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bin.py
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85 lines (71 loc) · 2.47 KB
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class Solution(object):
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
# Binary search is Half n Half.
# * Always check for half and choose most promising
# 2 pointer technique - to memorize the indexes as we move through the array without altering the array to perform the search should be apt...
# Init
left = 0
right = len(nums)-1
# Dividing by 2 and setting pointers as we proceed
while left < right:
mid = (left + right)//2
print left, mid, right
if nums[mid] == target:
return mid
elif nums[mid] < target:
left = mid
else:
right = mid
# Smallest array condition (when the array is of length 2)
if (right - left) == 1:
if nums[right] == target:
return right
elif nums[left] == target:
return left
else:
return -1
# Equal or single element condition
if left == right and nums[left] == target:
return left
# no match condition
return -1
"""
# Binary search works in the below logic, existence returned
# * Maintaining the index is not done here
while nums:
n = len(nums)
mid = n//2
if nums[mid] == target:
return True
elif nums[mid] < target:
nums = nums[mid:]
else:
nums = nums[:mid]
return False
"""
""""
# Altering same logic as above (for existence to crack)
# * Maintaining the index left and right is a challenge
start_index = 0
if nums[start_index] == target:
return 0
while len(nums) > 1:
n = len(nums)
mid = n//2
if nums[mid] == target:
if (start_index + mid) < start_index:
return start_index - mid - 1
else:
return start_index + mid
elif nums[mid] < target:
nums = nums[mid:]
else:
nums = nums[:mid]
start_index += mid
return -1
"""