<p>You have planned some train traveling one year in advance. The days of the year in which you will travel are given as an integer array <code>days</code>. Each day is an integer from <code>1</code> to <code>365</code>.</p>

<p>Train tickets are sold in <strong>three different ways</strong>:</p>

<ul>
	<li>a <strong>1-day</strong> pass is sold for <code>costs[0]</code> dollars,</li>
	<li>a <strong>7-day</strong> pass is sold for <code>costs[1]</code> dollars, and</li>
	<li>a <strong>30-day</strong> pass is sold for <code>costs[2]</code> dollars.</li>
</ul>

<p>The passes allow that many days of consecutive travel.</p>

<ul>
	<li>For example, if we get a <strong>7-day</strong> pass on day <code>2</code>, then we can travel for <code>7</code> days: <code>2</code>, <code>3</code>, <code>4</code>, <code>5</code>, <code>6</code>, <code>7</code>, and <code>8</code>.</li>
</ul>

<p>Return <em>the minimum number of dollars you need to travel every day in the given list of days</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> days = [1,4,6,7,8,20], costs = [2,7,15]
<strong>Output:</strong> 11
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
<strong>Output:</strong> 17
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= days.length &lt;= 365</code></li>
	<li><code>1 &lt;= days[i] &lt;= 365</code></li>
	<li><code>days</code> is in strictly increasing order.</li>
	<li><code>costs.length == 3</code></li>
	<li><code>1 &lt;= costs[i] &lt;= 1000</code></li>
</ul>

class Solution {
    vector<int> costs, days;
    map<int, int> mp;
public:
    int f(int n){
        if(n<=0) return 0;
        int ans = INT_MAX;
        if(mp[n] !=-1){
            cout<<n<<" "<<mp[n]<<endl;
            return mp[n];
        }
        if(!binary_search(days.begin(), days.end(), n)){
            ans = f(n-1);
        }
        ans = min({ans, f(n-1)+costs[0], f(n-7)+costs[1], f(n-30)+costs[2]});
        return mp[n] = ans;
    }
    int mincostTickets(vector<int>& days, vector<int>& costs) {
        this->costs = costs, this->days = days;
        for(int i = 1; i<=365; ++i){
            mp[i] = -1;
        }
        return f(365);
    }
};

// Initial approch:
// Brute force: There are 3 ways always. So what can we do for each days we can try all the 3 approch and move to next day. so the over all coplecity will be O(3^n)
// Lets write this solution in recurrance
// if(nth day not in list) f(n) = f(n-1)
//      f(n) = min(cost to travell for all the n days)
// else
//      f(n) = INF
// f(n) = min(f(n), f(n-7)+cost7, f(n-1)+cost1, f(n-30)+cost30)
// f(i) = 0 if(i<=0)          base case



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