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 Kotlin DSA version keeps the same data and final output as every other DSA book in this wave.

Basic Implementation

basic.kt
Replay: real traced execution (multi-file project)
fun search(arr: IntArray, target: Int, lo: Int, hi: Int): Int {
	if (lo > hi) return -1
	val mid = lo + (hi - lo) / 2
	if (arr[mid] == target) return mid
	if (arr[mid] < target) return search(arr, target, mid + 1, hi)
	return search(arr, target, lo, mid - 1)
}

fun main() {
	val arr = intArrayOf(1, 3, 5, 7, 9, 11, 13)
	val target = 11
	println(search(arr, target, 0, arr.size - 1))
}

  1. lo ← 0, hi ← 6, target ← 11

    1fun search(arr: IntArray, target: Int, lo: Int, hi: Int): Int {2	if (lo > hi) return -1
    values this step0lo6hi11target

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

    2if (lo > hi) return -13val mid = lo + (hi - lo) / 24if (arr[mid] == target) return mid
    values this step3mid7arr[mid](4, 6)next call0lo6hi

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

    2if (lo > hi) return -13val mid = lo + (hi - lo) / 24if (arr[mid] == target) return mid
    values this step5mid11arr[mid]5result4lo6hi

  4. stdout ← 5

    11	val target = 1112	println(search(arr, target, 0, arr.size - 1))13}
    values this step5stdout5result

Complexity

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

Implementation notes

  • Kotlin stores the sorted input as a primitive IntArray; arr is a val reference and the search only reads indexed Int values.
  • The recursive signature is search(arr: IntArray, target: Int, lo: Int, hi: Int): Int, so each call receives copied Int bounds and returns an index sentinel.
  • if (lo > hi) return -1 is the base case for a miss, avoiding nullable Int? result state.
  • The midpoint uses val mid = lo + (hi - lo) / 2, keeping the index arithmetic in Int while avoiding the direct lo + hi form.
  • A match returns immediately with return mid; otherwise arr[mid] < target recurses right with mid + 1, and the final branch recurses left with mid - 1.
  • The trace shows the first call (lo=0, hi=6), then mid=3 with value 7 recursing right to (4, 6), then mid=5 with value 11 returning 5.
  • println(search(arr, target, 0, arr.size - 1)) prints the returned index as 5.