Wait, Is This Actually DP? A Conversation About Memoization

January 16, 2025 · 10 min read

Software Developer

A casual conversation between two software engineers about the subtle differences between recursion, memoization, and "real" DP.

The Setup

It's late afternoon at the office. Alex just finished implementing a memoized solution to Min Cost Climbing Stairs and walks over to Jamie's desk with their laptop.

Alex: Hey Jamie, can I bug you for a second? I'm confused about something with DP.

Jamie: [looks up from code review] Sure, what's up?

Alex: So I wrote this solution for the climbing stairs problem. I started with recursion, then added memoization like everyone says. But now I'm wondering... is this actually DP? Or is it still just recursion?

The Code

Alex: Here's what I wrote:

class Solution {
    int[] arr; // memoization array

    public int minCostClimbingStairs(int[] cost) {
        int n = cost.length;
        arr = new int[n + 1];

        // Initialize memoization array with -1
        for (int i = 0; i <= n; i++) {
            arr[i] = -1;
        }

        // Base cases
        arr[0] = 0;
        arr[1] = 0;

        return func(n, cost);
    }

    int func(int i, int[] cost) {
        // Base case: if we've already computed this
        if (arr[i] != -1) {
            return arr[i];
        }

        int fromPrevStep;
        int fromTwoStepsBack;

        if (arr[i - 1] != -1) {
            fromPrevStep = arr[i - 1];
        } else {
            fromPrevStep = func(i - 1, cost);
        }
        fromPrevStep += cost[i - 1];

        if (arr[i - 2] != -1) {
            fromTwoStepsBack = arr[i - 2];
        } else {
            fromTwoStepsBack = func(i - 2, cost);
        }
        fromTwoStepsBack += cost[i - 2];

        int min = Math.min(fromPrevStep, fromTwoStepsBack);
        arr[i] = min;
        return min;
    }
}

Alex: Is this top-down DP? I wrote the unoptimized recursive solution first, then just added memoization. Does it become top-down?

The Answer

Jamie: [smiling] YES! This is absolutely top-down DP! ✅

Alex: Wait, really? But it doesn't look like the DP solutions I see in textbooks...

Jamie: That's because you're thinking of bottom-up DP. Let me explain what top-down actually means.

Understanding Top-Down

Jamie: Top-down DP has a simple formula:

The Formula

Top-Down DP = Recursion + Memoization

That's it. That's the whole thing.

Jamie: Think about how your code works:

Start from the main problem (top) → func(n)
    ↓
Break into subproblems (going down) → func(n-1), func(n-2)
    ↓
Cache results as you solve them → arr[i] = min
    ↓
Reuse cached results when needed → if (arr[i] != -1) return arr[i]

That's literally what top-down means!

Alex: Oh! So the "top" is the final answer, and I'm working my way down to base cases?

Jamie: Exactly! And your code does all the right things:

✅ You call func(n, cost) (start from the top - the main problem)
✅ It recursively breaks down to func(i-1) and func(i-2) (go down to subproblems)
✅ You cache results in arr[]
✅ You check cache before computing

Alex: Okay, that makes sense. But here's what's really bugging me... [scrolls through code] ...I don't see something like dp[i] = ... in my solution. It's only func and recursive calls. I thought DP always had that dp[i] pattern?

Jamie: [leans back] Ah! NOW we're getting to the really interesting part. This is where most people get confused!

Alex: So... is it DP without dp[i] = ...?

Jamie: It IS DP, and you DO have dp[i] = ... - you just can't see it because it's disguised!

Alex: What do you mean "disguised"?

💡 The Hidden dp[i] Pattern - The Aha Moment!

Jamie: Look at your code again. What's this line?

int min = Math.min(fromPrevStep, fromTwoStepsBack);
arr[i] = min;  // ← What's this?
return min;

Alex: That's just... storing the result in the cache?

Jamie: Look closer:

The Hidden Truth

arr[i] = min;  // ← THIS IS dp[i] = ...!

Alex: [surprised] Wait, that's the DP formula?

Jamie: YES! That's exactly dp[i] = ....

The Critical Insight

The only difference is:

Bottom-up: dp[i] is computed ITERATIVELY in a loop
Top-down: dp[i] is computed RECURSIVELY in a function

But it's the SAME ASSIGNMENT!

Side-by-Side Comparison

Jamie: Let me show you both versions side by side:

Bottom-Up (What You Expected to See):

int[] dp = new int[n + 1];
dp[0] = 0;
dp[1] = 0;

// Iterative loop - compute in order
for (int i = 2; i <= n; i++) {
    dp[i] = Math.min(dp[i-1] + cost[i-1],
                     dp[i-2] + cost[i-2]);
    //^^^^^ Explicit assignment in loop
}

return dp[n];

Top-Down (Your Code):

int[] memo = new int[n + 1];
Arrays.fill(memo, -1);
memo[0] = 0;
memo[1] = 0;

int func(int i) {
    if (memo[i] != -1) return memo[i];

    memo[i] = Math.min(func(i-1) + cost[i-1],
                       func(i-2) + cost[i-2]);
    //^^^^^^^  Same assignment, but in recursion!
    return memo[i];
}

return func(n);

Alex: Oh wow... it's the same recurrence relation!

Jamie: Exactly! The math is identical:

dp[i] = min(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2])

The only difference is how you compute it:

Aspect	Bottom-Up	Top-Down
Structure	Loop (`for`)	Recursion (`func`)
Order	Sequential (0→n)	On-demand (n→0)
Assignment	`dp[i] = ...` in loop	`memo[i] = ...` in function
Visibility	✅ Obvious	😕 Hidden in recursion

The Aha Moment

Alex: So when I write:

int func(int i) {
    // ...
    arr[i] = Math.min(func(i-1) + cost[i-1],
                      func(i-2) + cost[i-2]);
    return arr[i];
}

This is the DP formula, just wrapped in a function instead of a loop?

Jamie: [snaps fingers] EXACTLY! You got it!

The Key Insight

Both approaches solve the SAME recurrence:

dp[i] = min(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2])

Just different execution strategies:

Bottom-up: Build solutions from base cases UP
Top-down: Start from goal and recurse DOWN (with caching)

Visual Understanding

Jamie: [grabs whiteboard] Let me draw this out for you:

Bottom-Up Execution:

Start from smallest: dp[0], dp[1], dp[2], ...

Loop: i = 0 → n
      dp[i] = ...  (compute using already-computed values)

Order: Explicit, sequential

Top-Down Execution:

Start from largest: func(n)

Recursion: func(n) needs func(n-1) and func(n-2)
           func(n-1) needs func(n-2) and func(n-3)
           ...
           Eventually reach base cases
           memo[i] = ...  (cache as you return)

Order: Implicit, on-demand

Alex: So in top-down, I'm building the same dp[i] values, just in a different order?

Jamie: And only the ones you actually need! That's one advantage of top-down.

The Call Stack Example

Jamie: Here's what happens when you call func(5):

func(5)
  ├─ Check: memo[5] == -1? Yes, compute it
  ├─ Need: func(4) and func(3)
  │
  ├─ func(4)
  │   ├─ Check: memo[4] == -1? Yes, compute it
  │   ├─ Need: func(3) and func(2)
  │   │
  │   ├─ func(3)
  │   │   ├─ ... eventually returns value
  │   │   └─ memo[3] = value  ← This is dp[3] = ...
  │   │
  │   ├─ func(2)
  │   │   └─ memo[2] = value  ← This is dp[2] = ...
  │   │
  │   └─ memo[4] = min(func(3), func(2))  ← dp[4] = ...
  │
  └─ memo[5] = min(func(4), func(3))  ← dp[5] = ...

Alex: So each memo[i] = ... is exactly like dp[i] = ... in bottom-up!

Jamie: Now you're getting it!

Why Both Are "DP"

Alex: Okay, so let me see if I understand. Both are DP because...

Jamie: Because DP isn't about loops or recursion. DP is about:

What Makes Something DP

Optimal Substructure: Solution can be built from optimal solutions to subproblems
Overlapping Subproblems: Same subproblems get solved multiple times

Both top-down and bottom-up:

✅ Use the same recurrence relation
✅ Solve the same subproblems
✅ Build optimal solution from subproblems
✅ Store results to avoid recalculation

Just different execution strategies!

Alex: So the pattern dp[i] = ... isn't what makes it DP - it's the recurrence relation and storing results?

Jamie: Exactly! Whether you compute it in a loop or in recursion doesn't matter. The math is the same.

The Mental Model Shift

Jamie: Let me give you the mental model shift you need:

❌ Old Thinking (Wrong):

"DP means I see dp[i] = ... in a loop"

✅ New Thinking (Correct):

"DP means I'm computing dp[i] using previously computed values"

Whether that's:

In a loop (bottom-up) ← More visible
In recursion with memoization (top-down) ← Less visible but same thing

Both are DP! Just different flavors.

Final Annotated Code

Jamie: Let me annotate your code to show you where the DP is:

int func(int i, int[] cost) {
    // Early return if already computed (memoization)
    if (arr[i] != -1) {
        return arr[i];
    }

    // Your explicit cache checks
    int fromPrevStep;
    int fromTwoStepsBack;

    if (arr[i - 1] != -1) {
        fromPrevStep = arr[i - 1];
    } else {
        fromPrevStep = func(i - 1, cost);
    }
    fromPrevStep += cost[i - 1];

    if (arr[i - 2] != -1) {
        fromTwoStepsBack = arr[i - 2];
    } else {
        fromTwoStepsBack = func(i - 2, cost);
    }
    fromTwoStepsBack += cost[i - 2];

    // ⭐ THE DP TRANSITION! (Same as dp[i] = ...) ⭐
    int min = Math.min(fromPrevStep, fromTwoStepsBack);
    arr[i] = min;  // ← THIS IS dp[i] = min(dp[i-1], dp[i-2])!

    return min;
}

Alex: [staring at code] I can't believe I didn't see it before! The arr[i] = min; line is my dp[i] = ... assignment!

Jamie: Yep! It's just:

Not in a loop
Hidden in a recursive function
Computed on-demand instead of sequentially

But mathematically? It's the EXACT same recurrence relation!

The Wrap-Up

Alex: So to answer my original questions:

"Is this top-down DP?" → YES! ✅
"Does adding memoization to recursion make it top-down?" → YES! ✅
"Is it still DP without visible dp[i] = ...?" → YES! ✅ (It's there, just hidden in the function!)

Jamie: You got it! And just to drive it home:

// Bottom-up DP
for (int i = 2; i <= n; i++) {
    dp[i] = min(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2]);
}

// Top-down DP (your code)
int func(int i) {
    arr[i] = min(func(i-1) + cost[i-1], func(i-2) + cost[i-2]);
    return arr[i];
}

Same assignment, different wrapper!

Alex: This makes so much sense now. I was looking for loops and explicit dp[i] patterns, when the real pattern was the recurrence relation all along.

Jamie: [closing laptop] Exactly. The dp[i] = ... pattern exists in BOTH approaches - it's just explicit in loops (bottom-up) and implicit in recursion (top-down). Both are valid DP!

Alex: Thanks Jamie! Now I understand why my approach is actually proper top-down DP!

Jamie: [smiles] That's right! And hey, next time you see a recursive solution with memoization, you'll know - that's top-down DP, even if it doesn't "look" like textbook DP!

Key Takeaways

What We Learned

For the Learner (Alex):

✅ Top-Down DP = Recursion + Memoization (That's it!)
✅ The dp[i] = ... pattern exists in top-down too (Just hidden in the function)
✅ Both top-down and bottom-up solve the same recurrence (Different execution, same math)
✅ Your code IS valid DP (Even if it doesn't look like textbook examples)

The Universal Truth:

DP isn't about loops vs recursion. It's about:

Breaking problems into subproblems
Storing results to avoid recalculation
Building optimal solutions from optimal subproblems

Whether you do that in a loop or with recursion doesn't change the fundamental DP nature!

The Setup​

The Code​

The Answer​

Understanding Top-Down​

💡 The Hidden dp[i] Pattern - The Aha Moment!​

Side-by-Side Comparison​

The Aha Moment​

Visual Understanding​

The Call Stack Example​

Why Both Are "DP"​

The Mental Model Shift​

Final Annotated Code​

The Wrap-Up​

Key Takeaways​

Further Reading​