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MinimumCostForTickets.cpp
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MinimumCostForTickets.cpp
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// Source : https://leetcode.com/problems/minimum-cost-for-tickets/
// Author : Hao Chen
// Date : 2019-01-29
/*****************************************************************************************************
*
* In a country popular for train travel, you have planned some train travelling one year in advance.
* The days of the year that you will travel is given as an array days. Each day is an integer from 1
* to 365.
*
* Train tickets are sold in 3 different ways:
*
* a 1-day pass is sold for costs[0] dollars;
* a 7-day pass is sold for costs[1] dollars;
* a 30-day pass is sold for costs[2] dollars.
*
* The passes allow that many days of consecutive travel. For example, if we get a 7-day pass on day
* 2, then we can travel for 7 days: day 2, 3, 4, 5, 6, 7, and 8.
*
* Return the minimum number of dollars you need to travel every day in the given list of days.
*
* Example 1:
*
* Input: days = [1,4,6,7,8,20], costs = [2,7,15]
* Output: 11
* Explanation:
* For example, here is one way to buy passes that lets you travel your travel plan:
* On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
* On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9.
* On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
* In total you spent $11 and covered all the days of your travel.
*
* Example 2:
*
* Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
* Output: 17
* Explanation:
* For example, here is one way to buy passes that lets you travel your travel plan:
* On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, ..., 30.
* On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31.
* In total you spent $17 and covered all the days of your travel.
*
* Note:
*
* 1 <= days.length <= 365
* 1 <= days[i] <= 365
* days is in strictly increasing order.
* costs.length == 3
* 1 <= costs[i] <= 1000
*
******************************************************************************************************/
class Solution {
private:
int min(int x, int y){
return x < y ? x : y;
}
int min(int x, int y, int z) {
return min(min(x, y), z);
}
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
// Dynamic Programming
vector<int> dp(days.size(), INT_MAX);
// dp[i] is the minimal cost from Days[0] to Days[i]
dp[0] = costs[0];
for (int i = 1; i< days.size(); i ++) {
// the currnet day need at least 1-day pass cost
int OneDayPass = dp[i-1] + costs[0];
// Seprating the array to two parts.
// days[0] -> days[j] -> day[i]
//
// calculate the day[i] - day[j] whether can use 7-day pass or 30-day pass
//
// Traking the minimal costs, then can have dp[i] minimal cost
int SevenDayPass = INT_MAX, ThrityDayPass = INT_MAX;
for (int j=i-1; j>=0; j--){
if (days[i] - days[j] < 7 ) {
SevenDayPass = dp[j-1] + costs[1];
} else if (days[i] - days[j] < 30 ) {
ThrityDayPass = dp[j-1] + costs[2];
} else {
break;
}
int m = min(OneDayPass, SevenDayPass, ThrityDayPass);
if ( dp[i] > m ) dp[i] = m;
}
}
return dp[dp.size()-1];
}
};