Fenwick Tree Solution
Problem​
The problem requires efficiently processing a series of queries that involve updating and retrieving maximum values within a range. The solution uses the Fenwick Tree (Binary Indexed Tree) data structure to achieve this efficiently.
Solution​
A Fenwick Tree is used to keep track of the maximum values in the range. Each query is processed in constant or logarithmic time, making the solution efficient for large inputs.
Fenwick Tree Structure​
The Fenwick Tree is implemented with the following methods:
update(int i, int val): Updates the tree with the valuevalat indexi.get(int i): Retrieves the maximum value from the start to indexi.
Steps:​
- Initialization: Initialize the Fenwick Tree and a sorted set to keep track of obstacles.
- Update Queries: Process update queries to add obstacles.
- Compute Intervals: Compute intervals between obstacles and update the tree.
- Result Queries: Process result queries to check if the interval length is sufficient.
Code in Different Languages​
C++​
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
class FenwickTree {
public:
FenwickTree(int n) : vals(n + 1) {}
void update(int i, int val) {
while (i < vals.size()) {
vals[i] = max(vals[i], val);
i += lowbit(i);
}
}
int get(int i) const {
int res = 0;
while (i > 0) {
res = max(res, vals[i]);
i -= lowbit(i);
}
return res;
}
private:
vector<int> vals;
static int lowbit(int i) {
return i & -i;
}
};
class Solution {
public:
vector<bool> getResults(vector<vector<int>>& queries) {
const int n = min(50000, static_cast<int>(queries.size()) * 3);
vector<bool> ans;
FenwickTree tree(n + 1);
set<int> obstacles{0, n}; // sentinel values
for (const vector<int>& query : queries) {
const int type = query[0];
if (type == 1) {
const int x = query[1];
obstacles.insert(x);
}
}
for (auto it = obstacles.begin(); std::next(it) != obstacles.end(); ++it) {
const int x1 = *it;
const int x2 = *std::next(it);
tree.update(x2, x2 - x1);
}
for (int i = queries.size() - 1; i >= 0; --i) {
const int type = queries[i][0];
const int x = queries[i][1];
if (type == 1) {
const auto it = obstacles.find(x);
if (next(it) != obstacles.end()) // x is not the last element.
tree.update(*next(it), *next(it) - *prev(it));
obstacles.erase(it);
} else {
const int sz = queries[i][2];
const auto it = obstacles.upper_bound(x);
const int prev = *std::prev(it);
ans.push_back(tree.get(prev) >= sz || x - prev >= sz);
}
}
return {ans.rbegin(), ans.rend()};
}
};
Java​
import java.util.*;
class FenwickTree {
private int[] vals;
public FenwickTree(int n) {
vals = new int[n + 1];
}
public void update(int i, int val) {
while (i < vals.length) {
vals[i] = Math.max(vals[i], val);
i += lowbit(i);
}
}
public int get(int i) {
int res = 0;
while (i > 0) {
res = Math.max(res, vals[i]);
i -= lowbit(i);
}
return res;
}
private static int lowbit(int i) {
return i & -i;
}
}
class Solution {
public List<Boolean> getResults(int[][] queries) {
final int n = Math.min(50000, queries.length * 3);
List<Boolean> ans = new ArrayList<>();
FenwickTree tree = new FenwickTree(n + 1);
TreeSet<Integer> obstacles = new TreeSet<>(Arrays.asList(0, n)); // sentinel values
for (int[] query : queries) {
final int type = query[0];
if (type == 1) {
final int x = query[1];
obstacles.add(x);
}
}
Iterator<Integer> it = obstacles.iterator();
int x1 = it.next();
while (it.hasNext()) {
final int x2 = it.next();
tree.update(x2, x2 - x1);
x1 = x2;
}
for (int i = queries.length - 1; i >= 0; --i) {
final int type = queries[i][0];
final int x = queries[i][1];
if (type == 1) {
final Integer next = obstacles.higher(x);
final Integer prev = obstacles.lower(x);
if (next != null)
tree.update(next, next - prev);
obstacles.remove(x);
} else {
final int sz = queries[i][2];
final int prev = obstacles.floor(x);
ans.add(tree.get(prev) >= sz || x - prev >= sz);
}
}
Collections.reverse(ans);
return ans;
}
}
Python​
from sortedcontainers import SortedList
from typing import List
import itertools
class FenwickTree:
def __init__(self, n: int):
self.vals = [0] * (n + 1)
def update(self, i: int, val: int) -> None:
while i < len(self.vals):
self.vals[i] = max(self.vals[i], val)
i += FenwickTree.lowtree(i)
def get(self, i: int) -> int:
res = 0
while i > 0:
res = max(res, self.vals[i])
i -= FenwickTree.lowtree(i)
return res
@staticmethod
def lowtree(i: int) -> int:
return i & -i
class Solution:
def getResults(self, queries: List[List[int]]) -> List[bool]:
n = min(50000, len(queries) * 3)
ans = []
tree = FenwickTree(n + 1)
obstacles = SortedList([0, n]) # sentinel values
for query in queries:
type = query[0]
if type == 1:
x = query[1]
obstacles.add(x)
for x1, x2 in itertools.pairwise(obstacles):
tree.update(x2, x2 - x1)
for query in reversed(queries):
type = query[0]
x = query[1]
if type == 1:
i = obstacles.index(x)
next = obstacles[i + 1]
prev = obstacles[i - 1]
obstacles.remove(x)
tree.update(next, next - prev)
else:
sz = query[2]
i = obstacles.bisect_right(x)
prev = obstacles[i - 1]
ans.append(tree.get(prev) >= sz or x - prev >= sz)
return ans[::-1]