What is the brute force approach and complexity for Min Cost Climbing Stairs?

Recursive best(i) = cost[i] + min(best(i−1), best(i−2)), recomputed from scratch with time complexity O(2^N) and space complexity O(N).

What is the optimal approach and complexity for Min Cost Climbing Stairs?

Keep the cheapest totals for the last two steps in two variables and slide them forward with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Min Cost Climbing Stairs?

Look for key signals in the prompt: "minimum total cost" indicates We need the least total, so each step keeps only the cheapest way to reach it.; "climb one or two steps" indicates A step is reached from the one or two steps before it, so the answer for step i uses steps i − 1 and i − 2.; "cost of each step" indicates The price is added when you arrive at a step, so dp[i] = cost[i] + the best of its two arrivals..

Min Cost Climbing Stairs

Easy·LeetCode #746 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You get an array cost. cost[i] is the price of step i, and you pay that price each time you stand on step i.

From any step you can climb 1 step or 2 steps. You may begin on step 0 or on step 1, and you pay the price of the step you begin on.

Find the least total you must pay to reach the top. The top is one step past the last step (position cost.length). It costs nothing to stand there.

Input

An array cost, one price per step.

Output

One integer: the least total cost to reach the top.

Example 1

Input: cost = [10, 15, 20]

Output: 15

Begin on step 1 and pay 15, then climb 2 steps to the top. Every other route costs more.

Example 2

Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]

Output: 6

Begin on step 0 and pay 1. Then land on steps 2, 4 and 6, each costing 1. Step on to 7 and 9, each costing 1, and finish at the top. Six steps are paid, and each costs 1.

Constraints

• 2 ≤ cost.length ≤ 1000
• 0 ≤ cost[i] ≤ 999

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“minimum total cost”

→ We need the least total, so each step keeps only the cheapest way to reach it.

“climb one or two steps”

→ A step is reached from the one or two steps before it, so the answer for step i uses steps i − 1 and i − 2.

“cost of each step”

→ The price is added when you arrive at a step, so dp[i] = cost[i] + the best of its two arrivals.

Approach Complexity Ladder

Brute Force — recursion on the cheapest way to each stepO(2^N)O(N)

Recursive best(i) = cost[i] + min(best(i−1), best(i−2)), recomputed from scratch

Better — memoization (remember best for each step)O(N)O(N)

The same recursion, storing each best(i) the first time it is computed

Optimal — two totals, sliding up the stepsO(N)O(1)

Keep the cheapest totals for the last two steps in two variables and slide them forward

Brute Force Intuition

Let best(i) be the least total to stand on step i, price included. To stand on step i you came from step i − 1 or step i − 2, so best(i) = cost[i] + min(best(i − 1), best(i − 2)). The first two steps are the starting points. The answer is the smaller of best(n − 1) and best(n − 2). The method is correct, but it solves the same step many times.

Recognition rule for this sub-pattern:

dp[i] depends on dp[i−1] and dp[i−2]; keep just two variables.

Remember this · DP — 1D Linear

Trigger

"Number of ways", "maximum/minimum up to i", choices at each step that depend on the previous one or two.

Think

If I knew the answer for every smaller prefix, how would I get the answer for i?

Key question

What does dp[i] mean in one sentence?

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