Coin Path

Problem Statement

In this problem, you are given an integer array named coins, which is 1-indexed and represents the cost of landing on each index. Additionally, you are provided with an integer maxJump which determines the maximum range of indices you can jump forward from your current position. You start your journey from the index 1, which is guaranteed to be accessible (coins[1] != -1).

Your goal is to determine the sequence of indices you can follow to reach the last index (n) of the array with the minimum possible cost. It is crucial to note that you can only jump to an index if its value is not -1, as this represents an inaccessible position.

The result should be the path of indices, denoted as an integer array, that reflects the minimal cost route to index n. If multiple paths yield the same cost, you must return the sequence that is lexicographically smallest. In case reaching the final index is impossible, the function should return an empty array.

Examples

Example 1

Input:

coins = [1,2,4,-1,2], maxJump = 2

Output:

[1,3,5]

Example 2

Input:

coins = [1,2,4,-1,2], maxJump = 1

Output:

[]

Constraints

• 1 <= coins.length <= 1000
• -1 <= coins[i] <= 100
• coins[1] != -1
• 1 <= maxJump <= 100

Approach and Intuition

To solve this problem, an understanding of dynamic programming in the context of path-finding and cost minimization is crucial. Additionally, the preference for lexicographically smallest paths when costs are equivalent introduces a layer of complexity that must be carefully managed. Here’s a step-by-step breakdown of the approach:

1. Understand the problem and constraints:

   • You start at index 1 and aim to reach the last index with minimum cost.
   • You can jump forward from an index i to i+k, where 1 ≤ k ≤ maxJump if i+k is within bounds and coins[i+k] != -1.
   • You need to ensure the path obtained is the lexicographically smallest one in cases with similar costs.

2. Initial Setup:

   • Use a dynamic programming array dp where dp[i] keeps track of the minimum cost to reach index i from index 1.
   • Use another array path to store the actual path (sequence of indices) taken to reach every index.

3. Dynamic Programming Transition:

   • For each index i, if it's accessible (i.e., coins[i] != -1), update dp[i+k] for all 1 ≤ k ≤ maxJump, ensuring that you're not going out of bounds.
   • Update the cost and path based on comparisons: if moving to i+k gives a lesser cost or if the costs are equal but the new path is lexicographically smaller.

4. Lexicographical Consideration:

   • While updating paths, always ensure to choose the path that sorts lexicographically smaller in cases where multiple paths result in the same cost.

5. Result Compilation:

   • After filling the dp table, the minimum cost to reach the last index will be at dp[n]. The path leading to this result will be stored in path[n].

This approach essentially maps out optimal paths while keeping track of costs and ensures that in case of ties in cost, the path chosen favors lexicographical precedence. This kind of problem is typical in route optimization scenarios often found in network routing, logistics, and even game development.

Solutions

public class Solution {
    public List<Integer> findCheapestPath(int[] costs, int maxStep) {
        int[] nextStep = new int[costs.length];
        long[] minCosts = new long[costs.length];
        Arrays.fill(nextStep, -1);
        List<Integer> path = new ArrayList<>();
        for (int i = costs.length - 2; i >= 0; i--) {
            long globalMin = Integer.MAX_VALUE;
            for (int j = i + 1; j <= Math.min(i + maxStep, costs.length - 1); j++) {
                if (costs[j] >= 0) {
                    long currentCost = costs[i] + minCosts[j];
                    if (currentCost < globalMin) {
                        globalMin = currentCost;
                        nextStep[i] = j;
                    }
                }
            }
            minCosts[i] = globalMin;
        }
        int pos;
        for (pos = 0; pos < costs.length && nextStep[pos] > 0; pos = nextStep[pos])
            path.add(pos + 1);
        if (pos == costs.length - 1 && costs[pos] >= 0)
            path.add(costs.length);
        else
            return new ArrayList<Integer>();
        return path;
    }
}