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