746. Min Cost Climbing Stairs

At a Glance

Topic: dp-1d
Pattern: Dynamic Programming (linear minimum cost)
Difficulty: Easy
Companies: Amazon, Google, Adobe, Uber, Bloomberg
Frequency: High
LeetCode: 746

Each stair index has a non-negative cost[index]. Pay that cost to step onto it; start at index 0 or 1. Reach past the last index with minimum total cost. Input: cost array. Output: minimum sum.

Recognition Cues

“Minimum cost” + “from step i you can go to i+1 or i+2”
Start position choice (first or second stair)
Top is beyond last index (pay once then exit)

Diagram

At-a-glance flow (replace with problem-specific Mermaid as you refine this note). camelCase node IDs; no spaces in IDs.

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Approaches

Brute force — enumerate all paths — exponential time.
Better — top-down memo from each index — O(n) time / O(n) space.
Optimal — bottom-up min of two predecessors — O(n) time / O(1) space. <- pick this in interview.

Optimal Solution

Go

package main

func minCostClimbingStairsTable(cost []int) int {
	length := len(cost)
	dp := make([]int, length+1)
	dp[0], dp[1] = 0, 0
	for index := 2; index <= length; index++ {
		payCurrent := dp[index-1] + cost[index-1]
		payPrevious := dp[index-2] + cost[index-2]
		if payCurrent < payPrevious {
			dp[index] = payCurrent
		} else {
			dp[index] = payPrevious
		}
	}
	return dp[length]
}

func minCostClimbingStairs(cost []int) int {
	length := len(cost)
	prevPrev := 0
	prev := 0
	for index := 2; index <= length; index++ {
		payCurrent := prev + cost[index-1]
		payPrevious := prevPrev + cost[index-2]
		next := payCurrent
		if payPrevious < payCurrent {
			next = payPrevious
		}
		prevPrev = prev
		prev = next
	}
	return prev
}

JavaScript

function minCostClimbingStairsTable(cost) {
	const length = cost.length;
	const dp = new Array(length + 1).fill(0);
	for (let index = 2; index <= length; index++) {
		dp[index] = Math.min(
			dp[index - 1] + cost[index - 1],
			dp[index - 2] + cost[index - 2]
		);
	}
	return dp[length];
}

function minCostClimbingStairs(cost) {
	const length = cost.length;
	let prevPrev = 0;
	let prev = 0;
	for (let index = 2; index <= length; index++) {
		const next = Math.min(
			prev + cost[index - 1],
			prevPrev + cost[index - 2]
		);
		prevPrev = prev;
		prev = next;
	}
	return prev;
}