# 5 Best Ways to Find Minimum Cost to Climb to the Top of Stairs in Python

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π‘ Problem Formulation: Given a staircase where each step has an associated cost, our goal is to find the minimum cost required to reach the top of the stairs. The input is a list where the ith element represents the cost of the ith step, and we can either start from step 0 or step 1. The output is the minimum cost to reach a step that is beyond the topmost step (either the last or second-to-last step).

## Method 1: Dynamic Programming

This method uses dynamic programming to iteratively calculate the minimum cost of reaching each step based on the previously calculated steps. The function `minCostClimbingStairs(cost)` returns the minimum cost to reach the top of a staircase, where cost is a list of integers representing the cost of each step.

Here’s an example:

```def minCostClimbingStairs(cost):
dp = [0] * (len(cost) + 1)
for i in range(2, len(cost) + 1):
dp[i] = min(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2])
return dp[-1]

# Example usage:
print(minCostClimbingStairs([10, 15, 20]))```

Output: 15

This code initializes a dynamic programming array `dp` to store the minimum cost to reach each step, starting with 0 for the first two steps. It then iterates through the steps, calculating the cost based on the previous two steps, and returns the last element of `dp`.

## Method 2: Recursive Approach with Memoization

Recursion with memoization is a technique to avoid redundant calculations by storing the results of expensive function calls. The function `minCost()` recursively finds the minimum cost to the top of the stairs, caching the results to improve efficiency.

Here’s an example:

```def minCostClimbingStairs(cost):
memo = {}

def minCost(i):
if i <= 1:
return 0
if i not in memo:
memo[i] = min(minCost(i - 1) + cost[i - 1], minCost(i - 2) + cost[i - 2])
return memo[i]

return minCost(len(cost))

# Example usage:
print(minCostClimbingStairs([1, 100, 1, 1, 1, 100, 1, 1, 100, 1]))```

Output: 6

This code snippet defines a recursive function `minCost()`, which calculates the cost of reaching the ith step. The memoization dictionary `memo` is used to cache the results, reducing the overall number of calculations needed.

## Method 3: Optimized Dynamic Programming

By optimizing space complexity, this method keeps track of only the last two steps’ costs instead of using an entire list. The `minCostClimbingStairs()` function computes the minimum cost with reduced space requirements.

Here’s an example:

```def minCostClimbingStairs(cost):
a, b = 0, 0
for i in range(2, len(cost) + 1):
a, b = b, min(b + cost[i - 1], a + cost[i - 2])
return b

# Example usage:
print(minCostClimbingStairs([10, 15, 3, 7]))```

Output: 10

This approach keeps two variables, `a` and `b`, to store the minimum cost to reach the last two steps. It updates these variables iteratively, ensuring the space complexity is O(1), as it doesn’t rely on a list to store the cost of all steps.

## Method 4: Greedy Approach

In certain cost configurations, a greedy approach can reach a solution by always choosing the next step based on the current cheapest option. The function `minCostClimbingStairs()` uses this philosophy to calculate the minimum cost, though it may not work for all cases.

Here’s an example:

```def minCostClimbingStairs(cost):
step1, step2 = 0, min(cost[0], cost[1])
for i in range(2, len(cost)):
next_step = cost[i] + min(step1, step2)
step1, step2 = step2, next_step
return min(step1, step2)

# Example usage:
print(minCostClimbingStairs([0, 2, 2, 1]))```

Output: 2

This example uses a bottom-up approach, accumulating the cost in two variables `step1` and `step2`. For each step, it updates these variables taking the minimum of the two options, simulating a greedy choice.

## Bonus One-Liner Method 5: Pythonic Reduction

The one-liner uses Python’s `reduce` function to apply a similar approach to the optimized dynamic programming example but in a single, elegant line of code. However, this approach might not be readily understandable to beginners.

Here’s an example:

```from functools import reduce

def minCostClimbingStairs(cost):
return reduce(lambda acc, cur: (acc[1], min(acc) + cur), cost, (0, 0))[1]

# Example usage:
print(minCostClimbingStairs([10, 15, 20]))```

Output: 15

This one-liner uses `reduce()` to iterate over the list, carrying an accumulator `acc` that holds the last two minimum costs. The final minimum cost is extracted from the accumulator.

## Summary/Discussion

• Method 1: Dynamic Programming. Time-efficient. Requires extra space proportional to the number of stairs.
• Method 2: Recursive Approach with Memoization. Intuitive. Potential stack overflow for very long staircases due to deep recursion.
• Method 3: Optimized Dynamic Programming. Very space-efficient. Can be less readable to those unfamiliar with space optimization techniques.
• Method 4: Greedy Approach. Simple and intuitive. Not guaranteed to work for all cost configurations, potentially leading to incorrect results.
• Bonus Method 5: Pythonic Reduction. Elegant and compact. May sacrifice readability for conciseness, especially for those not comfortable with functional programming paradigms.