Min Cost Climbing Stairs

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.

Examples

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.

Solution

The greedy will not work as sometimes we will not always take 2 steps. Sometimes, 1 step also makes sense. See example 2. This problem is like an extension to [Climbing Stairs - Take atmost 2 Steps](climbing-stairs-take-atmost-2-steps).

Method 1 - Recursion

We start at either step 0 or step 1. The target is to reach either last or second last step, whichever is minimum.

Recurrence Relation

mincost(i) = cost[i]+min(mincost(i-1), mincost(i-2))

Base case

• mincost(0) = cost[0]
• mincost(1) = cost[1]

Code

Java

public int minCostClimbingStairs(int[] cost) {
	int n = cost.length;
	return Math.min(minCost(cost, n - 1), minCost(cost, n - 2));
}


private int minCost(int[] cost, int idx) {
	if (idx < 0) {
		return 0;
	}

	if (idx == 0 || idx == 1) {
		return cost[i];
	}

	return cost[idx] + Math.min(minCost(cost, idx - 1), min)
}

Complexity

• ⏰ Time complexity: O(2^n)
• 🧺 Space complexity: O(n)

Method 2 - Top Down DP with memoization

Code

Java

public int minCostClimbingStairs(int[] cost) {
	int n = cost.length;
	int dp = new int[n];
	return Math.min(minCost(cost, n - 1, dp), minCost(cost, n - 2, dp));
}


private int minCost(int[] cost, int idx, int[] dp) {
	if (idx < 0) {
		return 0;
	}

	if (idx == 0 || idx == 1) {
		return cost[i];
	}

	if (dp[idx] != 0) {
		return dp[idx];
	}

	dp[idx] = cost[idx] + Math.min(minCost(cost, idx - 1), min)

	return dp[n];
}

Complexity

• ⏰ Time complexity: O(n)
• 🧺 Space complexity: O(n)

Method 2 - Bottom UP DP

Code

Java

public int minCostClimbingStairs(int[] cost) {
	int n = cost.length;
	int[] dp = new int[n];
	dp[0] = cost[0];
	dp[1] = cost[1];

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

	return Math.min(dp[n - 1], dp[n - 2]);
}

Complexity

• ⏰ Time complexity: O(n)
• 🧺 Space complexity: O(n)

Method 3 - Bottom up DB without using extra space

Code

Java

public int minCostClimbingStairs(int[] cost) {
	int n = cost.length;
	int f1 = cost[0], f2 = cost[1];

	if (n <= 2) {
		return Math.min(f1, f2);
	}

	for (int i = 2; i < n; i++) {
		int curr = cost[i] + Math.min(f1, f2);
		f1 = f2;
		f2 = temp;
	}

	return Math.min(f1, f2);
}

Complexity

• ⏰ Time complexity: O(n)
• 🧺 Space complexity: O(1)