README_EN.md

1964. Find the Longest Valid Obstacle Course at Each Position

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Description

You want to build some obstacle courses. You are given a 0-indexed integer array obstacles of length n, where obstacles[i] describes the height of the ith obstacle.

For every index i between 0 and n - 1 (inclusive), find the length of the longest obstacle course in obstacles such that:

• You choose any number of obstacles between 0 and i inclusive.
• You must include the ith obstacle in the course.
• You must put the chosen obstacles in the same order as they appear in obstacles.
• Every obstacle (except the first) is taller than or the same height as the obstacle immediately before it.

Return an array ans of length n, where ans[i] is the length of the longest obstacle course for index i as described above.

 

Example 1:

Input: obstacles = [1,2,3,2]
Output: [1,2,3,3]
Explanation: The longest valid obstacle course at each position is:
- i = 0: [1], [1] has length 1.
- i = 1: [1,2], [1,2] has length 2.
- i = 2: [1,2,3], [1,2,3] has length 3.
- i = 3: [1,2,3,2], [1,2,2] has length 3.

Example 2:

Input: obstacles = [2,2,1]
Output: [1,2,1]
Explanation: The longest valid obstacle course at each position is:
- i = 0: [2], [2] has length 1.
- i = 1: [2,2], [2,2] has length 2.
- i = 2: [2,2,1], [1] has length 1.

Example 3:

Input: obstacles = [3,1,5,6,4,2]
Output: [1,1,2,3,2,2]
Explanation: The longest valid obstacle course at each position is:
- i = 0: [3], [3] has length 1.
- i = 1: [3,1], [1] has length 1.
- i = 2: [3,1,5], [3,5] has length 2. [1,5] is also valid.
- i = 3: [3,1,5,6], [3,5,6] has length 3. [1,5,6] is also valid.
- i = 4: [3,1,5,6,4], [3,4] has length 2. [1,4] is also valid.
- i = 5: [3,1,5,6,4,2], [1,2] has length 2.

 

Constraints:

• n == obstacles.length
• 1 <= n <= 105
• 1 <= obstacles[i] <= 107

Solutions

Binary Indexed Tree.

Python3

class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, val):
        while x <= self.n:
            self.c[x] = max(self.c[x], val)
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        s = 0
        while x > 0:
            s = max(s, self.c[x])
            x -= BinaryIndexedTree.lowbit(x)
        return s


class Solution:
    def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
        s = sorted(set(obstacles))
        m = {v: i for i, v in enumerate(s, 1)}
        tree = BinaryIndexedTree(len(m))
        ans = []
        for v in obstacles:
            x = m[v]
            ans.append(1 + tree.query(x))
            tree.update(x, ans[-1])
        return ans

Java

class Solution {
    public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
        TreeSet<Integer> ts = new TreeSet();
        for (int v : obstacles) {
            ts.add(v);
        }
        int idx = 1;
        Map<Integer, Integer> m = new HashMap<>();
        for (int v : ts) {
            m.put(v, idx++);
        }
        BinaryIndexedTree tree = new BinaryIndexedTree(m.size());
        int n = obstacles.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int v = obstacles[i];
            int x = m.get(v);
            ans[i] = tree.query(x) + 1;
            tree.update(x, ans[i]);
        }
        return ans;
    }
}

class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public void update(int x, int val) {
        while (x <= n) {
            c[x] = Math.max(c[x], val);
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s = Math.max(s, c[x]);
            x -= lowbit(x);
        }
        return s;
    }

    public static int lowbit(int x) {
        return x & -x;
    }
}

C++

class BinaryIndexedTree {
public:
    int n;
    vector<int> c;

    BinaryIndexedTree(int _n)
        : n(_n)
        , c(_n + 1) { }

    void update(int x, int val) {
        while (x <= n) {
            c[x] = max(c[x], val);
            x += lowbit(x);
        }
    }

    int query(int x) {
        int s = 0;
        while (x > 0) {
            s = max(s, c[x]);
            x -= lowbit(x);
        }
        return s;
    }

    int lowbit(int x) {
        return x & -x;
    }
};

class Solution {
public:
    vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
        set<int> s(obstacles.begin(), obstacles.end());
        int idx = 1;
        unordered_map<int, int> m;
        for (int v : s) m[v] = idx++;
        BinaryIndexedTree* tree = new BinaryIndexedTree(m.size());
        int n = obstacles.size();
        vector<int> ans(n);
        for (int i = 0; i < n; ++i) {
            int v = obstacles[i];
            int x = m[v];
            ans[i] = 1 + tree->query(x);
            tree->update(x, ans[i]);
        }
        return ans;
    }
};

Go

type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
	return x & -x
}

func (this *BinaryIndexedTree) update(x, val int) {
	for x <= this.n {
		if this.c[x] < val {
			this.c[x] = val
		}
		x += this.lowbit(x)
	}
}

func (this *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		if s < this.c[x] {
			s = this.c[x]
		}
		x -= this.lowbit(x)
	}
	return s
}

func longestObstacleCourseAtEachPosition(obstacles []int) []int {
	s := make(map[int]bool)
	for _, v := range obstacles {
		s[v] = true
	}
	var t []int
	for v, _ := range s {
		t = append(t, v)
	}
	sort.Ints(t)
	m := make(map[int]int)
	for i, v := range t {
		m[v] = i + 1
	}
	n := len(obstacles)
	ans := make([]int, n)
	tree := newBinaryIndexedTree(len(m))
	for i, v := range obstacles {
		x := m[v]
		ans[i] = 1 + tree.query(x)
		tree.update(x, ans[i])
	}
	return ans
}

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