Climbing Stairs Problem in DSA C

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 a staircase where you can take either one step or two steps at once. Counting all possible ways to reach the top is like counting all paths you can take by choosing 1-step or 2-step moves.

Start -> [Step 0]
          ↓
        Step 1
          ↓
        Step 2
          ↓
        Step 3
          ↓
        ...
          ↓
        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: Initialize ways to reach step 0 and step 1 as 1 each

ways[0] = 1, ways[1] = 1

Why: Base cases: only 1 way to stand still or take first step

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

ways[2] = 1 + 1 = 2

Why: You can get to step 2 from step 1 or step 0

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

ways[3] = 2 + 1 = 3

Why: You can get to step 3 from step 2 or step 1

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

ways[4] = 3 + 2 = 5

Why: You can get to step 4 from step 3 or step 2

Result:

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

Annotated Code

DSA C

#include <stdio.h>

// Function to calculate number of ways to climb n stairs
int climbStairs(int n) {
    if (n <= 1) return 1; // Base case: 1 way to climb 0 or 1 step

    int ways[n+1];
    ways[0] = 1; // 1 way to stand at step 0
    ways[1] = 1; // 1 way to reach step 1

    for (int i = 2; i <= n; i++) {
        ways[i] = ways[i-1] + ways[i-2]; // sum ways from previous two steps
    }

    return ways[n];
}

int main() {
    int stairs = 4;
    int result = climbStairs(stairs);
    printf("Number of ways to climb %d stairs: %d\n", stairs, result);
    return 0;
}

if (n <= 1) return 1; // Base case: 1 way to climb 0 or 1 step

Handle smallest stairs cases directly

ways[0] = 1; // 1 way to stand at step 0

Initialize ways to stand still

ways[1] = 1; // 1 way to reach step 1

Initialize ways to reach first step

for (int i = 2; i <= n; i++) {

Iterate through steps from 2 to n

ways[i] = ways[i-1] + ways[i-2]; // sum ways from previous two steps

Calculate ways to current step by adding ways to previous two steps

return ways[n];

Return total ways to reach top step

Output

Number of ways to climb 4 stairs: 5

Complexity Analysis

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

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

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

Edge Cases

n = 0 (no stairs)

Returns 1 because standing still counts as one way

DSA C

if (n <= 1) return 1;

n = 1 (one stair)

Returns 1 because only one step to climb

DSA C

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 sum ways from previous reachable points.

Common Mistakes

Mistake: Forgetting base cases and starting from step 2 without initializing ways[0] and ways[1]
Fix: Always initialize ways[0] and ways[1] to 1 before loop

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

Summary

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

Use when counting paths with limited step sizes or moves.

The number of ways to reach a step equals sum of ways to reach the two previous steps.

Practice

1. What does the Climbing Stairs problem ask you to find?
Given n steps, how many distinct ways can you climb to the top if you can take 1 or 2 steps at a time?

easy

A. The number of steps you skip

B. The total number of distinct ways to reach the top

C. The maximum number of steps you can take at once

D. The minimum number of steps to reach the top

Solution

1. Step 1: Understand the problem statement

   The problem asks for the count of all possible ways to climb to the top using steps of size 1 or 2.

2. Step 2: Identify what is being counted

   We count distinct sequences of 1-step and 2-step moves that sum to n steps.

3. Final Answer:

   The total number of distinct ways to reach the top -> Option B

4. Quick Check:

   Counting ways = total distinct sequences [OK]

Hint: Count sequences of 1 or 2 steps summing to n [OK]

Common Mistakes:

• Confusing number of ways with minimum steps
• Thinking about maximum step size instead of counting ways
• Ignoring the step size options

2. Which of the following is the correct base case initialization for the Climbing Stairs problem in C?

easy

A. int ways[1] = 1; int ways[2] = 2;

B. int ways[0] = 1; int ways[1] = 2;

C. int ways[0] = 0; int ways[1] = 1;

D. int ways[1] = 0; int ways[2] = 1;

Solution

1. Step 1: Identify base cases for climbing stairs

   For 1 step, ways = 1 (only one way). For 2 steps, ways = 2 (1+1 or 2).

2. Step 2: Match base cases with options

   int ways[1] = 1; int ways[2] = 2; correctly sets ways[1] = 1 and ways[2] = 2.

3. Final Answer:

   int ways[1] = 1; int ways[2] = 2; -> Option A

4. Quick Check:

   Base cases: ways[1]=1, ways[2]=2 [OK]

Hint: Remember ways(1)=1 and ways(2)=2 base cases [OK]

Common Mistakes:

• Setting ways[0] incorrectly
• Confusing ways for 1 and 2 steps
• Using zero for ways[1]

3. What is the output of the following C code snippet for n = 4?

int climbStairs(int n) {
    if (n <= 2) return n;
    int a = 1, b = 2, c;
    for (int i = 3; i <= n; i++) {
        c = a + b;
        a = b;
        b = c;
    }
    return b;
}

printf("%d", climbStairs(4));

medium

A. 7

B. 6

C. 5

D. 4

Solution

1. Step 1: Trace the loop for n=4

   Initial: a=1, b=2 i=3: c=1+2=3, a=2, b=3 i=4: c=2+3=5, a=3, b=5

2. Step 2: Return value after loop

   Return b=5, which is the number of ways to climb 4 steps.

3. Final Answer:

   5 -> Option C

4. Quick Check:

   Ways(4) = 5 [OK]

Hint: Use Fibonacci-like addition for steps > 2 [OK]

Common Mistakes:

• Returning a instead of b
• Off-by-one errors in loop
• Incorrect initial values

4. Identify the bug in this C function for climbing stairs:

int climbStairs(int n) {
    int ways[n+1];
    ways[1] = 1;
    ways[2] = 2;
    for (int i = 3; i <= n; i++) {
        ways[i] = ways[i-1] + ways[i-2];
    }
    return ways[n];
}

medium

A. Array ways is not initialized for ways[0]

B. No bug, code works correctly

C. Accessing ways[2] before initialization

D. Loop should start from i = 2

Solution

1. Step 1: Check array initialization

   Array ways is declared but ways[0] is not initialized, which may cause undefined behavior if accessed.

2. Step 2: Analyze usage of ways[0]

   Though ways[0] is not used in the loop, some compilers or logic might expect it initialized; best practice is to initialize ways[0] = 0.

3. Final Answer:

   Array ways is not initialized for ways[0] -> Option A

4. Quick Check:

   Uninitialized ways[0] can cause issues [OK]

Hint: Always initialize all array elements before use [OK]

Common Mistakes:

• Ignoring ways[0] initialization
• Assuming no undefined behavior without ways[0]
• Starting loop incorrectly

5. You want to find the number of ways to climb 10 steps but can take 1, 2, or 3 steps at a time. Which modification to the standard climbing stairs approach is correct?

hard

A. Use ways[i] = ways[i-1] + ways[i-3] with base cases ways[0]=1, ways[1]=1, ways[2]=2

B. Use ways[i] = ways[i-1] * ways[i-2] * ways[i-3] with base cases ways[0]=1, ways[1]=1, ways[2]=2

C. Use ways[i] = ways[i-1] + ways[i-2] with base cases ways[0]=1, ways[1]=1

D. Use ways[i] = ways[i-1] + ways[i-2] + ways[i-3] with base cases ways[0]=1, ways[1]=1, ways[2]=2

Solution

1. Step 1: Understand step size options

   Since steps can be 1, 2, or 3, ways to reach step i include ways to reach i-1, i-2, and i-3.

2. Step 2: Formulate recurrence and base cases

   ways[i] = ways[i-1] + ways[i-2] + ways[i-3]; base cases: ways[0]=1 (one way to stand still), ways[1]=1, ways[2]=2.

3. Final Answer:

   Use ways[i] = ways[i-1] + ways[i-2] + ways[i-3] with base cases ways[0]=1, ways[1]=1, ways[2]=2 -> Option D

4. Quick Check:

   Sum previous 3 ways for steps 1,2,3 [OK]

Hint: Add ways for all allowed step sizes [OK]

Common Mistakes:

• Multiplying ways instead of adding
• Ignoring the 3-step option
• Using wrong base cases