#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/dynamic-programming/dynamic-programming-fundamentals/fibonacci-memoized-code

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░█████████████████████████████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/dynamic-programming/dynamic-programming-fundamentals/fibonacci-memoized-code

Repovive

LeetCode 509 Fibonacci - Memoized Code - Dynamic Programming Fundamentals | Dynamic Programming | Repovive

Repovive.

Dynamic Programming21 sections · 918 units60h total

Open in Course

LeetCode 509 Fibonacci - Memoized Code

Three extra lines

Here's the complete memoized Fibonacci:

function fib(n, dp)
    if dp[n] ≠ -1 then
        return dp[n]
    if n ≤ 1 then
        return n
    dp[n] := fib(n - 1, dp) + fib(n - 2, dp)
    return dp[n]

You added just three lines: the lookup check, the storage line, and passing dp through recursive calls. The rest is identical to your original recursive solution. Three extra lines. Before: if dinosaurs had started computing fib( $40$ ), we would still be waiting for the answer. After: finishes in the time it takes to blink. This same pattern appears in game AI, route planning, stock trading. You will use it for years.