# Dynamic Programming - Min Cost Climbing Stairs

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Difficulty: Easy

DP

## 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:**

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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:**

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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

First, let’s think in terms of a `dfs`

problem. We can use the **template 2** that we have already discussed here.

At every step we can call our `dfs()`

function twice and find which path leads to min cost to reach the top of the floor. This is very similar to the Climbing Stairs, similarly we will add an additional `0`

to `cost`

as we did in the Climbing Stairs problem for the base case to work.

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cost.append(0)

Define a `min_cost`

variable.

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min_cost=float('inf')

In the `dfs()`

function, we pass the current `index`

and `curr_cost`

. Our base condition is whenever `index==len(cost)`

, we can compare `curr_cost`

with `min_cost`

.

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def dfs(index,curr_cost):
nonlocal min_cost
if index == len(cost):
min_cost = min(min_cost,curr_cost)
return

Now we need to have a loop runs only twice.

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for j in range(index, min(index+2,len(cost))):

Inside the loop, we increment `curr_cost`

, call `dfs()`

again and backtrack `curr_cost`

.

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for j in range(index, min(index+2,len(cost))):
curr_cost+=cost[j]
dfs(j+1,curr_cost)
curr_cost-=cost[j]

At the end, call `dfs(0,0)`

and return min_cost. (Full code at the end)

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dfs(0,0)
return min_cost

Unfortunately even though the code is all good, the solution won’t pass the Leetcode as it will takes $O(2^n)$ time and Leetcode will expect the solution to be competed in $O(n)$ time.

### Dynamic Programming

Idea to solve using Dynamic Programming is to start from backward (This is a patten in general). Consider the first example, we need to evaluate to find out if `20`

or `15`

is a place to be in. What we will do it, just a `0`

at the end for the loop. Then for `15`

find the `min`

between `20`

and `0`

. Set that as the new cost for `15`

. Now do the same for `10`

. The updated cost for `10`

will be `10+15=25`

.

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cost.append(0)
for i in range(len(cost)-3,-1,-1):
cost[i]+= min(cost[i+1],cost[i+2])

Finally return the `min`

`cost`

from first or second location.

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return min(cost[0],cost[1])

## Final Code

Here is the full code.

### DFS/Backtrack Solution

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def min_cost_climbing_stairs(cost):
min_cost = float('inf')
cost.append(0)
def dfs(index, curr_cost):
nonlocal min_cost
if index == len(cost):
min_cost = min(min_cost, curr_cost)
return
for j in range(index, min(index+2, len(cost))):
curr_cost += cost[j]
dfs(j+1, curr_cost)
curr_cost -= cost[j]
dfs(0, 0)
return min_cost

### Dynamic Programming Solution

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def min_cost_climbing_stairs(cost):
cost.append(0)
for i in range(len(cost)-3,-1,-1):
cost[i]+= min(cost[i+1],cost[i+2])
return min(cost[0],cost[1])