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Min Cost Climbing Stairs - Two Valid Approaches

The Puzzle​

Wait, Both Solutions Work?

We solved Min Cost Climbing Stairs with this approach:

dp[0] = 0;  // Start free
dp[1] = 0;  // Start free
dp[i] = Math.min(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2]);
return dp[n];  // Reach the top

But there's another solution that also works:

dp[0] = cost[0];  // Pay at step 0
dp[1] = cost[1];  // Pay at step 1
dp[i] = Math.min(dp[i-1], dp[i-2]) + cost[i];
return Math.min(dp[n-1], dp[n-2]);  // Leave from last steps

How can both be correct? 🤔

This reveals a fundamental DP insight: There are multiple valid state definitions for the same problem!

Layer 1: What's the Difference in State Definition?​

Our Solution​

State Definition: dp[i] = minimum cost to REACH position i (haven't paid cost[i] yet)

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

    dp[0] = 0;  // Reach step 0 for free
    dp[1] = 0;  // Reach step 1 for free

    for (int i = 2; i <= n; i++) {
        dp[i] = Math.min(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2]);
    }

    return dp[n];  // Cost to reach the top
}

Alternative Solution​

State Definition: dp[i] = minimum cost to REACH position i AND PAY cost[i] (ready to leave)

public int minCostClimbingStairs(int[] cost) {
    int n = cost.length;
    int[] dp = new int[n];

    dp[0] = cost[0];  // Reach step 0 AND pay cost[0]
    dp[1] = cost[1];  // Reach step 1 AND pay cost[1]

    for (int i = 2; i < n; i++) {
        dp[i] = Math.min(dp[i-1], dp[i-2]) + cost[i];
    }

    // Can start from step 0 or 1, then jump to top
    return Math.min(dp[n-1], dp[n-2]);
}
The Key Difference

  • Our approach: "Track cost to arrive at each position"

    • Pay costs when LEAVING steps
    • dp[n] = cost to reach the top

  • Alternative approach: "Track cost to leave each position"

    • Pay costs when ARRIVING at steps
    • min(dp[n-1], dp[n-2]) = cost to be ready to jump to top

Layer 2: Designer's Brain - Why Both Work​

Chronological Discovery:

Our Approach: "Track cost to arrive at each position"​

  1. dp[0] = 0 (arrive at step 0 for free)
  2. dp[3] = cost to reach the top
  3. Pay costs when LEAVING steps
  4. The top is an explicit position (n)

Alternative Approach: "Track cost to leave each position"​

  1. dp[0] = cost[0] (arrive AND pay to be ready to leave)
  2. dp[n-1] or dp[n-2] = cost to leave the last positions
  3. From there, you can reach the top with a free jump
  4. The top is reached from either n-1 or n-2
The Fundamental Insight

Both solutions calculate the same total cost, just at different accounting moments!

  • Our solution: Pay costs during transitions
  • Alternative: Pay costs when arriving

Same total, different bookkeeping! 📊

Layer 3: Tracing Both Solutions Side-by-Side​

Example: cost = [10, 15, 20]​

Our Solution (dp = cost to reach)​

dp[0] = 0       // reach step 0 free
dp[1] = 0       // reach step 1 free
dp[2] = min(0+15, 0+10) = 10       // reach step 2
dp[3] = min(10+20, 0+15) = 15      // reach TOP ✓

return dp[3] = 15

Visual:

Position:  0    1    2    TOP(3)
dp value:  0    0    10   15

Alternative Solution (dp = cost to leave)​

dp[0] = 10      // reach step 0 AND pay cost[0]
dp[1] = 15      // reach step 1 AND pay cost[1]
dp[2] = min(10, 15) + 20 = 30      // reach step 2 AND pay cost[2]

return min(dp[1], dp[2]) = min(15, 30) = 15 ✓

Visual:

Position:  0    1    2
dp value:  10   15   30

Can jump to top from step 1 (cost 15) or step 2 (cost 30)
Choose step 1: 15 ✓
Both Arrive at 15!

Same answer, different paths through the DP table.

  • Our solution explicitly calculates reaching position 3
  • Alternative calculates being ready to leave from positions 1 or 2

Layer 4: Where Is "The Top"?​

Our Solution: The Top is Position n​

Our Mental Model

The top is position n (one past the last step).

We calculate: "What's the cost to reach position n?"

  • From step n-1: pay cost[n-1] and jump to n
  • From step n-2: pay cost[n-2] and jump to n
Steps:    [0]  [1]  [2]  → TOP(n=3)
          pay  pay  pay
          when leaving

Alternative Solution: The Top is Wherever You Land After Leaving​

Alternative Mental Model

The top is wherever you land AFTER leaving the staircase.

We calculate: "What's the cost to be ready to leave from the last steps?"

  • dp[n-1] = cost to leave from step n-1 → one jump reaches top (free)
  • dp[n-2] = cost to leave from step n-2 → two jumps reaches top (free)

Both can reach the top, so take the minimum!

Steps:    [0]  [1]  [2]
          pay  pay  pay
          when arriving

Then: free jump to TOP
The Critical Difference
  • Our solution: Top is an explicit position in the DP array (dp[n])
  • Alternative: Top is implicit - reached by free jump from n-1 or n-2

Same destination, different representation!

Layer 5: The Recurrence Relations Compared​

Our Solution​

// To REACH position i (without paying cost[i]):
dp[i] = Math.min(
    dp[i-1] + cost[i-1],  // reach i-1, pay to leave, arrive at i
    dp[i-2] + cost[i-2]   // reach i-2, pay to leave, arrive at i
)

Cost is added BEFORE reaching the next position (during transition)

Alternative Solution​

// To REACH position i AND PAY cost[i]:
dp[i] = Math.min(
    dp[i-1],  // already paid cost[i-1] and left
    dp[i-2]   // already paid cost[i-2] and left
) + cost[i]  // now pay cost[i] to be ready to leave i

Cost is added AFTER reaching the current position (at arrival)

Notice the Pattern

Our solution: dp[i-1] + cost[i-1] → add cost during transition

Alternative: min(dp[i-1], dp[i-2]) + cost[i] → add cost at destination

Same total cost, different accounting moment!

Layer 6: Visual Model - Following the Money​

Our Solution (pay when leaving)​

Path 1:
Start → [Step 0] --pay 10--> [Step 2] --pay 20--> [TOP]
         free       +10        +20      = 30

Path 2:
Start → [Step 1] -----------pay 15-----------> [TOP]
         free                +15       = 15 ✓

Cost tracking: We track cumulative cost to reach each position

Alternative Solution (pay when arriving)​

Path 1:
Start → [Step 0+pay 10] → (free jump) → [TOP]
           dp[0]=10          1 or 2 steps

Need to check: Can we reach top from here?
Yes! From step 0, jump 2 steps → top (but cost is 10)

Path 2:
Start → [Step 1+pay 15] → (free jump) → [TOP]
           dp[1]=15          1 or 2 steps

Can we reach top from here?
Yes! From step 1, jump 2 steps → top (cost is 15) ✓

Cost tracking: We track cost to be ready to leave each position

The Money Trail

Both solutions track the same costs, just at different checkpoints:

  • Our solution: Cost accumulates as you enter positions
  • Alternative: Cost accumulates as you prepare to leave positions

The total is identical! 💰

Layer 7: Why the Return Statements Differ​

Our Solution: return dp[n]​

Our Return Logic
  • We calculated cost to reach the top directly
  • The top is position n
  • dp[n] contains the answer
// Array size: n+1 (positions 0 to n)
// Return: dp[n] (the top position)

Alternative Solution: return Math.min(dp[n-1], dp[n-2])​

Alternative Return Logic
  • We calculated cost to be ready to leave each step
  • From step n-1: one free jump to top
  • From step n-2: two free jumps to top
  • Both can reach top, so choose the cheaper starting point
// Array size: n (positions 0 to n-1)
// Return: min(dp[n-1], dp[n-2]) (cheapest exit point)
Why Different Sizes?
  • Our solution: Array of size n+1 (includes the top at position n)
  • Alternative: Array of size n (top is reached via free jump)

Different array sizes because of different state definitions!

Layer 8: Edge Case Validator​

Example: cost = [10, 15]​

Our Solution​

dp[0] = 0
dp[1] = 0
dp[2] = min(0+15, 0+10) = 10 ✓

return dp[2] = 10

Interpretation: Start at step 0 (free), pay 10 to jump to top

Alternative Solution​

dp[0] = 10
dp[1] = 15

return min(dp[0], dp[1]) = min(10, 15) = 10 ✓

Interpretation: Reach step 0 and pay 10, then free jump to top

Both Give 10! ✓

Same answer, proving both approaches are correct!

Layer 9: The Meta-Lesson​

Why Multiple Valid Approaches Exist​

State Definition Flexibility

State can be defined at different "checkpoints":

  1. Before paying (our solution)
  2. After paying (alternative)

As long as you're consistent:

  • If you track "before paying", add costs when transitioning
  • If you track "after paying", add costs when arriving

The total cost is the same:

  • You can't avoid paying any cost
  • Just matters WHEN you account for it in your DP table

The Designer's Choice Framework​

When you see two solutions, ask:

  1. What does dp[i] represent in each?

    • Ours: cost to arrive at i
    • Alternative: cost to arrive and pay at i

  2. Where is the "finish line"?

    • Ours: position n (explicit top)
    • Alternative: positions n-1 or n-2 (can reach top from either)

  3. When are costs added?

    • Ours: during transition (dp[i-1] + cost[i-1])
    • Alternative: at current state (min(...) + cost[i])
The Profound Insight

DP lets you choose your accounting checkpoint, as long as you're consistent about when costs are paid!

This is why DP is an art - multiple valid state definitions can solve the same problem.

Which Is "Better"?​

Neither! Both are correct.

Choose based on mental model preference:

  • Our solution is more intuitive if you think:

    "What does it cost to get TO the top?"

  • Alternative solution is more intuitive if you think:

    "What does it cost to be ready to LEAVE each step?"

The choice is about mental model preference, not correctness.

Comparison Table​

AspectOur SolutionAlternative Solution
State DefinitionCost to reach position iCost to reach AND pay at position i
Array Sizen + 1n
Base Casesdp[0] = 0, dp[1] = 0dp[0] = cost[0], dp[1] = cost[1]
Recurrencemin(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2])min(dp[i-1], dp[i-2]) + cost[i]
Return Valuedp[n]min(dp[n-1], dp[n-2])
Mental Model"Cost to arrive""Cost to be ready to leave"
Cost TimingPay during transitionPay at arrival

Key Takeaways​

  1. ✅ Multiple state definitions can solve the same DP problem

  2. ✅ Accounting moment matters:

    • Pay when leaving (our solution)
    • Pay when arriving (alternative)

  3. ✅ Consistency is key:

    • Choose a state definition and stick to it
    • Adjust base cases and transitions accordingly

  4. ✅ Same total cost:

    • Both solutions calculate identical final answers
    • Different bookkeeping, same mathematics

  5. ✅ DP is flexible:

    • State definition is a design choice
    • Pick what makes most sense for your mental model
The Beautiful Truth

DP problems often have multiple valid solutions with different state definitions. Understanding why different approaches work deepens your grasp of the problem structure.

This flexibility is a feature, not a bug! 🎨