Climbing Stairs Problem in DSA Typescript

Mental Model

You want to find how many ways you can reach the top of stairs by taking 1 or 2 steps at a time.

Analogy: Imagine climbing stairs where you can take either one step or jump two steps at once. Counting all possible ways to reach the top is like counting all different paths you can take.

Start -> [Step 0]
Step 1 -> [Step 1]
Step 2 -> [Step 2]
...
Step n -> [Top]

Dry Run Walkthrough

Input: stairs: 4 steps

Goal: Find how many distinct ways to reach step 4 by taking 1 or 2 steps at a time

Step 1: At step 0, ways to reach = 1 (starting point)

ways[0] = 1

Why: We start at the bottom, so there is exactly one way to be there

Step 2: Calculate ways to reach step 1: ways[1] = ways[0] = 1

ways = [1, 1]

Why: From step 0, only one way to step 1 by taking one step

Step 3: Calculate ways to reach step 2: ways[2] = ways[1] + ways[0] = 1 + 1 = 2

ways = [1, 1, 2]

Why: Step 2 can be reached from step 1 or step 0

Step 4: Calculate ways to reach step 3: ways[3] = ways[2] + ways[1] = 2 + 1 = 3

ways = [1, 1, 2, 3]

Why: Step 3 can be reached from step 2 or step 1

Step 5: Calculate ways to reach step 4: ways[4] = ways[3] + ways[2] = 3 + 2 = 5

ways = [1, 1, 2, 3, 5]

Why: Step 4 can be reached from step 3 or step 2

Result:

ways = [1, 1, 2, 3, 5]
Answer: 5 ways to climb 4 steps

Annotated Code

DSA Typescript

class ClimbingStairs {
  static climbStairs(n: number): number {
    if (n <= 1) return 1
    const ways: number[] = new Array(n + 1).fill(0)
    ways[0] = 1 // base case: 1 way to stand at step 0
    ways[1] = 1 // base case: 1 way to reach step 1
    for (let i = 2; i <= n; i++) {
      ways[i] = ways[i - 1] + ways[i - 2] // sum ways from previous two steps
    }
    return ways[n]
  }
}

// Driver code
const stairs = 4
const result = ClimbingStairs.climbStairs(stairs)
console.log(result)

if (n <= 1) return 1

handle base cases where stairs are 0 or 1

ways[0] = 1 // base case: 1 way to stand at step 0

initialize ways to stand at the bottom

ways[1] = 1 // base case: 1 way to reach step 1

initialize ways to reach first step

for (let i = 2; i <= n; i++) { ways[i] = ways[i - 1] + ways[i - 2] }

calculate ways to reach step i by summing ways to reach previous two steps

return ways[n]

return total ways to reach top step n

Output

5

Complexity Analysis

Time: O(n) because we calculate ways for each step from 2 to n once

Space: O(n) because we store ways for all steps from 0 to n

vs Alternative: Compared to recursive approach without memoization which is exponential, this dynamic programming approach is efficient and avoids repeated calculations

Edge Cases

n = 0 (no stairs)

Returns 1 because standing at the bottom counts as one way

DSA Typescript

if (n <= 1) return 1

n = 1 (one stair)

Returns 1 because only one step to climb

DSA Typescript

if (n <= 1) return 1

When to Use This Pattern

When you see a problem asking for number of ways to reach a point with fixed step sizes, use dynamic programming to build solutions from smaller steps.

Common Mistakes

Mistake: Not handling base cases n=0 or n=1 correctly, causing wrong answers or errors
Fix: Add explicit checks for n <= 1 returning 1

Mistake: Using recursion without memoization leading to slow exponential time
Fix: Use bottom-up dynamic programming with an array to store intermediate results

Summary

Calculates how many ways to climb n stairs taking 1 or 2 steps at a time.

Use when counting distinct paths with fixed step increments.

The number of ways to reach step n equals sum of ways to reach steps n-1 and n-2.