Searching

Binary Search (Recursive)

Use the same binary-search window as the iterative lesson, but pass lo and hi through recursive calls.

Algorithm

execution replay The checked-in replay follows the language-neutral state table for `search-binary-recursive`.
cross-language comparison This Python DSA version keeps the same data and final output as every other DSA book in this wave.

Basic Implementation

basic.py
Replay: real traced execution (multi-file project)
arr = [1, 3, 5, 7, 9, 11, 13]
target = 11

def search(lo, hi):
    if lo > hi:
        return -1
    mid = lo + (hi - lo) // 2
    if arr[mid] == target:
        return mid
    if arr[mid] < target:
        return search(mid + 1, hi)
    return search(lo, mid - 1)

print(search(0, len(arr) - 1))

  1. lo ← 0, hi ← 6, target ← 11

    1arr = [1, 3, 5, 7, 9, 11, 13]2target = 11
    values this step0lo6hi11target

  2. mid ← 3, arr[mid] ← 7, next call ← (4, 6)

    6    return -17mid = lo + (hi - lo) // 28if arr[mid] == target:
    values this step3mid7arr[mid](4, 6)next call0lo6hi

  3. mid ← 5, arr[mid] ← 11, result ← 5

    6    return -17mid = lo + (hi - lo) // 28if arr[mid] == target:
    values this step5mid11arr[mid]5result4lo6hi

  4. stdout ← 5

    14print(search(0, len(arr) - 1))
    values this step5stdout5result

Complexity

  • Time: O(log n)
  • Space: O(log n) call stack

Implementation notes

  • Python keeps arr as one list of references to immutable int objects. The recursive helper passes only lo and hi; it does not slice or allocate sublists.
  • Each call creates a normal Python stack frame with local integer bindings for lo, hi, and mid. The base case is lo > hi, and the midpoint is computed as lo + (hi - lo) // 2 before reading arr[mid].
  • The search compares Python integers, recursing right with search(mid + 1, hi) for this fixture and returning 5 when arr[5] == 11. The list is not mutated, and GC is not part of the visible replay state.