Memoization and tabulation are the two ways to actually implement dynamic programming, and it's easy to mix them up because they solve the exact same problems and reach the exact same time complexity — the difference is entirely in direction. Memoization works top-down, starting from the original question and recursing toward simpler ones. Tabulation works bottom-up, starting from the simplest cases and iteratively building toward the original question.

If you've read our guide on dynamic programming, you've already seen both of these in action — this post focuses specifically on how to choose between them.

Memoization: Top-Down

Memoization keeps the natural recursive structure of the problem — you write the function exactly as you'd think about it recursively — and simply adds a cache so no subproblem is ever solved twice.

def climb_stairs_memo(n, cache=None):
    if cache is None:
        cache = {}
    if n <= 2:
        return n
    if n in cache:
        return cache[n]

    cache[n] = climb_stairs_memo(n - 1, cache) + climb_stairs_memo(n - 2, cache)
    return cache[n]

print(climb_stairs_memo(5))  # ways to climb 5 stairs, taking 1 or 2 steps at a time

Output:

8

Advantage: it's usually easier to write and reason about, because it mirrors how you'd naturally think through the recursive problem — you only compute the subproblems that are actually needed to answer the original question, which matters if many subproblems are never actually reached.

Downside: every call still goes through the call stack, so very deep recursion (large n) can hit a language's recursion limit, and function-call overhead makes it somewhat slower in practice than the equivalent loop.

Tabulation: Bottom-Up

Tabulation drops recursion entirely. You build a table starting from the smallest subproblems and iteratively work your way up, so by the time you reach n, every smaller value it depends on has already been computed.

def climb_stairs_tabulation(n):
    if n <= 2:
        return n

    table = [0] * (n + 1)
    table[1], table[2] = 1, 2
    for i in range(3, n + 1):
        table[i] = table[i - 1] + table[i - 2]

    return table[n]

print(climb_stairs_tabulation(5))

Output:

8

Advantage: no recursion, so no stack-overflow risk and no call-overhead — usually the faster option in practice. It also makes it easy to spot a further optimization: since this particular table only ever looks back two steps, you can often shrink it down to two rolling variables instead of a full array, cutting space from O(n) to O(1).

Downside: it computes every subproblem up to n even if the final answer doesn't actually depend on all of them, and the iterative structure can be less intuitive to write than the recursive version, especially for problems with a more complex dependency shape (like a 2D grid, as in our memoized Fibonacci program, which contrasts the naive recursive approach directly against the cached version).

Which Should You Use?

As a practical default: start by writing the memoized (top-down) version, since it's usually faster to get correct — you're just adding a cache to code you'd write anyway. If performance or recursion depth becomes a real concern, convert it to tabulation once the recursive structure (and therefore the table's dependencies) is clear. In an interview setting, either is generally acceptable, but being able to explain the tradeoff between the two — which this post just walked through — often matters as much as the code itself.