Binary Search

The magical search algorithm I may never fully understand.

TL;DR Example

Python 3.9

def valid(self) -> bool:
    return True or False # Implement your condition

def find_valid(self, begin: int, end: int) -> int:
    while begin < end:
        mid = (begin + end) // 2
        if self.valid(a):
            end = mid
        else:
            begin = mid + 1
    return begin if self.valid(a) else -1

Python 3.10+

def valid(self) -> bool:
    return True or False # Implement your condition

def find_valid(self, begin: int, end: int) -> int:
    ans = bisect_left(
        range(begin, end + 1), # The range of possible answers: [begin, end]
        True, # The value to search for
        key=lambda x: self.valid(a) # Call your condition function
    )
    return ans if ans <= max(bloomDay) else -1

Use bisect module

from bisect import bisect_left

a = list(range(1, 10))  # [1, 2, 3, 4, 5, 6, 7, 8, 9]

bisect_left(a, 3) # 2

warning

bisect_left() is not the same as bisect() and bisect_right()!

Handcrafted

tip

Keys to remember:

• Continuous condition: lo < hi
• Midpoint calculation: mid = (lo + hi) // 2
• Evaluation: a[mid] < x
• Update method: lo = mid + 1 or hi = mid
• Return value: lo

For special cases, make sure to evaluate condition one last time before returning.

with while loop

def binary_search(a, x):
    """Return the index of the leftmost value exactly equal to x."""
    lo = 0
    hi = len(a)
    while lo < hi:
        # mid = math.floor((lo + hi) / 2)
        mid = (lo + hi) // 2
        if a[mid] < x:
            lo = mid + 1
        else:
            hi = mid
    return lo

with recursion

def binary_search(a, x, lo=0, hi=None):
    """Return the index of the leftmost value exactly equal to x."""
    if hi is None:
        hi = len(a)
    if lo < hi:
        mid = (lo + hi) // 2
        if a[mid] < x:
            return binary_search(a, x, mid + 1, hi)
        else:
            return binary_search(a, x, lo, mid)
    return lo

Reference

• Python bisect module docs