Solved leetcode problem 3161- Block Placement Queries

dsa-solutions/lc-solutions/3100-3199/3161-Block Placement Queries.md

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+---
+id: fenwick-tree-solution
+level: Hard
+title: Fenwick Tree Solution
+sidebar_label: Fenwick Tree Solution
+tags:
+- Fenwick Tree
+- Data Structure
+- C++
+- Java
+- Python
+description: "This document provides solutions to a problem using the Fenwick Tree (Binary Indexed Tree) data structure, implemented in C++, Java, and Python."
+---
+
+## 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 value `val` at index `i`.
+- `get(int i)`: Retrieves the maximum value from the start to index `i`.
+
+### Steps:
+1. **Initialization**: Initialize the Fenwick Tree and a sorted set to keep track of obstacles.
+2. **Update Queries**: Process update queries to add obstacles.
+3. **Compute Intervals**: Compute intervals between obstacles and update the tree.
+4. **Result Queries**: Process result queries to check if the interval length is sufficient.
+
+## Code in Different Languages
+
+### C++
+```cpp
+#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
+```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
+```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]
+```
+
+# Complexity Analysis
+## Time Complexity: $O(log N)$ for update and get methods.
+### Reason: Each update or get operation involves a logarithmic number of steps due to the nature of the Fenwick Tree.
+# Space Complexity: $O(N)$
+### Reason: We store values in the Fenwick Tree and maintain a sorted list of obstacles.