## Question

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 can’t 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

#### Approach #1 Brute Force[Time Limit Exceeded]

In this approach, we make use of a array of size . Here, refers to the size of the given array. The array is used such that is used to store the minimum number of coins needed to jump till the end of the array , starting from the index .

We start by filling the array with all -1's. Then, in order to fill this array, we make use of a recursive function jump(A, B, i, next) which fills the array starting from the index onwards, given as the coins array and as the largest jump value.

With as the current index, we can consider every possible index from to as the next place to be jumped to. For every such next index, , if this place can be jumped to, we determine the cost of reaching the end of the array starting from the index , and with as the next index jumped from , as . If this cost is lesser than the minimum cost required till now, for the same starting index, we can update the minimum cost and the value of as well.

For every such function call, we also need to return this minimum cost.

At the end, we traverse over the array starting from the index 1. At every step, we add the current index to the list to be returned and also jump/move to the index pointed by , since this refers to the next index for the minimum cost. We continue in the same manner till we reach the end of the array .

Java

public class Solution {
public List < Integer > cheapestJump(int[] A, int B) {
int[] next = new int[A.length];
Arrays.fill(next, -1);
jump(A, B, 0, next);
List < Integer > res = new ArrayList();
int i;
for (i = 0; i < A.length && next[i] > 0; i = next[i])
if (i == A.length - 1 && A[i]>= 0)
else
return new ArrayList < Integer > ();
return res;
}
public long jump(int[] A, int B, int i, int[] next) {
if (i == A.length - 1 && A[i] >= 0)
return A[i];
long min_cost = Integer.MAX_VALUE;
for (int j = i + 1; j <= i + B && j < A.length; j++) {
if (A[j] >= 0) {
long cost = A[i] + jump(A, B, j, next);
if (cost < min_cost) {
min_cost = cost;
next[i] = j;
}
}
}
return min_cost;
}
}


Complexity Analysis

• Time complexity : . The size of the recursive tree can grow upto in the worst case. This is because, we have possible branches at every step. Here, refers to the limit of the largest jump and refers to the size of the given array.

• Space complexity : . The depth of the recursive tree can grow upto . array of size is used.

#### Approach #2 Using Memoization [Accepted]

Algorithm

In the recursive solution just discussed, a lot of duplicate function calls are made, since we are considering the same index through multiple paths. To remove this redundancy, we can make use of memoization.

We keep a array, such that is used to store the minimum cost of jumps to reach the end of the array . Whenever the value for any index is calculated once, it is stored in its appropriate location. Thus, next time whenever the same function call is made, we can return the result directly from this array, pruning the search space to a great extent.

Java

public class Solution {
public List < Integer > cheapestJump(int[] A, int B) {
int[] next = new int[A.length];
Arrays.fill(next, -1);
long[] memo = new long[A.length];
jump(A, B, 0, next, memo);
List < Integer > res = new ArrayList();
int i;
for (i = 0; i < A.length && next[i] > 0; i = next[i])
if (i == A.length - 1 && A[i] >= 0)
else
return new ArrayList < Integer > ();
return res;
}
public long jump(int[] A, int B, int i, int[] next, long[] memo) {
if (memo[i] > 0)
return memo[i];
if (i == A.length - 1 && A[i] >= 0)
return A[i];
long min_cost = Integer.MAX_VALUE;
for (int j = i + 1; j <= i + B && j < A.length; j++) {
if (A[j] >= 0) {
long cost = A[i] + jump(A, B, j, next, memo);
if (cost < min_cost) {
min_cost = cost;
next[i] = j;
}
}
}
memo[i] = min_cost;
return min_cost;
}
}


Complexity Analysis

• Time complexity : . array of size is filled only once. We also do a traversal over the array, which will go upto steps. Here, refers to the number of nodes in the given tree.

• Space complexity : . The depth of the recursive tree can grow upto . array of size is used.

#### Approach #3 Using Dynamic Programming [Accepted]

Algorithm

From the solutions discussed above, we can observe that the cost of jumping till the end of the array starting from the index is only dependent on the elements following the index and not the ones before it. This inspires us to make use of Dynamic Programming to solve the current problem.

We again make use of a array to store the next jump locations. We also make use of a with the same size as that of the given array. is used to store the minimum cost of jumping till the end of the array , starting from the index . We start with the last index as the current index and proceed backwards for filling the and array.

With as the current index, we consider all the next possible positions from , ,..., , and determine the position, , which leads to a minimum cost of reaching the end of , which is given by . We update with this corresponding index. We also update with the minimum cost, to be used by the previous indices' cost calculations.

At the end, we again jump over the indices as per the array and put these indices in the array to be returned.

Java

public class Solution {
public List < Integer > cheapestJump(int[] A, int B) {
int[] next = new int[A.length];
long[] dp = new long[A.length];
Arrays.fill(next, -1);
List < Integer > res = new ArrayList();
for (int i = A.length - 2; i >= 0; i--) {
long min_cost = Integer.MAX_VALUE;
for (int j = i + 1; j <= i + B && j < A.length; j++) {
if (A[j] >= 0) {
long cost = A[i] + dp[j];
if (cost < min_cost) {
min_cost = cost;
next[i] = j;
}
}
}
dp[i] = min_cost;
}
int i;
for (i = 0; i < A.length && next[i] > 0; i = next[i])
if (i == A.length - 1 && A[i] >= 0)
else
return new ArrayList < Integer > ();
return res;
}
}


Complexity Analysis

• Time complexity : . We need to consider all the possible positions for every current index considered in the array. Here, refers to the number of elements in .

• Space complexity : . and array of size are used.

Analysis written by: @vinod23