# Coin Path in C++

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Suppose we have an array A (index starts at 1) with N numbers: A1, A2, ..., AN and another integer B. The integer B denotes that from any index is i in the array A, we can jump to any one of the places in the array A indexed i+1, i+2, …, i+B if this place can be jumped to. Also, if we step on the index i, we have to pay Ai amount of coins. If Ai is -1, it means we can’t jump to the place indexed i in the array.

Now, when we start from the place indexed 1 in the array A, and our aim is to reach the place indexed N using the minimum coins. We have to return the path of indexes (starting from 1 to N) in the array we should take to get to the place indexed N using minimum coins. If we have multiple paths with the same cost, we have to find the lexicographically smallest such path. And if we have no such possible path to reach the place indexed N then we will return an empty array.

So, if the input is like [1,2,4,-1,2], 2, then the output will be [1,3,5]

To solve this, we will follow these steps −

• n := size of A

• Define an array ret

• Define an array cost of size n, and fill this with inf

• Define an array next of size n, and fill this with -1

• if not n is non-zero or A[n - 1] is same as -1, then −

• endPoint := n - 1

• cost[n - 1] = A[n - 1]

• for initialize i := n - 2, when i >= 0, update (decrease i by 1), do −

• if A[i] is same as -1, then −

• Ignore following part, skip to the next iteration

• for j in range i + 1 to minimum of (n - 1) and i + B, increase j by 1 −

• if cost[j] + A[i] < cost[i], then −

• cost[i] := cost[j] + A[i]

• next[i] := j

• endPoint := i

• if endPoint is not equal to 0, then −

• return empty array

• for endPoint is not equal to -1, update endPoint = next[endPoint], do −

• insert endPoint + 1 at the end of ret

• return ret

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<int> cheapestJump(vector<int>& A, int B) {
int n = A.size();
vector <int> ret;
vector <int> cost(n, 1e9);
vector <int> next(n, -1);
if(!n || A[n - 1] == -1) return {};
int endPoint = n - 1;
cost[n - 1] = A[n - 1];
for(int i = n - 2; i >= 0; i--){
if(A[i] == -1) continue;
for(int j = i + 1 ; j <= min(n - 1, i + B); j++){
if(cost[j] + A[i] < cost[i]){
cost[i] = cost[j] + A[i];
next[i] = j;
endPoint = i;
}
}
}
if(endPoint != 0) return {};
for(;endPoint != - 1; endPoint = next[endPoint]){
ret.push_back(endPoint + 1);
}
return ret;
}
};
main(){
Solution ob;
vector<int> v = {1,2,4,-1,2};
print_vector(ob.cheapestJump(v, 2));
}

## Input

{1,2,4,-1,2}, 2

## Output

[1, 3, 5, ]