Minimum Number of Coins for Fruits II
<p>You are at a fruit market with different types of exotic fruits on display.</p>

<p>You are given a <strong>1-indexed</strong> array <code>prices</code>, where <code>prices[i]</code> denotes the number of coins needed to purchase the <code>i<sup>th</sup></code> fruit.</p>

<p>The fruit market has the following offer:</p>

<ul>
	<li>If you purchase the <code>i<sup>th</sup></code> fruit at <code>prices[i]</code> coins, you can get the next <code>i</code> fruits for free.</li>
</ul>

<p><strong>Note</strong> that even if you <strong>can</strong> take fruit <code>j</code> for free, you can still purchase it for <code>prices[j]</code> coins to receive a new offer.</p>

<p>Return <em>the <strong>minimum</strong> number of coins needed to acquire all the fruits</em>.</p>

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

<pre>
<strong>Input:</strong> prices = [3,1,2]
<strong>Output:</strong> 4
<strong>Explanation:</strong> You can acquire the fruits as follows:
- Purchase the 1<sup>st</sup> fruit with 3 coins, and you are allowed to take the 2<sup>nd</sup> fruit for free.
- Purchase the 2<sup>nd</sup> fruit with 1 coin, and you are allowed to take the 3<sup>rd</sup> fruit for free.
- Take the 3<sup>rd</sup> fruit for free.
Note that even though you were allowed to take the 2<sup>nd</sup> fruit for free, you purchased it because it is more optimal.
It can be proven that 4 is the minimum number of coins needed to acquire all the fruits.
</pre>

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

<pre>
<strong>Input:</strong> prices = [1,10,1,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> You can acquire the fruits as follows:
- Purchase the 1<sup>st</sup> fruit with 1 coin, and you are allowed to take the 2<sup>nd</sup> fruit for free.
- Take the 2<sup>nd</sup> fruit for free.
- Purchase the 3<sup>rd</sup> fruit for 1 coin, and you are allowed to take the 4<sup>th</sup> fruit for free.
- Take the 4<sup>t</sup><sup>h</sup> fruit for free.
It can be proven that 2 is the minimum number of coins needed to acquire all the fruits.
</pre>

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

<ul>
	<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= prices[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Queue; Array; Dynamic Programming; Monotonic Queue; Heap (Priority Queue)
func minimumCoins(prices []int) int {
	n := len(prices)
	q := Deque{}
	for i := n; i > 0; i-- {
		for q.Size() > 0 && q.Front() > i*2+1 {
			q.PopFront()
		}
		if i <= (n-1)/2 {
			prices[i-1] += prices[q.Front()-1]
		}
		for q.Size() > 0 && prices[q.Back()-1] >= prices[i-1] {
			q.PopBack()
		}
		q.PushBack(i)
	}
	return prices[0]
}

// template
type Deque struct{ l, r []int }

func (q Deque) Empty() bool {
	return len(q.l) == 0 && len(q.r) == 0
}

func (q Deque) Size() int {
	return len(q.l) + len(q.r)
}

func (q *Deque) PushFront(v int) {
	q.l = append(q.l, v)
}

func (q *Deque) PushBack(v int) {
	q.r = append(q.r, v)
}

func (q *Deque) PopFront() (v int) {
	if len(q.l) > 0 {
		q.l, v = q.l[:len(q.l)-1], q.l[len(q.l)-1]
	} else {
		v, q.r = q.r[0], q.r[1:]
	}
	return
}

func (q *Deque) PopBack() (v int) {
	if len(q.r) > 0 {
		q.r, v = q.r[:len(q.r)-1], q.r[len(q.r)-1]
	} else {
		v, q.l = q.l[0], q.l[1:]
	}
	return
}

func (q Deque) Front() int {
	if len(q.l) > 0 {
		return q.l[len(q.l)-1]
	}
	return q.r[0]
}

func (q Deque) Back() int {
	if len(q.r) > 0 {
		return q.r[len(q.r)-1]
	}
	return q.l[0]
}

func (q Deque) Get(i int) int {
	if i < len(q.l) {
		return q.l[len(q.l)-1-i]
	}
	return q.r[i-len(q.l)]
}