746, Min Cost Climbing Stairs

I Problem

You are given an integer array cost where cost[i] is the cost of iᵗʰ step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index 1.

Return the minimum cost to reach the top of the floor.

Example 1
Input: cost = [10, 15, 20]
Output: 15
Explanation: You will start at index 1.

• Pay 15 and climb two steps to reach the top.
  The total cost is 15.

Example 2
Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
Output: 6
Explanation: You will start at index 0.

• Pay 1 and climb two steps to reach index 2.
• Pay 1 and climb two steps to reach index 4.
• Pay 1 and climb two steps to reach index 6.
• Pay 1 and climb one step to reach index 7.
• Pay 1 and climb two steps to reach index 9.
• Pay 1 and climb one step to reach the top.
  The total cost is 6.

Constraints

• 2 <= cost.length <= 1000
• 0 <= cost[i] <= 999

Related Topics

• Array
• Dynamic Programming

II Solution

Approach 1: Dynamic Programming

pub fn min_cost_climbing_stairs(cost: Vec<i32>) -> i32 {
    //Self::dp(cost)
    Self::optimize_dp(cost)
}

/// Time Complexity: O(n)
/// Space Complexity: O(n)
fn dp(cost: Vec<i32>) -> i32 {
    let len = cost.len();
    let mut dp = vec![0; len + 1];

    for i in 2..=len {
        dp[i] = std::cmp::min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
    }

    dp[len]
}

/// Time Complexity: O(n)
/// Space Complexity: O(1)
fn optimize_dp(cost: Vec<i32>) -> i32 {
    let (mut prev, mut curr) = (0, 0);

    for i in 2..=cost.len() {
        let next = std::cmp::min(curr + cost[i - 1], prev + cost[i - 2]);
        (prev, curr) = (curr, next);
    }

    curr
}

public int minCostClimbingStairs(int[] cost) {
    //return this.dp(cost);
    return this.optimizeDP(cost);
}

int dp(int[] cost) {
    int len = cost.length;
    int[] dp = new int[len + 1];

    for (int i = 2; i <= len; i++) {
        dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
    }

    return dp[len];
}

int optimizeDP(int[] cost) {
    int prev = 0, curr = 0;

    for (int i = 2; i <= cost.length; i++) {
        int next = Math.min(curr + cost[i - 1], prev + cost[i - 2]);
        prev = curr;
        curr = next;
    }

    return curr;
}

func minCostClimbingStairs(cost []int) int {
    //return dp(cost)
    return optimizeDP(cost)
}

func dp(cost []int) int {
    size := len(cost)
    dp := make([]int, size+1)
    
    for i := 2; i <= size; i++ {
        dp[i] = min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2])
    }

    return dp[size]
}

func optimizeDP(cost []int) int {
    prev, curr := 0, 0
    
    for i, size := 2, len(cost); i <= size; i++ {
        next := min(curr+cost[i-1], prev+cost[i-2])
        prev, curr = curr, next
    }

    return curr
}