Climbing Stairs
An easy Dynamic Programming problem included in Apna College, Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Dynamic Programming
- Sheets
- 3
- Core for
- 9 roles
- Platform
- LeetCode
The problem
You are climbing a staircase that takes n steps to reach the top. Each time you can climb either 1 or 2 steps. Determine the number of distinct ways to reach the top.
Example 1
- Input
- n = 2
- Output
- 2
- Why
- There are two distinct ways: taking two 1-step moves (1 + 1) or taking a single 2-step move (2).
Example 2
- Input
- n = 3
- Output
- 3
- Why
- There are three distinct ways: taking three 1-step moves (1 + 1 + 1), taking 1 step then 2 steps (1 + 2), or taking 2 steps then 1 step (2 + 1).
Constraints
- 1 <= n <= 45
How to think about it
Updated 2026-09-09Arriving at step n requires a final leap from either step n - 1 with a single stride or step n - 2 with a double stride. Because those two previous landings are mutually exclusive, the total paths to step n is the sum of ways to reach those two predecessors, reducing the entire climb to a Fibonacci recurrence.
Approaches, worst first
Exhaustive branching
time O(2^n) · space O(n)
Branch into steps of size 1 and 2 from the current stair and sum the valid completions. It explores every permutation of strides individually, repeatedly recalculating identical suffixes and choking on moderate heights.
Linear table accumulation
time O(n) · space O(n)
Allocate an array of size n + 1 where dp[i] records distinct paths to step i. Filling entries sequentially from dp[1] = 1 and dp[2] = 2 removes redundant subtrees, though retaining every historical step consumes auxiliary memory.
Two variable progressionWrite this one
time O(n) · space O(1)
Compute the next step using two rolling registers that track ways to the prior two steps. Advancing these pointers step-by-step reaches the target with minimal state.
Where people lose marks · 3
- Handling n = 1 by indexing dp[2] before allocating space causes an out-of-bounds crash unless guarded or pre-allocated to accommodate low stair counts.
- Using 32-bit signed integers when extending n past 45 triggers integer overflow because Fibonacci values grow exponentially.
- Conflating 0 steps with an invalid climb: reaching ground level requires exactly one empty sequence of moves, whereas treating base step 0 as zero ways breaks the recurrence.
Full solution
Two rolling variables (the 'two variable progression' approach): each step's answer is the sum of the previous two, so there is no need to keep a full dp array.
Python
def climb_stairs(n: int) -> int:
# Two rolling registers: ways to reach the previous two steps.
prev2, prev1 = 1, 1 # ways to reach step 0, step 1
for _ in range(2, n + 1):
prev2, prev1 = prev1, prev2 + prev1
return prev1
JavaScript
function climbStairs(n) {
let prev2 = 1, prev1 = 1;
for (let i = 2; i <= n; i++) {
[prev2, prev1] = [prev1, prev2 + prev1];
}
return prev1;
}
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
What each operation costs
| Operation | Time |
|---|---|
| fill dynamic programming table of n states | O(n) |
| solve two-dimensional grid of m by n states | O(m * n) |
| space-optimized state transition keeping one row | O(n) |
What usually goes wrong with Dynamic Programming
- Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
- Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
- Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.
Which roles need this problem
Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.
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