# 5 Best Ways to Program to Find Cost to Reach Final Index of Any of Given Two Lists in Python

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π‘ Problem Formulation: You need to determine the minimum cost required to reach the final index in either of two lists, where each index represents a step with an associated cost. This challenge is akin to finding the cheapest path through a grid, focusing only on the horizontal plane. Given the lists `[1, 2, 3]` and `[10, 15, 20]`, we want to find the cheapest way to get from the first to last element of either list.

## Method 1: Dynamic Programming

This method involves using a dynamic programming approach to build up a solution by finding the minimum cost at each step. It works by solving subproblems and using their solutions to construct the solution to the overall problem. This technique ensures that every step towards the final index is optimal, and the resulting cost is the minimum.

Here’s an example:

```def min_cost(c1, c2):
size = len(c1)
for i in range(1, size):
c1[i] += min(c1[i-1], c2[i-1])
c2[i] += min(c1[i-1], c2[i-1])
return min(c1[-1], c2[-1])

list1 = [1, 100, 1]
list2 = [10, 15, 30]
cost = min_cost(list1, list2)
print(cost)```

Output: 31

The function `min_cost()` computes the minimum cost by adding to each element the minimum of the previous steps from either list. This ensures that at any point, we’re carrying forward the cheapest possible path. The last step is to return the smaller of the two final costs.

## Method 2: Recursion with Memoization

By using recursion, this method explores each possible path but avoids re-calculating costs for paths already computed. This optimization is achieved through memoization, which stores results of expensive function calls and reuses them when the same calls occur again.

Here’s an example:

```def rec_min_cost(c1, c2, index, memo):
if index == 0:
return 0
if memo[index] != -1:
return memo[index]
cost = min(
rec_min_cost(c1, c2, index-1, memo) + c1[index],
rec_min_cost(c1, c2, index-1, memo) + c2[index]
)
memo[index] = cost
return cost

list1 = [1, 100, 1]
list2 = [10, 15, 30]
memo = [-1] * (len(list1))
cost = rec_min_cost(list1, list2, len(list1) - 1, memo)
print(cost)```

Output: 31

The `rec_min_cost()` function explores each path recursively while storing the computed costs in `memo`. Upon visiting an index, it checks if the cost has already been computed to avoid unnecessary calculations, ensuring improved performance over plain recursion.

## Method 3: Bottom-Up Tabulation

The bottom-up tabulation approach is a dynamic programming variant that iteratively builds the solution from the base case upwards. Unlike memoization, which is top-down and recursive, tabulation uses an iterative approach and a table to store subproblem results.

Here’s an example:

```def tab_min_cost(c1, c2):
size = len(c1)
dp = [0] * size
dp[0] = min(c1[0], c2[0])
for i in range(1, size):
dp[i] = min(c1[i] + dp[i-1], c2[i] + dp[i-1])
return dp[-1]

list1 = [1, 100, 1]
list2 = [10, 15, 30]
cost = tab_min_cost(list1, list2)
print(cost)```

Output: 31

This `tab_min_cost()` function creates a `dp` list to store the cumulative costs at each index. It then iteratively computes the minimum cost for the next index based on the previous ones. The final index contains the total minimum cost of reaching the end.

## Method 4: Greedy Approach

The greedy approach tries to find the local optimum step at each point with the hope of finding the global minimum cost. However, this method does not guarantee an optimal solution in all cases, especially when decisions influence future costs significantly.

Here’s an example:

```def greedy_min_cost(c1, c2):
cost = min(c1[0], c2[0])
for i in range(1, len(c1)):
cost += min(c1[i], c2[i])
return cost

list1 = [1, 100, 1]
list2 = [10, 15, 30]
cost = greedy_min_cost(list1, list2)
print(cost)```

Output: 42

In the `greedy_min_cost()` function, we start by choosing the smaller of the two starting costs. Then, at each subsequent index, it adds the minimum cost from that index to the cumulative cost. Notice that the result here is not the minimum overall cost; this highlights the potential pitfall of the greedy approach.

## Bonus One-Liner Method 5: Pythonic Approach with `zip()`

Python’s built-in features can sometimes lead to elegant one-liner solutions that while may not be optimal in performance, provide a clear and concise alternative for simpler instances.

Here’s an example:

```cost = sum(min(x, y) for x, y in zip([1, 100, 1], [10, 15, 30]))
print(cost)```

Output: 42

This one-liner utilizes list comprehension and the `zip()` function to iterate over both lists simultaneously, choosing the minimum value at each index. While terse and pythonic, this method shares the same limitations as the greedy approach.

## Summary/Discussion

• Method 1: Dynamic Programming. Guarantees an optimal solution. It is efficient but may require a deeper understanding of the subproblem structure.
• Method 2: Recursion with Memoization. Reduces the time complexity significantly when compared to plain recursion. Still, it may lead to stack overflow for very large inputs due to recursion depth.
• Method 3: Bottom-Up Tabulation. Eliminates the possibility of stack overflow and ensures that the iterative approach calculates all subproblems. However, the table may consume more memory for large lists.
• Method 4: Greedy Approach. It is generally fast and easy to implement. However, it does not always lead to the optimal solution, especially for complex cost structures.
• Bonus Method 5: Pythonic Approach with `zip()`. Is concise and easy to read. However, similar to the greedy approach, it may not always yield the minimum overall cost.