<p>There is a special kind of apple tree that grows apples every day for <code>n</code> days. On the <code>i<sup>th</sup></code> day, the tree grows <code>apples[i]</code> apples that will rot after <code>days[i]</code> days, that is on day <code>i + days[i]</code> the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by <code>apples[i] == 0</code> and <code>days[i] == 0</code>.</p>

<p>You decided to eat <strong>at most</strong> one apple a day (to keep the doctors away). Note that you can keep eating after the first <code>n</code> days.</p>

<p>Given two integer arrays <code>days</code> and <code>apples</code> of length <code>n</code>, return <em>the maximum number of apples you can eat.</em></p>

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

<pre>
<strong>Input:</strong> apples = [1,2,3,5,2], days = [3,2,1,4,2]
<strong>Output:</strong> 7
<strong>Explanation:</strong> You can eat 7 apples:
- On the first day, you eat an apple that grew on the first day.
- On the second day, you eat an apple that grew on the second day.
- On the third day, you eat an apple that grew on the second day. After this day, the apples that grew on the third day rot.
- On the fourth to the seventh days, you eat apples that grew on the fourth day.
</pre>

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

<pre>
<strong>Input:</strong> apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
<strong>Output:</strong> 5
<strong>Explanation:</strong> You can eat 5 apples:
- On the first to the third day you eat apples that grew on the first day.
- Do nothing on the fouth and fifth days.
- On the sixth and seventh days you eat apples that grew on the sixth day.
</pre>

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

<ul>
	<li><code>n == apples.length == days.length</code></li>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= apples[i], days[i] &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>days[i] = 0</code> if and only if <code>apples[i] = 0</code>.</li>
</ul>

// Idea: Greedy with a PQ wrt the earliest expiration date.
// wrong idea: intervals +/-1 on timeline is not suitable for this problem
class Solution {
public:
    // Greedy with PQ or ordered map
    int eatenApples(vector<int>& apples, vector<int>& days) {
        const int n = days.size();
        // using Item = pair<int, int>; // expiration date, appleCount
        // priority_queue<Item, vector<Item>, greater<>> pq;
        map<int, int> m; // expiration date --> apple count
        int res = 0;
        for (int d = 0; d < n || !m.empty(); ++d) {
            if (d < n) {
                m[d + days[d]] += apples[d];
            }
            // remove the expired apples if any
            while (!m.empty() && m.begin()->first <= d) {
                m.erase(m.begin());
            }
            // eat an apple
            if (!m.empty()) {
                assert(m.begin()->second > 0);
                m.begin()->second--;
                ++res;
                if (m.begin()->second == 0) {
                    m.erase(m.begin());
                }
            }
        }
        return res;
    }

    int eatenApples_timeline_interval(vector<int>& apples, vector<int>& days) {
        const int n = days.size();
        int nd = 0;
        for (int i = 0; i < n; ++i) {
            nd = max(nd, i + days[i] + 1);
        }
        vector<int> line(nd);
        for (int i = 0; i < n; ++i) {
            line[i] += apples[i];
            line[i + days[i]] -= apples[i];
        }
        int eaten = 0;
        for (int i = 0, accum = 0; i < nd; ++i) {
            accum += line[i];
            accum = max(0, accum);
            if (accum > 0) {
                eaten++;
                accum--;
            }
        }
        return eaten;
    }
};
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