Coin Path

Hard

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ArraysDynamic Programming

You are given an array of integers coins representing coin denominations and an integer maxJump representing the maximum length of a jump. There is also a cost array cost where cost[i] is the cost to step on the i^th coin.

You want to reach the last coin starting from the first coin. You can jump to any coin from the current coin if the following conditions are met:

You can jump at most maxJump coins at a time.
You can only jump to a coin that has a positive value (coins[i] > 0).
If you land on a coin with value coins[i], it will cost you cost[i] to step on that coin.

Return a list of the indices of the coins you will step on to reach the last coin with the minimum cost. If there are multiple possible paths with the same minimum cost, return the lexicographically smallest path. If it is not possible to reach the last coin, return an empty list.

Example 1:

Input: coins = [1,2,4,-1,2], cost = [4,3,2,1,0], maxJump = 2
Output: [0,2,4]
Explanation:
We can reach the last coin with a minimum cost of 6 by jumping from coin 0 to coin 2 and then to coin 4. The path is [0,2,4].

Example 2:

Input: coins = [1,2,4,-1,2], cost = [4,3,2,1,0], maxJump = 1
Output: []
Explanation:
It is not possible to reach the last coin because you cannot jump over coin 3 which has a value of -1.

Example 3:

Input: coins = [1,2,4,-1,2], cost = [4,3,2,1,0], maxJump = 3
Output: [0,2,4]
Explanation:
We can reach the last coin with a minimum cost of 6 by jumping from coin 0 to coin 2 and then to coin 4. The path is [0,2,4].
Note that we can also jump directly from coin 0 to coin 4, but the cost will be 4 + 0 = 4 which is not the minimum cost.

Constraints:

1 <= coins.length == cost.length <= 1000
coins[i] is in the range [-100, 100].
cost[i] is in the range [0, 100].
1 <= maxJump <= 100

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected behavior if no path exists that reaches the end (i.e., no combination of coin values allows me to reach the last cell)? Should I return an empty array, null, or something else?
Are the coin values guaranteed to be non-negative integers? Can a coin have a value of 0?
Is the input array guaranteed to have at least one element?
If multiple paths with the minimum cost exist, can I return any one of them, or is there a specific path I should prefer (e.g., lexicographically smallest)?
What is the maximum possible value for a coin, and what is the maximum possible length of the input array?

Brute Force Solution

Approach

The brute force approach to finding the cheapest coin path involves exploring every single possible path from the starting point to the end. We essentially try out every combination of jumps to see which one costs the least. We don't worry about being efficient, just about finding all possible solutions and picking the best one.

Here's how the algorithm would work step-by-step:

Start at the beginning, which is the first spot.
From that spot, consider every possible jump you are allowed to make.
For each of those jumps, see where you land and remember the cost of making that jump.
From each of those new spots, again consider every jump you are allowed to make.
Keep doing this, building every possible path, until you either reach the end or you get stuck somewhere with no valid jumps left.
Once you reach the end with a path, calculate the total cost of all the jumps you made along that path.
Keep track of the cheapest path you've found so far.
Once you've explored every single possible path, the cheapest path you recorded is your answer.

Code Implementation

def coin_path_brute_force(coins, max_jump):
    number_of_coins = len(coins)
    cheapest_path = None
    cheapest_path_cost = float('inf')

    def find_paths(current_index, current_path, current_cost):
        nonlocal cheapest_path, cheapest_path_cost

        # If we've reached the end, check if this path is cheaper
        if current_index == number_of_coins - 1:
            if current_cost < cheapest_path_cost:
                cheapest_path_cost = current_cost
                cheapest_path = current_path[:]
            return

        # Explore possible jumps from the current index
        for jump_length in range(1, min(max_jump + 1, number_of_coins - current_index)):
            next_index = current_index + jump_length

            # Make sure the next coin is valid
            if coins[next_index] != -1:

                # Add to the path and recurse
                new_path = current_path + [next_index + 1]
                new_cost = current_cost + coins[next_index]
                find_paths(next_index, new_path, new_cost)

    # Initiate the search from the starting index
    if coins[0] != -1:

        #Start the recursive search
        find_paths(0, [1], coins[0])

    return cheapest_path

Big(O) Analysis

Time Complexity

O(m^n) – The described brute force approach explores all possible paths by considering every possible jump from each position. Assuming the maximum jump distance is 'm' (as we consider every possible jump we are allowed to make), and the total number of positions is 'n', at each step we have 'm' choices. Because we can potentially make 'n' jumps (in the case where jumps of length 1 will lead to the target), this creates a branching factor of 'm' at each of the 'n' steps. Therefore, the total number of paths explored grows exponentially with 'n', leading to a time complexity of O(m^n).

Space Complexity

O(N^J) – The brute force approach explores every possible path using recursion. In the worst-case, the algorithm could explore a tree where each node has J children (representing the maximum allowed jump), and the depth of the tree could be up to N (the number of coins). This leads to the creation of a call stack proportional to the number of paths explored. Therefore, the space complexity is driven by the recursion stack, which could grow to O(N^J), where N is the number of coins and J is the maximum jump distance, as the algorithm potentially explores every possible path of length N with up to J branches at each step. It stores all paths into the recursion stack. Consequently, the space complexity is O(N^J).

Optimal Solution

Approach

The goal is to find the cheapest path from the start to the end, given that you can only move to reachable spots. We use a method similar to finding the shortest route on a map, where the cost of each step is determined by the coin value at that position. This avoids exploring paths that are guaranteed to be more expensive.

Here's how the algorithm would work step-by-step:

Start at the beginning position.
Keep track of the cheapest way to reach each position from the start.
From each reachable position, consider all possible next moves (based on the maximum jump distance).
Calculate the total cost to reach the next position by adding the current cost and the coin value at the next position.
If the cost to reach the next position is cheaper than the previously recorded cost, update the cheapest cost for that position and remember the path taken to get there.
Repeat steps 3-5 until you reach the end position or you've explored all reachable positions.
If you reach the end position, trace back the path from the end to the start using the remembered paths to find the cheapest path.
If you cannot reach the end, there is no valid path.

Code Implementation

def coin_path(coins, max_jump):
    number_of_coins = len(coins)
    cheapest_costs = [float('inf')] * number_of_coins
    cheapest_costs[0] = coins[0] if coins[0] != -1 else float('inf')
    previous_positions = [None] * number_of_coins

    reachable_positions = [0]

    while reachable_positions:
        current_position = reachable_positions.pop(0)

        for next_jump in range(1, max_jump + 1):
            next_position = current_position + next_jump

            if next_position < number_of_coins and coins[next_position] != -1:
                cost_to_next_position = cheapest_costs[current_position] + coins[next_position]

                # Update cost if we found a cheaper path
                if cost_to_next_position < cheapest_costs[next_position]:
                    cheapest_costs[next_position] = cost_to_next_position
                    previous_positions[next_position] = current_position

                    # Explore this position later
                    if next_position not in reachable_positions:
                        reachable_positions.append(next_position)

    # No path to the end.
    if cheapest_costs[number_of_coins - 1] == float('inf'):
        return []

    # Reconstruct the path.
    path = []
    current_position = number_of_coins - 1

    while current_position is not None:
        path.append(current_position + 1)
        current_position = previous_positions[current_position]

    return path[::-1]

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm explores reachable positions using a process similar to Dijkstra's algorithm. In the worst-case scenario, we might need to visit each of the n positions in the coin array. For each of these n positions, we iterate up to m possible next positions (jump distance), checking if the new path is cheaper. Thus, the overall time complexity is bounded by O(n*m), where n is the number of coins and m is the maximum jump distance.

Space Complexity

O(N) – The algorithm maintains a 'cheapest way to reach each position' which necessitates an array of size N, where N is the number of coin values. Additionally, 'remembered paths' from each reachable position to the start requires another array of size N to store the preceding node in the optimal path. Therefore, the auxiliary space is proportional to the number of positions (N), resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
costs array is null or empty	Return an empty path list because there are no coins available to form a path.
cost of the starting coin (costs[0]) is -1	Return an empty path list since the starting point is unreachable.
costs array has only one element (n=1)	If costs[0] is not -1, return a list containing only the index 0; otherwise, return an empty list.
costs array contains only -1 values except for costs[0]	Return an empty path list, as there is no path to the end.
costs array contains large values, potentially leading to integer overflow in path costs	Use a `long` data type to store path costs to prevent integer overflow.
beam is 1, meaning only adjacent coins can be reached.	This limits the possible paths, potentially making the end unreachable and must be considered during neighbor generation.
No path exists to the last coin (costs[n-1])	The algorithm should return an empty path list to indicate no solution is found.
Multiple valid paths exist; find the lexicographically smallest.	When multiple paths have the same cost, prioritize paths with smaller indices earlier in the path using proper comparators in the priority queue.

costs array is null or empty

How to Handle:

Return an empty path list because there are no coins available to form a path.

cost of the starting coin (costs[0]) is -1

How to Handle:

Return an empty path list since the starting point is unreachable.

costs array has only one element (n=1)

How to Handle:

If costs[0] is not -1, return a list containing only the index 0; otherwise, return an empty list.

costs array contains only -1 values except for costs[0]

How to Handle:

Return an empty path list, as there is no path to the end.

costs array contains large values, potentially leading to integer overflow in path costs

How to Handle:

Use a `long` data type to store path costs to prevent integer overflow.

beam is 1, meaning only adjacent coins can be reached.

How to Handle:

This limits the possible paths, potentially making the end unreachable and must be considered during neighbor generation.

No path exists to the last coin (costs[n-1])

How to Handle:

The algorithm should return an empty path list to indicate no solution is found.

Multiple valid paths exist; find the lexicographically smallest.

How to Handle:

When multiple paths have the same cost, prioritize paths with smaller indices earlier in the path using proper comparators in the priority queue.