💡 Problem Formulation: The ‘Min Cost Climbing Stairs’ problem asks for the minimum cost to reach the top of a staircase, where each step has a certain non-negative cost attached to it. You can either start from the first or second step and climb one or two steps at a time. Given an input array cost where cost[i] is the cost of step i, the desired output is the minimal total cost to reach the top of the stairs.
Method 1: Dynamic Programming with Iteration
Dynamic programming is a method where the solution is built up from the base cases using previously computed values. For the ‘Min Cost Climbing Stairs’ problem, we can use an iterative approach that goes from the second step to the top, determining the minimum cost for each step based on the two previous steps.
Here’s an example:
def minCostClimbingStairs(cost):
f1, f2 = 0, 0
for x in reversed(cost):
f1, f2 = x + min(f1, f2), f1
return min(f1, f2)
print(minCostClimbingStairs([10, 15, 20]))
print(minCostClimbingStairs([1, 100, 1, 1, 1, 100, 1, 1, 100, 1]))The output of this code snippet:
15 6
This code defines a function minCostClimbingStairs() that iterates through the cost array in reverse to use the minimum cost to reach the top from the current step. Here, f1 and f2 store the minimum costs from the current and previous steps respectively. The final result is the smaller of the two minimum costs for the first two steps.
Method 2: Dynamic Programming with Memoization
In this method, we use an auxiliary array to store computed minimum costs to avoid redundant calculations. This approach, known as memoization, can improve efficiency, especially for larger staircases.
Here’s an example:
def minCostClimbingStairs(cost):
n = len(cost)
memo = [0] * (n+1)
for i in range(2, n+1):
stepOne = memo[i-1] + cost[i-1]
stepTwo = memo[i-2] + cost[i-2]
memo[i] = min(stepOne, stepTwo)
return memo[n]
print(minCostClimbingStairs([10, 15, 20]))
print(minCostClimbingStairs([1, 100, 1, 1, 1, 100, 1, 1, 100, 1]))The output of this code snippet:
15 6
This code snippet introduces a memoization array to store the minimum cost of reaching each step. The function minCostClimbingStairs() iteratively computes the minimum cost of each step considering the cost to climb either from one step or two steps behind. At the end, memo[n] holds the minimum cost to reach the top.
Method 3: Using Recursion with Memoization
Recursion can also be used with memoization to solve the ‘Min Cost Climbing Stairs’ problem. This approach involves a top-down perspective, where we calculate the necessary values by diving deep with recursion and then use memoization to avoid duplicate work.
Here’s an example:
def minCostClimbingStairs(cost):
def minCost(i):
if i <= 1:
return 0
if memo[i] == -1:
downOne = minCost(i-1) + cost[i-1]
downTwo = minCost(i-2) + cost[i-2]
memo[i] = min(downOne, downTwo)
return memo[i]
memo = [-1] * (len(cost) + 1)
return minCost(len(cost))
print(minCostClimbingStairs([10, 15, 20]))
print(minCostClimbingStairs([1, 100, 1, 1, 1, 100, 1, 1, 100, 1]))The output of this code snippet:
15 6
This version of minCostClimbingStairs() employs a recursive helper function minCost() that includes memoization to avoid recalculating the minimum cost of a step multiple times. The base cases handle the first two steps, and the memo array initialized with -1 is updated with the calculated minimum cost of each step during recursive calls.
Method 4: Optimized Dynamic Programming
By noticing that we only need the last two values to compute the cost for the next step, we can further optimize the space complexity of dynamic programming. This method results in constant space usage while still providing the solution efficiently.
Here’s an example:
def minCostClimbingStairs(cost):
two_back = cost[0]
one_back = cost[1]
for i in range(2, len(cost)):
current = cost[i] + min(two_back, one_back)
two_back, one_back = one_back, current
return min(two_back, one_back)
print(minCostClimbingStairs([10, 15, 20]))
print(minCostClimbingStairs([1, 100, 1, 1, 1, 100, 1, 1, 100, 1]))The output of this code snippet:
15 6
The function minCostClimbingStairs() only uses two variables, two_back and one_back, to store the minimum cost of the two previous steps. Once we iterate over the cost array, these variables are updated to reflect the current minimum costs. The final return statement provides the least cost to reach the top from the two starting positions.
Bonus One-Liner Method 5: Using functools and Recursion
Python’s functools module can be applied to efficiently use recursion with memoization. The lru_cache decorator can automatically take care of caching the results of expensive function calls.
Here’s an example:
from functools import lru_cache
@lru_cache(maxsize=None)
def minCostClimbingStairs(cost, i=None):
if i is None:
i = len(cost) - 1
if i <= 1:
return cost[i]
return cost[i] + min(minCostClimbingStairs(cost, i-1), minCostClimbingStairs(cost, i-2))
cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
print(minCostClimbingStairs(tuple(cost)))The output of this code snippet:
6
In this elegant one-liner approach, the function minCostClimbingStairs() has been decorated with @lru_cache, which provides automatic memoization of the recursive calls. The result is a concise and clear recursion that performs effectively even for large inputs. Note that @lru_cache requires an immutable argument, hence the list cost is converted to a tuple.
Summary/Discussion
- Method 1: Iterative Dynamic Programming. Strengths: straightforward, in-place optimization. Weaknesses: mutable state within the function.
- Method 2: Dynamic Programming with Memoization. Strengths: clear separation of the task each step performs, easy to understand. Weaknesses: requires additional space for the memo array.
- Method 3: Recursive with Memoization. Strengths: clean recursive code structure, memoization optimizes performance. Weaknesses: overhead of recursive call stack, a bit more complex to trace than iterative solutions.
- Method 4: Optimized Dynamic Programming. Strengths: space-efficient linear approach using only two variables. Weaknesses: may be less intuitive as it obscures the underlying dynamic programming concept.
- Bonus Method 5: One-Liner with functools. Strengths: extremely concise, automatic caching. Weaknesses: requires understanding of decorators and caching, must convert inputs to immutable types.