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binary search
-
a binary search operates on a (sorted) contiguous sequence with a specified left and right index (this is called the search space).
-
binary searching is composed of 3 sections:
- pre-processing: sort if collection is unsorted
- binary search: using a loop or recursion to divide search space in half after each comparison (
O(log(N)
) - **post-processing`: determine viable candidates in the remaining space
-
there are 3 "templates" when writing a binary search:
while left < right
, withleft = mid + 1
andright = mid - 1
while left < right
, withleft = mid + 1
andright = mid
, andleft
is returnedwhile left + 1 < right
, withleft = mid
andright = mid
, andleft
andright
are returned
iterative
if lens(nums) == 0:
return False
lower, higher = 0, len(array)
while lower < higher:
mid = (higher + lower) // 2
if array[mid] == item:
return mid
elif array[mid] > item:
higher = mid - 1
else:
lower = mid + 1
return False
recursive
def binary_search_recursive(array, item, higher=None, lower=0):
higher = higher or len(array)
if higher < lower:
return False
mid = (higher + lower) // 2
if item == array[mid]:
return mid
elif item < array[mid]:
return binary_search_recursive(array, item, mid - 1, lower)
else:
return binary_search_recursive(array, item, higher, mid + 1)
in a matrix
def binary_search_matrix(matrix, item, lower=0, higher=None):
if not matrix:
return False
rows = len(matrix)
cols = len(matrix[0])
higher = higher or rows * cols
if higher > lower:
mid = (higher + lower) // 2
row = mid // cols
col = mid % cols
if item == matrix[row][col]:
return row, col
elif item < matrix[row][col]:
return binary_search_matrix(matrix, item, lower, mid - 1)
else:
return binary_search_matrix(matrix, item, mid + 1, higher)
return False
find the square root
def sqrt(x) -> int:
if x < 2:
return x
left, right = 2, x // 2
while left <= right:
mid = (right + left) // 2
num = mid * mid
if num > x:
right = mid - 1
elif num < x:
left = mid + 1
else:
return mid
return right
find min in a rotated array
def find_min(nums):
left, right = 0, len(nums) - 1
while nums[left] > nums[right]:
mid = (left + right) // 2
if nums[mid] < nums[right]:
right = mid
else:
left = mid + 1
return nums[left]
find a peak element
- a peak element is an element that is strictly greater than its neighbors.
def peak_element(nums):
left, right = 0, len(nums) - 1
while left < right:
mid = (left + right) // 2
if nums[mid + 1] < nums[mid]:
right = mid
else:
left = mid + 1
return left