3013. Divide an Array Into Subarrays With Minimum Cost II

Description

You are given a 0-indexed array of integers nums of length n, and two positive integers k and dist.

The cost of an array is the value of its first element. For example, the cost of [1,2,3] is 1 while the cost of [3,4,1] is 3.

You need to divide nums into k disjoint contiguous subarrays, such that the difference between the starting index of the second subarray and the starting index of the kth subarray should be less than or equal to dist. In other words, if you divide nums into the subarrays nums[0..(i1 - 1)], nums[i1..(i2 - 1)], ..., nums[ik-1..(n - 1)], then ik-1 - i1 <= dist.

Return the minimum possible sum of the cost of these subarrays.

Example 1:

Input: nums = [1,3,2,6,4,2], k = 3, dist = 3
Output: 5
Explanation: The best possible way to divide nums into 3 subarrays is: [1,3], [2,6,4], and [2]. This choice is valid because ik-1 - i1 is 5 - 2 = 3 which is equal to dist. The total cost is nums[0] + nums[2] + nums[5] which is 1 + 2 + 2 = 5.
It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 5.


Example 2:

Input: nums = [10,1,2,2,2,1], k = 4, dist = 3
Output: 15
Explanation: The best possible way to divide nums into 4 subarrays is: [10], [1], [2], and [2,2,1]. This choice is valid because ik-1 - i1 is 3 - 1 = 2 which is less than dist. The total cost is nums[0] + nums[1] + nums[2] + nums[3] which is 10 + 1 + 2 + 2 = 15.
The division [10], [1], [2,2,2], and [1] is not valid, because the difference between ik-1 and i1 is 5 - 1 = 4, which is greater than dist.
It can be shown that there is no possible way to divide nums into 4 subarrays at a cost lower than 15.


Example 3:

Input: nums = [10,8,18,9], k = 3, dist = 1
Output: 36
Explanation: The best possible way to divide nums into 4 subarrays is: [10], [8], and [18,9]. This choice is valid because ik-1 - i1 is 2 - 1 = 1 which is equal to dist.The total cost is nums[0] + nums[1] + nums[2] which is 10 + 8 + 18 = 36.
The division [10], [8,18], and [9] is not valid, because the difference between ik-1 and i1 is 3 - 1 = 2, which is greater than dist.
It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 36.


Constraints:

• 3 <= n <= 105
• 1 <= nums[i] <= 109
• 3 <= k <= n
• k - 2 <= dist <= n - 2

Solutions

Solution 1

• class Solution {
public long minimumCost(int[] nums, int k, int dist) {
long result = Long.MAX_VALUE, sum = 0L;
int n = nums.length;
TreeSet<Integer> set1
= new TreeSet<>((a, b) -> nums[a] == nums[b] ? a - b : nums[a] - nums[b]);
TreeSet<Integer> set2
= new TreeSet<>((a, b) -> nums[a] == nums[b] ? a - b : nums[a] - nums[b]);
for (int i = 1; i < n; i++) {
sum += nums[i];
if (set1.size() >= k) {
int x = set1.pollLast();
sum -= nums[x];
}
if (i - dist > 0) {
result = Math.min(result, sum);
int temp = i - dist;
if (set1.contains(temp)) {
set1.remove(temp);
sum -= nums[temp];
if (set2.size() > 0) {
int y = set2.pollFirst();
sum += nums[y];
}
} else {
set2.remove(i - dist);
}
}
}
return result + nums[0];
}
}


• class Solution {
public:
long long minimumCost(vector<int>& nums, int k, int dist) {
multiset<int> sml, big;
int sz = dist + 1;
long long sum = 0, ans = 0;
for (int i = 1; i <= sz; i++) {
sml.insert(nums[i]);
sum += nums[i];
}
while (sml.size() > k - 1) {
big.insert(*sml.rbegin());
sum -= *sml.rbegin();
sml.erase(sml.find(*sml.rbegin()));
}
ans = sum;
for (int i = sz + 1; i < nums.size(); i++) {
sum += nums[i];
sml.insert(nums[i]);
if (big.find(nums[i - sz]) != big.end()) {
big.erase(big.find(nums[i - sz]));
} else {
sum -= nums[i - sz];
sml.erase(sml.find(nums[i - sz]));
}

while (sml.size() > k - 1) {
sum -= *sml.rbegin();
big.insert(*sml.rbegin());
sml.erase(sml.find(*sml.rbegin()));
}
while (sml.size() < k - 1) {
sum += *big.begin();
sml.insert(*big.begin());
big.erase(big.begin());
}
while (!sml.empty() && !big.empty() && *sml.rbegin() > *big.begin()) {
sum -= *sml.rbegin() - *big.begin();
sml.insert(*big.begin());
big.insert(*sml.rbegin());
sml.erase(sml.find(*sml.rbegin()));
big.erase(big.begin());
}
ans = min(ans, sum);
}
int p = 0;
return nums[0] + ans;
}
};


• from sortedcontainers import SortedList

class Solution:
def minimumCost(self, nums: List[int], k: int, dist: int) -> int:
n = len(nums)

sl = SortedList()
y = nums[0]
ans = float("inf")
i = 1
running_sum = 0

for j in range(1, n):
pos = bisect.bisect_left(sl, nums[j])

if pos < k - 1:
running_sum += nums[j]
if len(sl) > k - 1:
running_sum -= sl[k - 1]

while j - i > dist:
removed_pos = sl.index(nums[i])
removed_element = nums[i]
sl.remove(removed_element)

if removed_pos < k - 1:
running_sum -= removed_element
if len(sl) >= k - 1:
running_sum += sl[k - 2]
i += 1

if j - i + 1 >= k - 1:
ans = min(ans, running_sum)

return ans + y


• func minimumCost(nums []int, k int, dist int) int64 {
res := nums[0] + slices.Min(windowTopKSum(nums[1:], dist+1, k-1, true))
return int64(res)
}

func windowTopKSum(nums []int, windowSize, k int, min bool) []int {
n := len(nums)
ts := NewTopKSum(k, min)
res := []int{}
for right := 0; right < n; right++ {
if right >= windowSize {
}
if right >= windowSize-1 {
res = append(res, ts.Query())
}
}
return res
}

type TopKSum struct {
sum     int
k       int
in      *Heap
out     *Heap
dIn     *Heap
dOut    *Heap
counter map[int]int
}

func NewTopKSum(k int, min bool) *TopKSum {
var less func(a, b int) bool
if min {
less = func(a, b int) bool { return a < b }
} else {
less = func(a, b int) bool { return a > b }
}
return &TopKSum{
k:       k,
in:      NewHeap(less),
out:     NewHeap(less),
dIn:     NewHeap(less),
dOut:    NewHeap(less),
counter: map[int]int{},
}
}

func (t *TopKSum) Query() int {
return t.sum
}

func (t *TopKSum) Add(x int) {
t.counter[x]++
t.in.Push(-x)
t.sum += x
t.modify()
}

func (t *TopKSum) Discard(x int) bool {
if t.counter[x] == 0 {
return false
}
t.counter[x]--
if t.in.Len() > 0 && -t.in.Top() == x {
t.sum -= x
t.in.Pop()
} else if t.in.Len() > 0 && -t.in.Top() > x {
t.sum -= x
t.dIn.Push(-x)
} else {
t.dOut.Push(x)
}
t.modify()
return true
}

func (t *TopKSum) SetK(k int) {
t.k = k
t.modify()
}

func (t *TopKSum) GetK() int {
return t.k
}

func (t *TopKSum) Len() int {
return t.in.Len() + t.out.Len() - t.dIn.Len() - t.dOut.Len()
}

func (t *TopKSum) Has(x int) bool {
return t.counter[x] > 0
}

func (t *TopKSum) modify() {
for t.out.Len() > 0 && (t.in.Len()-t.dIn.Len() < t.k) {
p := t.out.Pop()
if t.dOut.Len() > 0 && p == t.dOut.Top() {
t.dOut.Pop()
} else {
t.sum += p
t.in.Push(-p)
}
}

for t.in.Len()-t.dIn.Len() > t.k {
p := -t.in.Pop()
if t.dIn.Len() > 0 && p == -t.dIn.Top() {
t.dIn.Pop()
} else {
t.sum -= p
t.out.Push(p)
}
}

for t.dIn.Len() > 0 && t.in.Top() == t.dIn.Top() {
t.in.Pop()
t.dIn.Pop()
}
}

type H = int

func NewHeap(less func(a, b H) bool, nums ...H) *Heap {
nums = append(nums[:0:0], nums...)
heap := &Heap{less: less, data: nums}
heap.heapify()
return heap
}

type Heap struct {
data []H
less func(a, b H) bool
}

func (h *Heap) Push(value H) {
h.data = append(h.data, value)
h.pushUp(h.Len() - 1)
}

func (h *Heap) Pop() (value H) {
if h.Len() == 0 {
panic("heap is empty")
}

value = h.data[0]
h.data[0] = h.data[h.Len()-1]
h.data = h.data[:h.Len()-1]
h.pushDown(0)
return
}

func (h *Heap) Top() (value H) {
value = h.data[0]
return
}

func (h *Heap) Len() int { return len(h.data) }

func (h *Heap) heapify() {
n := h.Len()
for i := (n >> 1) - 1; i > -1; i-- {
h.pushDown(i)
}
}

func (h *Heap) pushUp(root int) {
for parent := (root - 1) >> 1; parent >= 0 && h.less(h.data[root], h.data[parent]); parent = (root - 1) >> 1 {
h.data[root], h.data[parent] = h.data[parent], h.data[root]
root = parent
}
}

func (h *Heap) pushDown(root int) {
n := h.Len()
for left := (root<<1 + 1); left < n; left = (root<<1 + 1) {
right := left + 1
minIndex := root

if h.less(h.data[left], h.data[minIndex]) {
minIndex = left
}

if right < n && h.less(h.data[right], h.data[minIndex]) {
minIndex = right
}

if minIndex == root {
return
}
h.data[root], h.data[minIndex] = h.data[minIndex], h.data[root]
root = minIndex
}
}