1. Small stairs
There are n ways for 1 or 2 stairs.
if (n <= 2) return nTyporithm · easy
Climbing Stairs
▷ Interactive algorithm labclimbing-stairs-dpdynamic-programmingsequencespace-optimized
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Full lesson · 8 learning steps Open any exercise for focused practice
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Step 1 of 8 · intro card
Learn Climbing Stairs
Dynamic Programming
Climbing Stairs is Fibonacci with a story: each step can be reached from one or two steps below.
The state is the number of ways to reach a stair.
The transition adds the previous two states.
Only two variables are needed because older states are never read again.
Preparing algorithm lab…
01
State: ways to reach step i.
02
Base: step 1 has 1 way, step 2 has 2 ways.
03
Transition: ways[i-1] + ways[i-2].
04
Space can be O(1).
function climbStairs(n) {
if (n <= 2) return n;
let oneStepBefore = 2;
let twoStepsBefore = 1;
for (let step = 3; step <= n; step++) {
const current = oneStepBefore + twoStepsBefore;
twoStepsBefore = oneStepBefore;
oneStepBefore = current;
}
return oneStepBefore;
}Read the implementation
Connect each part of the code to the concept before you type.
2. Previous state
Ways to reach step 2.
let oneStepBefore = 23. Build upward
Compute each next stair count from earlier counts.
for (let step = 3; step <= n; step++)4. Ways recurrence
Reach current from one step below or two steps below.
const current = oneStepBefore + twoStepsBeforeA complete learning loop
01
Understand
Read the concept map and follow the code breakdown.
02
Reconstruct
Type the implementation, repair mistakes and work through every recall exercise.
03
Verify
Use the final check without hints; revisit weak concepts in short practice sessions.