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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 <= 1051 <= nums[i] <= 1093 <= k <= nk - 2 <= dist <= n - 2
Solutions
Solution 1
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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++) { set1.add(i); sum += nums[i]; if (set1.size() >= k) { int x = set1.pollLast(); sum -= nums[x]; set2.add(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]; set1.add(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]) sl.add(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++ { ts.Add(nums[right]) if right >= windowSize { ts.Discard(nums[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 } } -
function minimumCost(nums: number[], k: number, dist: number): number { --k; const l = new TreapMultiSet<number>((a, b) => a - b); const r = new TreapMultiSet<number>((a, b) => a - b); let s = nums[0]; for (let i = 1; i < dist + 2; ++i) { s += nums[i]; l.add(nums[i]); } const l2r = () => { const x = l.pop()!; s -= x; r.add(x); }; const r2l = () => { const x = r.shift()!; l.add(x); s += x; }; while (l.size > k) { l2r(); } let ans = s; for (let i = dist + 2; i < nums.length; ++i) { const x = nums[i - dist - 1]; if (l.has(x)) { l.delete(x); s -= x; } else { r.delete(x); } const y = nums[i]; if (y < l.last()!) { l.add(y); s += y; } else { r.add(y); } while (l.size < k) { r2l(); } while (l.size > k) { l2r(); } ans = Math.min(ans, s); } return ans; } type CompareFunction<T, R extends 'number' | 'boolean'> = ( a: T, b: T, ) => R extends 'number' ? number : boolean; interface ITreapMultiSet<T> extends Iterable<T> { add: (...value: T[]) => this; has: (value: T) => boolean; delete: (value: T) => void; bisectLeft: (value: T) => number; bisectRight: (value: T) => number; indexOf: (value: T) => number; lastIndexOf: (value: T) => number; at: (index: number) => T | undefined; first: () => T | undefined; last: () => T | undefined; lower: (value: T) => T | undefined; higher: (value: T) => T | undefined; floor: (value: T) => T | undefined; ceil: (value: T) => T | undefined; shift: () => T | undefined; pop: (index?: number) => T | undefined; count: (value: T) => number; keys: () => IterableIterator<T>; values: () => IterableIterator<T>; rvalues: () => IterableIterator<T>; entries: () => IterableIterator<[number, T]>; readonly size: number; } class TreapNode<T = number> { value: T; count: number; size: number; priority: number; left: TreapNode<T> | null; right: TreapNode<T> | null; constructor(value: T) { this.value = value; this.count = 1; this.size = 1; this.priority = Math.random(); this.left = null; this.right = null; } static getSize(node: TreapNode<any> | null): number { return node?.size ?? 0; } static getFac(node: TreapNode<any> | null): number { return node?.priority ?? 0; } pushUp(): void { let tmp = this.count; tmp += TreapNode.getSize(this.left); tmp += TreapNode.getSize(this.right); this.size = tmp; } rotateRight(): TreapNode<T> { // eslint-disable-next-line @typescript-eslint/no-this-alias let node: TreapNode<T> = this; const left = node.left; node.left = left?.right ?? null; left && (left.right = node); left && (node = left); node.right?.pushUp(); node.pushUp(); return node; } rotateLeft(): TreapNode<T> { // eslint-disable-next-line @typescript-eslint/no-this-alias let node: TreapNode<T> = this; const right = node.right; node.right = right?.left ?? null; right && (right.left = node); right && (node = right); node.left?.pushUp(); node.pushUp(); return node; } } class TreapMultiSet<T = number> implements ITreapMultiSet<T> { private readonly root: TreapNode<T>; private readonly compareFn: CompareFunction<T, 'number'>; private readonly leftBound: T; private readonly rightBound: T; constructor(compareFn?: CompareFunction<T, 'number'>); constructor(compareFn: CompareFunction<T, 'number'>, leftBound: T, rightBound: T); constructor( compareFn: CompareFunction<T, any> = (a: any, b: any) => a - b, leftBound: any = -Infinity, rightBound: any = Infinity, ) { this.root = new TreapNode<T>(rightBound); this.root.priority = Infinity; this.root.left = new TreapNode<T>(leftBound); this.root.left.priority = -Infinity; this.root.pushUp(); this.leftBound = leftBound; this.rightBound = rightBound; this.compareFn = compareFn; } get size(): number { return this.root.size - 2; } get height(): number { const getHeight = (node: TreapNode<T> | null): number => { if (node == null) return 0; return 1 + Math.max(getHeight(node.left), getHeight(node.right)); }; return getHeight(this.root); } /** * * @complexity `O(logn)` * @description Returns true if value is a member. */ has(value: T): boolean { const compare = this.compareFn; const dfs = (node: TreapNode<T> | null, value: T): boolean => { if (node == null) return false; if (compare(node.value, value) === 0) return true; if (compare(node.value, value) < 0) return dfs(node.right, value); return dfs(node.left, value); }; return dfs(this.root, value); } /** * * @complexity `O(logn)` * @description Add value to sorted set. */ add(...values: T[]): this { const compare = this.compareFn; const dfs = ( node: TreapNode<T> | null, value: T, parent: TreapNode<T>, direction: 'left' | 'right', ): void => { if (node == null) return; if (compare(node.value, value) === 0) { node.count++; node.pushUp(); } else if (compare(node.value, value) > 0) { if (node.left) { dfs(node.left, value, node, 'left'); } else { node.left = new TreapNode(value); node.pushUp(); } if (TreapNode.getFac(node.left) > node.priority) { parent[direction] = node.rotateRight(); } } else if (compare(node.value, value) < 0) { if (node.right) { dfs(node.right, value, node, 'right'); } else { node.right = new TreapNode(value); node.pushUp(); } if (TreapNode.getFac(node.right) > node.priority) { parent[direction] = node.rotateLeft(); } } parent.pushUp(); }; values.forEach(value => dfs(this.root.left, value, this.root, 'left')); return this; } /** * * @complexity `O(logn)` * @description Remove value from sorted set if it is a member. * If value is not a member, do nothing. */ delete(value: T): void { const compare = this.compareFn; const dfs = ( node: TreapNode<T> | null, value: T, parent: TreapNode<T>, direction: 'left' | 'right', ): void => { if (node == null) return; if (compare(node.value, value) === 0) { if (node.count > 1) { node.count--; node?.pushUp(); } else if (node.left == null && node.right == null) { parent[direction] = null; } else { // 旋到根节点 if ( node.right == null || TreapNode.getFac(node.left) > TreapNode.getFac(node.right) ) { parent[direction] = node.rotateRight(); dfs(parent[direction]?.right ?? null, value, parent[direction]!, 'right'); } else { parent[direction] = node.rotateLeft(); dfs(parent[direction]?.left ?? null, value, parent[direction]!, 'left'); } } } else if (compare(node.value, value) > 0) { dfs(node.left, value, node, 'left'); } else if (compare(node.value, value) < 0) { dfs(node.right, value, node, 'right'); } parent?.pushUp(); }; dfs(this.root.left, value, this.root, 'left'); } /** * * @complexity `O(logn)` * @description Returns an index to insert value in the sorted set. * If the value is already present, the insertion point will be before (to the left of) any existing values. */ bisectLeft(value: T): number { const compare = this.compareFn; const dfs = (node: TreapNode<T> | null, value: T): number => { if (node == null) return 0; if (compare(node.value, value) === 0) { return TreapNode.getSize(node.left); } else if (compare(node.value, value) > 0) { return dfs(node.left, value); } else if (compare(node.value, value) < 0) { return dfs(node.right, value) + TreapNode.getSize(node.left) + node.count; } return 0; }; return dfs(this.root, value) - 1; } /** * * @complexity `O(logn)` * @description Returns an index to insert value in the sorted set. * If the value is already present, the insertion point will be before (to the right of) any existing values. */ bisectRight(value: T): number { const compare = this.compareFn; const dfs = (node: TreapNode<T> | null, value: T): number => { if (node == null) return 0; if (compare(node.value, value) === 0) { return TreapNode.getSize(node.left) + node.count; } else if (compare(node.value, value) > 0) { return dfs(node.left, value); } else if (compare(node.value, value) < 0) { return dfs(node.right, value) + TreapNode.getSize(node.left) + node.count; } return 0; }; return dfs(this.root, value) - 1; } /** * * @complexity `O(logn)` * @description Returns the index of the first occurrence of a value in the set, or -1 if it is not present. */ indexOf(value: T): number { const compare = this.compareFn; let isExist = false; const dfs = (node: TreapNode<T> | null, value: T): number => { if (node == null) return 0; if (compare(node.value, value) === 0) { isExist = true; return TreapNode.getSize(node.left); } else if (compare(node.value, value) > 0) { return dfs(node.left, value); } else if (compare(node.value, value) < 0) { return dfs(node.right, value) + TreapNode.getSize(node.left) + node.count; } return 0; }; const res = dfs(this.root, value) - 1; return isExist ? res : -1; } /** * * @complexity `O(logn)` * @description Returns the index of the last occurrence of a value in the set, or -1 if it is not present. */ lastIndexOf(value: T): number { const compare = this.compareFn; let isExist = false; const dfs = (node: TreapNode<T> | null, value: T): number => { if (node == null) return 0; if (compare(node.value, value) === 0) { isExist = true; return TreapNode.getSize(node.left) + node.count - 1; } else if (compare(node.value, value) > 0) { return dfs(node.left, value); } else if (compare(node.value, value) < 0) { return dfs(node.right, value) + TreapNode.getSize(node.left) + node.count; } return 0; }; const res = dfs(this.root, value) - 1; return isExist ? res : -1; } /** * * @complexity `O(logn)` * @description Returns the item located at the specified index. * @param index The zero-based index of the desired code unit. A negative index will count back from the last item. */ at(index: number): T | undefined { if (index < 0) index += this.size; if (index < 0 || index >= this.size) return undefined; const dfs = (node: TreapNode<T> | null, rank: number): T | undefined => { if (node == null) return undefined; if (TreapNode.getSize(node.left) >= rank) { return dfs(node.left, rank); } else if (TreapNode.getSize(node.left) + node.count >= rank) { return node.value; } else { return dfs(node.right, rank - TreapNode.getSize(node.left) - node.count); } }; const res = dfs(this.root, index + 2); return ([this.leftBound, this.rightBound] as any[]).includes(res) ? undefined : res; } /** * * @complexity `O(logn)` * @description Find and return the element less than `val`, return `undefined` if no such element found. */ lower(value: T): T | undefined { const compare = this.compareFn; const dfs = (node: TreapNode<T> | null, value: T): T | undefined => { if (node == null) return undefined; if (compare(node.value, value) >= 0) return dfs(node.left, value); const tmp = dfs(node.right, value); if (tmp == null || compare(node.value, tmp) > 0) { return node.value; } else { return tmp; } }; const res = dfs(this.root, value) as any; return res === this.leftBound ? undefined : res; } /** * * @complexity `O(logn)` * @description Find and return the element greater than `val`, return `undefined` if no such element found. */ higher(value: T): T | undefined { const compare = this.compareFn; const dfs = (node: TreapNode<T> | null, value: T): T | undefined => { if (node == null) return undefined; if (compare(node.value, value) <= 0) return dfs(node.right, value); const tmp = dfs(node.left, value); if (tmp == null || compare(node.value, tmp) < 0) { return node.value; } else { return tmp; } }; const res = dfs(this.root, value) as any; return res === this.rightBound ? undefined : res; } /** * * @complexity `O(logn)` * @description Find and return the element less than or equal to `val`, return `undefined` if no such element found. */ floor(value: T): T | undefined { const compare = this.compareFn; const dfs = (node: TreapNode<T> | null, value: T): T | undefined => { if (node == null) return undefined; if (compare(node.value, value) === 0) return node.value; if (compare(node.value, value) >= 0) return dfs(node.left, value); const tmp = dfs(node.right, value); if (tmp == null || compare(node.value, tmp) > 0) { return node.value; } else { return tmp; } }; const res = dfs(this.root, value) as any; return res === this.leftBound ? undefined : res; } /** * * @complexity `O(logn)` * @description Find and return the element greater than or equal to `val`, return `undefined` if no such element found. */ ceil(value: T): T | undefined { const compare = this.compareFn; const dfs = (node: TreapNode<T> | null, value: T): T | undefined => { if (node == null) return undefined; if (compare(node.value, value) === 0) return node.value; if (compare(node.value, value) <= 0) return dfs(node.right, value); const tmp = dfs(node.left, value); if (tmp == null || compare(node.value, tmp) < 0) { return node.value; } else { return tmp; } }; const res = dfs(this.root, value) as any; return res === this.rightBound ? undefined : res; } /** * @complexity `O(logn)` * @description * Returns the last element from set. * If the set is empty, undefined is returned. */ first(): T | undefined { const iter = this.inOrder(); iter.next(); const res = iter.next().value; return res === this.rightBound ? undefined : res; } /** * @complexity `O(logn)` * @description * Returns the last element from set. * If the set is empty, undefined is returned . */ last(): T | undefined { const iter = this.reverseInOrder(); iter.next(); const res = iter.next().value; return res === this.leftBound ? undefined : res; } /** * @complexity `O(logn)` * @description * Removes the first element from an set and returns it. * If the set is empty, undefined is returned and the set is not modified. */ shift(): T | undefined { const first = this.first(); if (first === undefined) return undefined; this.delete(first); return first; } /** * @complexity `O(logn)` * @description * Removes the last element from an set and returns it. * If the set is empty, undefined is returned and the set is not modified. */ pop(index?: number): T | undefined { if (index == null) { const last = this.last(); if (last === undefined) return undefined; this.delete(last); return last; } const toDelete = this.at(index); if (toDelete == null) return; this.delete(toDelete); return toDelete; } /** * * @complexity `O(logn)` * @description * Returns number of occurrences of value in the sorted set. */ count(value: T): number { const compare = this.compareFn; const dfs = (node: TreapNode<T> | null, value: T): number => { if (node == null) return 0; if (compare(node.value, value) === 0) return node.count; if (compare(node.value, value) < 0) return dfs(node.right, value); return dfs(node.left, value); }; return dfs(this.root, value); } *[Symbol.iterator](): Generator<T, any, any> { yield* this.values(); } /** * @description * Returns an iterable of keys in the set. */ *keys(): Generator<T, any, any> { yield* this.values(); } /** * @description * Returns an iterable of values in the set. */ *values(): Generator<T, any, any> { const iter = this.inOrder(); iter.next(); const steps = this.size; for (let _ = 0; _ < steps; _++) { yield iter.next().value; } } /** * @description * Returns a generator for reversed order traversing the set. */ *rvalues(): Generator<T, any, any> { const iter = this.reverseInOrder(); iter.next(); const steps = this.size; for (let _ = 0; _ < steps; _++) { yield iter.next().value; } } /** * @description * Returns an iterable of key, value pairs for every entry in the set. */ *entries(): IterableIterator<[number, T]> { const iter = this.inOrder(); iter.next(); const steps = this.size; for (let i = 0; i < steps; i++) { yield [i, iter.next().value]; } } private *inOrder(root: TreapNode<T> | null = this.root): Generator<T, any, any> { if (root == null) return; yield* this.inOrder(root.left); const count = root.count; for (let _ = 0; _ < count; _++) { yield root.value; } yield* this.inOrder(root.right); } private *reverseInOrder(root: TreapNode<T> | null = this.root): Generator<T, any, any> { if (root == null) return; yield* this.reverseInOrder(root.right); const count = root.count; for (let _ = 0; _ < count; _++) { yield root.value; } yield* this.reverseInOrder(root.left); } }