| comments | difficulty | edit_url | tags | |||||
|---|---|---|---|---|---|---|---|---|
true |
Hard |
|
You are at a fruit market with different types of exotic fruits on display.
You are given a 1-indexed array prices, where prices[i] denotes the number of coins needed to purchase the ith fruit.
The fruit market has the following offer:
- If you purchase the
ithfruit atprices[i]coins, you can get the nextifruits for free.
Note that even if you can take fruit j for free, you can still purchase it for prices[j] coins to receive a new offer.
Return the minimum number of coins needed to acquire all the fruits.
Example 1:
Input: prices = [3,1,2] Output: 4 Explanation: You can acquire the fruits as follows: - Purchase the 1st fruit with 3 coins, and you are allowed to take the 2nd fruit for free. - Purchase the 2nd fruit with 1 coin, and you are allowed to take the 3rd fruit for free. - Take the 3rd fruit for free. Note that even though you were allowed to take the 2nd 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.
Example 2:
Input: prices = [1,10,1,1] Output: 2 Explanation: You can acquire the fruits as follows: - Purchase the 1st fruit with 1 coin, and you are allowed to take the 2nd fruit for free. - Take the 2nd fruit for free. - Purchase the 3rd fruit for 1 coin, and you are allowed to take the 4th fruit for free. - Take the 4th fruit for free. It can be proven that 2 is the minimum number of coins needed to acquire all the fruits.
Constraints:
1 <= prices.length <= 1051 <= prices[i] <= 105
We define
The state transition equation is
In implementation, we calculate from back to front, and we can directly perform state transition on the array
The time complexity of the above method is
We observe the state transition equation and find that for each
We calculate from back to front, maintain a monotonically increasing queue
The time complexity is
class Solution:
def minimumCoins(self, prices: List[int]) -> int:
n = len(prices)
q = deque()
for i in range(n, 0, -1):
while q and q[0] > i * 2 + 1:
q.popleft()
if i <= (n - 1) // 2:
prices[i - 1] += prices[q[0] - 1]
while q and prices[q[-1] - 1] >= prices[i - 1]:
q.pop()
q.append(i)
return prices[0]class Solution {
public int minimumCoins(int[] prices) {
int n = prices.length;
Deque<Integer> q = new ArrayDeque<>();
for (int i = n; i > 0; --i) {
while (!q.isEmpty() && q.peek() > i * 2 + 1) {
q.poll();
}
if (i <= (n - 1) / 2) {
prices[i - 1] += prices[q.peek() - 1];
}
while (!q.isEmpty() && prices[q.peekLast() - 1] >= prices[i - 1]) {
q.pollLast();
}
q.offer(i);
}
return prices[0];
}
}class Solution {
public:
int minimumCoins(vector<int>& prices) {
int n = prices.size();
deque<int> q;
for (int i = n; i; --i) {
while (q.size() && q.front() > i * 2 + 1) {
q.pop_front();
}
if (i <= (n - 1) / 2) {
prices[i - 1] += prices[q.front() - 1];
}
while (q.size() && prices[q.back() - 1] >= prices[i - 1]) {
q.pop_back();
}
q.push_back(i);
}
return prices[0];
}
};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)]
}function minimumCoins(prices: number[]): number {
const n = prices.length;
const q = new Deque<number>();
for (let i = n; i; --i) {
while (q.getSize() && q.frontValue()! > i * 2 + 1) {
q.popFront();
}
if (i <= (n - 1) >> 1) {
prices[i - 1] += prices[q.frontValue()! - 1];
}
while (q.getSize() && prices[q.backValue()! - 1] >= prices[i - 1]) {
q.popBack();
}
q.pushBack(i);
}
return prices[0];
}
class Node<T> {
value: T;
next: Node<T> | null;
prev: Node<T> | null;
constructor(value: T) {
this.value = value;
this.next = null;
this.prev = null;
}
}
class Deque<T> {
private front: Node<T> | null;
private back: Node<T> | null;
private size: number;
constructor() {
this.front = null;
this.back = null;
this.size = 0;
}
pushFront(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.next = this.front;
this.front!.prev = newNode;
this.front = newNode;
}
this.size++;
}
pushBack(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.prev = this.back;
this.back!.next = newNode;
this.back = newNode;
}
this.size++;
}
popFront(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.front!.value;
this.front = this.front!.next;
if (this.front !== null) {
this.front.prev = null;
} else {
this.back = null;
}
this.size--;
return value;
}
popBack(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.back!.value;
this.back = this.back!.prev;
if (this.back !== null) {
this.back.next = null;
} else {
this.front = null;
}
this.size--;
return value;
}
frontValue(): T | undefined {
return this.front?.value;
}
backValue(): T | undefined {
return this.back?.value;
}
getSize(): number {
return this.size;
}
isEmpty(): boolean {
return this.size === 0;
}
}