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let twoStepsBefore = 0Typorithm · easy
Min Cost Climbing Stairs
▷ Interactive algorithm labmin-cost-climbing-stairs-dpdynamic-programmingminimizationspace-optimized
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Step 1 of 8 · intro card
Learn Min Cost Climbing Stairs
Dynamic Programming
Min Cost Climbing Stairs chooses the cheaper of reaching the top from one step below or two steps below.
The top is one position after the last cost.
Each state is the minimum cost to reach a position.
The transition compares paying from one step below versus two steps below.
Preparing algorithm lab…
01
State: min cost to reach position i.
02
Base: cost before first two positions is 0.
03
Transition: min(one-step, two-step).
04
Time O(n), space O(1).
function minCostClimbingStairs(cost) {
let twoStepsBefore = 0;
let oneStepBefore = 0;
for (let i = 2; i <= cost.length; i++) {
const takeOne = oneStepBefore + cost[i - 1];
const takeTwo = twoStepsBefore + cost[i - 2];
const current = Math.min(takeOne, takeTwo);
twoStepsBefore = oneStepBefore;
oneStepBefore = current;
}
return oneStepBefore;
}Read the implementation
Connect each part of the code to the concept before you type.
2. One-step option
Pay the cost of the stair one step below.
const takeOne = oneStepBefore + cost[i - 1]3. Two-step option
Pay the cost of the stair two steps below.
const takeTwo = twoStepsBefore + cost[i - 2]4. Min transition
Choose the cheaper way to reach this position.
const current = Math.min(takeOne, takeTwo)A 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.