Formatted question description: https://leetcode.ca/all/656.html
656. Coin Path
Level
Hard
Description
Given an array A
(index starts at 1
) consisting of N integers: A_{1}, A_{2}, …, A_{N} 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 A_{i} coins. If A_{i} 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:
 Path Pa_{1}, Pa_{2}, …, Pa_{n} is lexicographically smaller than Pb_{1}, Pb_{2}, …, Pb_{m}, if and only if at the first
i
where Pa_{i} and Pb_{i} differ, Pa_{i} < Pb_{i}; when no suchi
exists, thenn
<m
.  A_{1} >= 0. A_{2}, …, A_{N} (if exist) will in the range of [1, 100].
 Length of A is in the range of [1, 1000].
 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) { path.add(index + 1); index = next[index]; } return path; } }

Todo

print("Todo!")