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

Basic Implementation

basic.R
Replay: real traced execution (multi-file project)
arr <- c(1, 3, 5, 7, 9, 11, 13)
target <- 11
search <- function(lo, hi) {
  if (lo > hi) return(-1)
  mid <- lo + (hi - lo) %/% 2
  value <- arr[mid + 1]
  if (value == target) return(mid)
  if (value < target) return(search(mid + 1, hi))
  search(lo, mid - 1)
}
cat(search(0, length(arr) - 1), "\n", sep = "")

  1. lo ← 0, hi ← 6, target ← 11

    1arr <- c(1, 3, 5, 7, 9, 11, 13)2target <- 11
    values this step0lo6hi11target

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

    4if (lo > hi) return(-1)5mid <- lo + (hi - lo) %/% 26value <- arr[mid + 1]
    values this step3mid7arr[mid](4, 6)next call0lo6hi

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

    4if (lo > hi) return(-1)5mid <- lo + (hi - lo) %/% 26value <- arr[mid + 1]
    values this step5mid11arr[mid]5result4lo6hi

  4. stdout ← 5

    1arr <- c(1, 3, 5, 7, 9, 11, 13)2target <- 11
    values this step5stdout5result

Complexity

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

Implementation notes

  • arr <- c(1, 3, 5, 7, 9, 11, 13) creates the sorted numeric R vector.
  • target <- 11 is captured from the surrounding scope by search.
  • search <- function(lo, hi) passes only the search bounds through recursive calls.
  • The bounds and returned index are zero-based: the first call is search(0, length(arr) - 1), which is search(0, 6).
  • R vector access is still 1-based, so the probe reads value <- arr[mid + 1].
  • The base case is if (lo > hi) return(-1).
  • The midpoint uses integer division: mid <- lo + (hi - lo) %/% 2.
  • A match returns the zero-based mid.
  • If the value is too small, the code returns search(mid + 1, hi); otherwise it searches search(lo, mid - 1).

Replay steps

call search(0, 6), target = 11
mid=3, arr[4]=7: recurse right to search(4, 6)
mid=5, arr[6]=11: return 5
print 5
  • cat(search(0, length(arr) - 1), "\n", sep = "") prints the returned zero-based index 5.