Compute fib(n) recursively. Cache each fib(k) in a memo map so each subproblem is solved at most once.

Algorithm

Canonical input n = 6 produces fib(6) = 8. Replay highlights every memo write and every cache hit.

Basic Implementation

basic.kt
fun fib(n: Int, memo: HashMap<Int, Int>): Int {
	if (memo.containsKey(n)) {
		return memo[n]!!
	}
	if (n < 2) {
		memo[n] = n
		return n
	}
	val value = fib(n - 1, memo) + fib(n - 2, memo)
	memo[n] = value
	return value
}

fun main() {
	val memo = HashMap<Int, Int>()
	val result = fib(6, memo)
	println(result)
}

Complexity

  • Time: O(n) with memoization (vs. O(2^n) without)
  • Space: O(n) memo + O(n) call stack

Implementation notes

  • Kotlin: the recursion takes the memo as a HashMap<Int, Int> argument rather than a companion object field, which keeps state explicit without hiding the lesson behind a shared global. The containsKey + !! indexer pair stays parallel to the lesson spec instead of leaning on getOrPut.
  • The replay shows the call stack on one side and the memo map on the other so memo writes and cache hits are visually distinct.
memoization A `HashMap<Int, Int>` keyed by `n` stores each completed subproblem. Before recursing, check `memo.containsKey(n)`: a hit returns immediately, a miss descends.
explicit memo state The memo is threaded through the recursion as `memo: HashMap<Int, Int>` so the lesson stays about caching, not global state.