Formatted question description: https://leetcode.ca/all/656.html

# 656. Coin Path

Hard

## Description

Given an array A (index starts at 1) consisting of N integers: A1, A2, …, AN and an integer B. The integer B denotes that from any place (suppose the index is i) in the array A, you can jump to any one of the place in the array A indexed i+1, i+2, …, i+B if this place can be jumped to. Also, if you step on the index i, you have to pay Ai coins. If Ai is -1, it means you cant jump to the place indexed i in the array.

Now, you start from the place indexed 1 in the array A, and your aim is to reach the place indexed N using the minimum coins. You need to return the path of indexes (starting from 1 to N) in the array you should take to get to the place indexed N using minimum coins.

If there are multiple paths with the same cost, return the lexicographically smallest such path.

If it’s not possible to reach the place indexed N then you need to return an empty array.

Example 1:

Input: [1,2,4,-1,2], 2

Output: [1,3,5]

Example 2:

Input: [1,2,4,-1,2], 1

Output: []

Note:

1. Path Pa1, Pa2, …, Pan is lexicographically smaller than Pb1, Pb2, …, Pbm, if and only if at the first i where Pai and Pbi differ, Pai < Pbi; when no such i exists, then n < m.
2. A1 >= 0. A2, …, AN (if exist) will in the range of [-1, 100].
3. Length of A is in the range of [1, 1000].
4. B is in the range of [1, 100].

## Solution

Use dynamic programming. Create an array dp and an array next of length A.length, where dp[i] represents the total cost starting from index i to the end, and next[i] represents the next index of index i in the path with minimum cost.

Initialize dp such that all elements are Integer.MAX_VALUE and initialize next such that all elements are A.length. If the last element in A is not -1, then set the last element of dp to be the last element of A.

Loop over A backwards from index A.length - 2 to 0. At each index i, if A[i] is -1, then the place cannot be jumped to, so continue. For each index j from i + 1 to i + B (or A.length - 1 if i + B exceeds A.length - 1), if dp[j] is Integer.MAX_VALUE, continue since it is impossible to reach the end from j. Otherwise, if dp[j] + A[i] < dp[i], then update dp[i] = dp[j] + A[i] and next[i] = j.

If dp[0] is Integer.MAX_VALUE, then no path exists, so return an empty list. Otherwise, starting from index 0 to find a path that reaches the end, and return the path as a list.

• class Solution {
public List<Integer> cheapestJump(int[] A, int B) {
List<Integer> path = new ArrayList<Integer>();
int length = A.length;
int[] dp = new int[length];
Arrays.fill(dp, Integer.MAX_VALUE);
if (A[length - 1] != -1)
dp[length - 1] = A[length - 1];
int[] next = new int[length];
Arrays.fill(next, length);
for (int i = length - 2; i >= 0; i--) {
if (A[i] == -1)
continue;
int start = i + 1, end = Math.min(i + B, length - 1);
for (int j = start; j <= end; j++) {
if (dp[j] == Integer.MAX_VALUE)
continue;
if (dp[j] + A[i] < dp[i]) {
dp[i] = dp[j] + A[i];
next[i] = j;
}
}
}
if (dp[0] == Integer.MAX_VALUE)
return path;
int index = 0;
while (index < length) {
index = next[index];
}
return path;
}
}

• Todo

• print("Todo!")