4, 5, 10],6
Output: -1
Explanation:
Not exist 6 in array.
Challenge
If the count of numbers is bigger than 2^32, can your code work properly?
class Solution:
"""
@param nums: An integer array sorted in ascending order
@param target: An integer
@return: An integer
"""
def binarySearch(self, nums, target):
# corner cases
if not nums:
return -1
# init boundaries
up = len(nums) - 1
low = 0
# while up and low are not next to each other
while up > low + 1:
mid = int(low + (up - low)/2)
if nums[mid] == target:
# because 'first', mid may be smaller
up = mid
elif nums[mid] > target:
up = mid
else: # (nums[mid] < target)
low = mid
# check wether up or low
# because 'first', check the smaller one first
if nums[low] == target:
return low
if nums[up] == target:
return up
# if not found
return -1