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3578. Count Partitions With Max-Min Difference at Most K

Description

You are given an integer array nums and an integer k. Your task is to partition nums into one or more non-empty contiguous segments such that in each segment, the difference between its maximum and minimum elements is at most k.

Return the total number of ways to partition nums under this condition.

Since the answer may be too large, return it modulo 109 + 7.

 

Example 1:

Input: nums = [9,4,1,3,7], k = 4

Output: 6

Explanation:

There are 6 valid partitions where the difference between the maximum and minimum elements in each segment is at most k = 4:

  • [[9], [4], [1], [3], [7]]
  • [[9], [4], [1], [3, 7]]
  • [[9], [4], [1, 3], [7]]
  • [[9], [4, 1], [3], [7]]
  • [[9], [4, 1], [3, 7]]
  • [[9], [4, 1, 3], [7]]

Example 2:

Input: nums = [3,3,4], k = 0

Output: 2

Explanation:

There are 2 valid partitions that satisfy the given conditions:

  • [[3], [3], [4]]
  • [[3, 3], [4]]

 

Constraints:

  • 2 <= nums.length <= 5 * 104
  • 1 <= nums[i] <= 109
  • 0 <= k <= 109

Solutions

Solution 1: Dynamic Programming + Two Pointers + Ordered Set

We define $f[i]$ as the number of ways to partition the first $i$ elements. If an array satisfies that the difference between its maximum and minimum values does not exceed $k$, then any of its subarrays also satisfies this condition. Therefore, we can use two pointers to maintain a sliding window representing the current subarray.

When we reach the $r$-th element, we need to find the left pointer $l$ such that the subarray from $l$ to $r$ satisfies that the difference between the maximum and minimum values does not exceed $k$. We can use an ordered set to maintain the elements in the current window, so that we can quickly get the maximum and minimum values.

Each time we add a new element, we insert it into the ordered set and check the difference between the maximum and minimum values in the current window. If it exceeds $k$, we move the left pointer $l$ until the condition is satisfied. The number of partition ways ending at the $r$-th element is $f[l - 1] + f[l] + \ldots + f[r - 1]$. We can use a prefix sum array to quickly calculate this value.

The answer is $f[n]$, where $n$ is the length of the array.

The time complexity is $O(n \times \log n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$.

  • class Solution {
    public int countPartitions(int[] nums, int k) {
        final int mod = (int) 1e9 + 7;
        TreeMap<Integer, Integer> sl = new TreeMap<>();
        int n = nums.length;
        int[] f = new int[n + 1];
        int[] g = new int[n + 1];
        f[0] = 1;
        g[0] = 1;
        int l = 1;
        for (int r = 1; r <= n; r++) {
            int x = nums[r - 1];
            sl.merge(x, 1, Integer::sum);
            while (sl.lastKey() - sl.firstKey() > k) {
                if (sl.merge(nums[l - 1], -1, Integer::sum) == 0) {
                    sl.remove(nums[l - 1]);
                }
                ++l;
            }
            f[r] = (g[r - 1] - (l >= 2 ? g[l - 2] : 0) + mod) % mod;
            g[r] = (g[r - 1] + f[r]) % mod;
        }
        return f[n];
    }
}


  • class Solution {
public:
    int countPartitions(vector<int>& nums, int k) {
        const int mod = 1e9 + 7;
        multiset<int> sl;
        int n = nums.size();
        vector<int> f(n + 1, 0), g(n + 1, 0);
        f[0] = 1;
        g[0] = 1;
        int l = 1;
        for (int r = 1; r <= n; ++r) {
            int x = nums[r - 1];
            sl.insert(x);
            while (*sl.rbegin() - *sl.begin() > k) {
                sl.erase(sl.find(nums[l - 1]));
                ++l;
            }
            f[r] = (g[r - 1] - (l >= 2 ? g[l - 2] : 0) + mod) % mod;
            g[r] = (g[r - 1] + f[r]) % mod;
        }
        return f[n];
    }
};


  • class Solution:
    def countPartitions(self, nums: List[int], k: int) -> int:
        mod = 10**9 + 7
        sl = SortedList()
        n = len(nums)
        f = [1] + [0] * n
        g = [1] + [0] * n
        l = 1
        for r, x in enumerate(nums, 1):
            sl.add(x)
            while sl[-1] - sl[0] > k:
                sl.remove(nums[l - 1])
                l += 1
            f[r] = (g[r - 1] - (g[l - 2] if l >= 2 else 0) + mod) % mod
            g[r] = (g[r - 1] + f[r]) % mod
        return f[n]


  • func countPartitions(nums []int, k int) int {
	const mod int = 1e9 + 7
	sl := redblacktree.New[int, int]()
	merge := func(st *redblacktree.Tree[int, int], x, v int) {
		c, _ := st.Get(x)
		if c+v == 0 {
			st.Remove(x)
		} else {
			st.Put(x, c+v)
		}
	}
	n := len(nums)
	f := make([]int, n+1)
	g := make([]int, n+1)
	f[0], g[0] = 1, 1
	for l, r := 1, 1; r <= n; r++ {
		merge(sl, nums[r-1], 1)
		for sl.Right().Key-sl.Left().Key > k {
			merge(sl, nums[l-1], -1)
			l++
		}
		f[r] = g[r-1]
		if l >= 2 {
			f[r] = (f[r] - g[l-2] + mod) % mod
		}
		g[r] = (g[r-1] + f[r]) % mod
	}
	return f[n]
}

  • function countPartitions(nums: number[], k: number): number {
    const mod = 10 ** 9 + 7;
    const n = nums.length;
    const sl = new TreapMultiSet<number>((a, b) => a - b);
    const f: number[] = Array(n + 1).fill(0);
    const g: number[] = Array(n + 1).fill(0);
    f[0] = 1;
    g[0] = 1;
    for (let l = 1, r = 1; r <= n; ++r) {
        const x = nums[r - 1];
        sl.add(x);
        while (sl.last()! - sl.first()! > k) {
            sl.delete(nums[l - 1]);
            l++;
        }
        f[r] = (g[r - 1] - (l >= 2 ? g[l - 2] : 0) + mod) % mod;
        g[r] = (g[r - 1] + f[r]) % mod;
    }
    return f[n];
}

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);
    }
}


  • use std::collections::BTreeMap;

impl Solution {
    pub fn count_partitions(nums: Vec<i32>, k: i32) -> i32 {
        const mod_val: i32 = 1_000_000_007;
        let n = nums.len();
        let mut f = vec![0; n + 1];
        let mut g = vec![0; n + 1];
        f[0] = 1;
        g[0] = 1;
        let mut sl = BTreeMap::new();
        let mut l = 1;
        for r in 1..=n {
            let x = nums[r - 1];
            *sl.entry(x).or_insert(0) += 1;
            while sl.keys().last().unwrap() - sl.keys().next().unwrap() > k {
                let val = nums[l - 1];
                if let Some(cnt) = sl.get_mut(&val) {
                    *cnt -= 1;
                    if *cnt == 0 {
                        sl.remove(&val);
                    }
                }
                l += 1;
            }
            f[r] = (g[r - 1] - if l >= 2 { g[l - 2] } else { 0 } + mod_val) % mod_val;
            g[r] = (g[r - 1] + f[r]) % mod_val;
        }
        f[n]
    }
}

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