**Bus Fare - Amazon Top Interview Questions**

### Problem Statement :

You are given a list of sorted integers days , where you must take the bus for on each day. Return the lowest cost it takes to travel for all the days. There are 3 types of bus tickets. 1 day pass for 2 dollars 7 day pass for 7 dollars 30 day pass for 25 dollars Constraints n ≤ 100,000 where n is the length of days Example 1 Input days = [1, 3, 4, 5, 29] Output 9 Explanation The lowest cost can be achieved by purchasing a 7 day pass in the beginning and then a 1 day pass on the 29th day. Example 2 Input days = [1] Output 2

### Solution :

` ````
Solution in C++ :
const int TICKET_TYPE = 3;
const int price[TICKET_TYPE] = {2, 7, 25};
const int duration[TICKET_TYPE] = {1, 7, 30};
int solve(vector<int>& days) {
vector<int> pts(TICKET_TYPE);
int n = days.size();
vector<int> dp(n + 1, INT_MAX);
dp[0] = 0;
auto get_day = [&](int idx) { return idx == 0 ? 0 : days[idx - 1]; };
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < TICKET_TYPE; ++j) {
while (get_day(i) - get_day(pts[j] + 1) >= duration[j]) pts[j]++;
dp[i] = min(dp[i], dp[pts[j]] + price[j]);
}
}
return dp[n];
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] days) {
return lowest(new int[days.length], days, 0);
}
public int lowest(int[] dp, int[] days, int i) {
if (i >= days.length) {
return 0;
}
if (dp[i] != 0) {
return dp[i];
}
int beg = i + 1;
return dp[i] = min(2 + lowest(dp, days, search(days, beg, days[i] + 1)),
7 + lowest(dp, days, search(days, beg, days[i] + 7)),
25 + lowest(dp, days, search(days, beg, days[i] + 30)));
}
int search(int[] days, int beg, int day) {
while (beg < days.length && days[beg] < day) {
beg++;
}
return beg;
}
public int min(int... args) {
int min = args[0];
for (int i = 1; i < args.length; i++) {
min = Integer.min(min, args[i]);
}
return min;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, days):
dp = [0] * (len(days) + 1)
for i in range(len(days) - 1, -1, -1):
one = bisect.bisect_left(days, days[i] + 1)
seven = bisect.bisect_left(days, days[i] + 7)
thirty = bisect.bisect_left(days, days[i] + 30)
dp[i] = min(2 + dp[one], 7 + dp[seven], 25 + dp[thirty])
return dp[0]
```

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