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LintCode 14 First Position Of Target - Easy

14. First Position of Target

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14. First Position Of Target — Easy

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Problem

  1. First Position of Target

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 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?

Solution

class Solution:
    """
    @param nums: The integer array.
    @param target: Target to find.
    @return: The first position of target. Position starts from 0.
    """
    def binarySearch(self, nums, target):
        # write your code here
        if not nums:
            return -1
            
        start, end = 0, len(nums) - 1
        
        while start + 1 < end:
            
            mid = int((start + end) / 2)
            
            if nums[mid] == target:
                end = mid
            
            elif nums[mid] < target:
                start = mid + 1
            else:
                end = mid - 1
                
        ### out the loop, leave start, end, start + 1 = end:
        #since we are looking for first, so we should check the left part first:
        if nums[start] == target:
            return start
        
        if nums[end] == target:
            return end
        
        return -1

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