Why is a correct program still not good enough, and what can a clock reading taken twice tell you about the cost of a call?
Short drills on what this video just taught. Write the code, run the checks, and reveal the answer only if you are stuck.
The memoized Fibonacci from the video keeps every value it has already worked out in a dictionary, so nothing is ever computed twice. Your memo already holds {1: 1, 2: 1, 3: 2, 4: 3}. Get the next Fibonacci number out of the memo alone — no recursion, nothing recomputed. Assign it to new_value; the starter then stores it in memo under the key 5.