使用细分树查询给定索引范围内大于K的元素

给定一个由N个元素组成的数组arr []和多个查询，其中每个查询将包含三个整数L ， R和K。对于每个查询，任务是找到子数组arr [L…R]中大于K的元素数。
例子：

Input: arr[] = {7, 3, 9, 13, 5, 4}, q[] = {{0, 3, 6}, {1, 5, 8}} 
Output: 
3 
2 
Query 1: Only 7, 9 and 13 are greater 
than 6 in the subarray {7, 3, 9, 13}. 
Query 2: Only 9 and 13 are greater 
than 8 in the subarray {3, 9, 13, 5, 4}.

Input: arr[] = {0, 1, 2, 3, 4, 5, 6, 7}, q[] = {{0, 7, 3}, {4, 6, 10}} 
Output: 
4 
0 

先决条件：段树
天真的方法：只需从索引l遍历到r遍历数组即可找到每个查询的答案，并在数组元素大于k时继续向计数加1 。这种方法的时间复杂度将是O(n * q) 。
高效的方法：在每个节点上建立一个带有向量的分段树，其中包含按排序顺序包含子范围的所有元素。使用段树回答每个查询，其中可以使用二元搜索来计算子节点位于查询范围内大于K的每个节点中存在多少个数字。此方法的时间复杂度将为O(q * log(n)* log(n))
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Merge procedure to merge two
// vectors into a single vector
vector merge(vector& v1, vector& v2)
{
    int i = 0, j = 0;
 
    // Final vector to return
    // after merging
    vector v;
 
    // Loop continues until it reaches
    // the end of one of the vectors
    while (i < v1.size() && j < v2.size()) {
        if (v1[i] <= v2[j]) {
            v.push_back(v1[i]);
            i++;
        }
        else {
            v.push_back(v2[j]);
            j++;
        }
    }
 
    // Here, simply add the remaining
    // elements to the vector v
    for (int k = i; k < v1.size(); k++)
        v.push_back(v1[k]);
    for (int k = j; k < v2.size(); k++)
        v.push_back(v2[k]);
    return v;
}
 
// Procedure to build the segment tree
void buildTree(vector* tree, int* arr,
               int index, int s, int e)
{
 
    // Reached the leaf node
    // of the segment tree
    if (s == e) {
        tree[index].push_back(arr[s]);
        return;
    }
 
    // Recursively call the buildTree
    // on both the nodes of the tree
    int mid = (s + e) / 2;
    buildTree(tree, arr, 2 * index, s, mid);
    buildTree(tree, arr, 2 * index + 1, mid + 1, e);
 
    // Storing the final vector after merging
    // the two of its sorted child vector
    tree[index] = merge(tree[2 * index], tree[2 * index + 1]);
}
 
// Query procedure to get the answer
// for each query l and r are query range
int query(vector* tree, int index, int s,
          int e, int l, int r, int k)
{
 
    // out of bound or no overlap
    if (r < s || l > e)
        return 0;
 
    // Complete overlap
    // Query range completely lies in
    // the segment tree node range
    if (s >= l && e <= r) {
        // binary search to find index of k
        return (tree[index].size()
                - (lower_bound(tree[index].begin(),
                               tree[index].end(), k)
                   - tree[index].begin()));
    }
 
    // Partially overlap
    // Query range partially lies in
    // the segment tree node range
    int mid = (s + e) / 2;
    return (query(tree, 2 * index, s,
                  mid, l, r, k)
            + query(tree, 2 * index + 1, mid + 1,
                    e, l, r, k));
}
 
// Function to perform the queries
void performQueries(int L[], int R[], int K[],
                    int n, int q, vector tree[])
{
    for (int i = 0; i < q; i++) {
        cout << query(tree, 1, 0, n - 1,
                      L[i] - 1, R[i] - 1, K[i])
             << endl;
    }
}
 
// Driver code
int main()
{
    int arr[] = { 7, 3, 9, 13, 5, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    vector tree[4 * n + 1];
    buildTree(tree, arr, 1, 0, n - 1);
 
    // 1-based indexing
    int L[] = { 1, 2 };
    int R[] = { 4, 6 };
    int K[] = { 6, 8 };
 
    // Number of queries
    int q = sizeof(L) / sizeof(L[0]);
 
    performQueries(L, R, K, n, q, tree);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Merge procedure to merge two
    // vectors into a single vector
    static Vector merge(Vector v1,
                               Vector v2)
    {
        int i = 0, j = 0;
 
        // Final vector to return
        // after merging
        Vector v = new Vector<>();
 
        // Loop continues until it reaches
        // the end of one of the vectors
        while (i < v1.size() && j < v2.size())
        {
            if (v1.elementAt(i) <= v2.elementAt(j))
            {
                v.add(v1.elementAt(i));
                i++;
            }
            else
            {
                v.add(v2.elementAt(j));
                j++;
            }
        }
 
        // Here, simply add the remaining
        // elements to the vector v
        for (int k = i; k < v1.size(); k++)
            v.add(v1.elementAt(k));
        for (int k = j; k < v2.size(); k++)
            v.add(v2.elementAt(k));
        return v;
    }
 
    // Procedure to build the segment tree
    static void buildTree(Vector[] tree, int[] arr,
                        int index, int s, int e)
    {
 
        // Reached the leaf node
        // of the segment tree
        if (s == e)
        {
            tree[index].add(arr[s]);
            return;
        }
 
        // Recursively call the buildTree
        // on both the nodes of the tree
        int mid = (s + e) / 2;
        buildTree(tree, arr, 2 * index, s, mid);
        buildTree(tree, arr, 2 * index + 1, mid + 1, e);
 
        // Storing the final vector after merging
        // the two of its sorted child vector
        tree[index] = merge(tree[2 * index], tree[2 * index + 1]);
    }
 
    // Query procedure to get the answer
    // for each query l and r are query range
    static int query(Vector[] tree, int index, int s,
                    int e, int l, int r, int k)
    {
 
        // out of bound or no overlap
        if (r < s || l > e)
            return 0;
 
        // Complete overlap
        // Query range completely lies in
        // the segment tree node range
        if (s >= l && e <= r)
        {
             
            // binary search to find index of k
            return (tree[index].size() - lowerBound(tree[index],
                    tree[index].size(), k));
        }
 
        // Partially overlap
        // Query range partially lies in
        // the segment tree node range
        int mid = (s + e) / 2;
        return (query(tree, 2 * index, s, mid, l, r, k) +
                query(tree, 2 * index + 1, mid + 1, e, l, r, k));
    }
 
    // Function to perform the queries
    static void performQueries(int L[], int R[], int K[],
                        int n, int q, Vector tree[])
    {
        for (int i = 0; i < q; i++)
        {
            System.out.println(query(tree, 1, 0, n - 1,
                                    L[i] - 1, R[i] - 1, K[i]));
        }
    }
 
    static int lowerBound(Vector array,
                        int length, int value)
    {
        int low = 0;
        int high = length;
        while (low < high)
        {
            final int mid = (low + high) / 2;
            if (value <= array.elementAt(mid))
            {
                high = mid;
            }
            else
            {
                low = mid + 1;
            }
        }
        return low;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 7, 3, 9, 13, 5, 4 };
        int n = arr.length;
        @SuppressWarnings("unchecked")
        Vector[] tree = new Vector[4 * n + 1];
        for (int i = 0; i < (4 * n + 1); i++)
        {
            tree[i] = new Vector<>();
        }
 
        buildTree(tree, arr, 1, 0, n - 1);
 
        // 1-based indexing
        int L[] = { 1, 2 };
        int R[] = { 4, 6 };
        int K[] = { 6, 8 };
 
        // Number of queries
        int q = L.length;
 
        performQueries(L, R, K, n, q, tree);
    }
}
 
// This code is contributed by
// sanjeev2552

Python3

# Python3 implementation of the approach
from bisect import bisect_left as lower_bound
 
# Merge procedure to merge two
# vectors into a single vector
def merge(v1, v2):
    i = 0
    j = 0
 
    # Final vector to return
    # after merging
    v = []
 
    # Loop continues until it reaches
    # the end of one of the vectors
    while (i < len(v1) and j < len(v2)):
        if (v1[i] <= v2[j]):
            v.append(v1[i])
            i += 1
        else:
            v.append(v2[j])
            j += 1
 
    # Here, simply add the remaining
    # elements to the vector v
    for k in range(i, len(v1)):
        v.append(v1[k])
    for k in range(j, len(v2)):
        v.append(v2[k])
    return v
 
# Procedure to build the segment tree
def buildTree(tree,arr,index, s, e):
 
    # Reached the leaf node
    # of the segment tree
    if (s == e):
        tree[index].append(arr[s])
        return
 
    # Recursively call the buildTree
    # on both the nodes of the tree
    mid = (s + e) // 2
    buildTree(tree, arr, 2 * index, s, mid)
    buildTree(tree, arr, 2 * index + 1, mid + 1, e)
 
    # Storing the final vector after merging
    # the two of its sorted child vector
    tree[index] = merge(tree[2 * index], tree[2 * index + 1])
 
# Query procedure to get the answer
# for each query l and r are query range
def query(tree, index, s, e, l, r, k):
 
    # out of bound or no overlap
    if (r < s or l > e):
        return 0
 
    # Complete overlap
    # Query range completely lies in
    # the segment tree node range
    if (s >= l and e <= r):
         
        # binary search to find index of k
        return len(tree[index]) - (lower_bound(tree[index], k))
 
    # Partially overlap
    # Query range partially lies in
    # the segment tree node range
    mid = (s + e) // 2
    return (query(tree, 2 * index, s,mid, l, r, k)
            + query(tree, 2 * index + 1, mid + 1,e, l, r, k))
 
# Function to perform the queries
def performQueries(L, R, K,n, q,tree):
    for i in range(q):
        print(query(tree, 1, 0, n - 1,L[i] - 1, R[i] - 1, K[i]))
 
# Driver code
if __name__ == '__main__':
    arr = [7, 3, 9, 13, 5, 4]
    n = len(arr)
    tree = [[] for i in range(4 * n + 1)]
    buildTree(tree, arr, 1, 0, n - 1)
 
    # 1-based indexing
    L = [1, 2]
    R = [4, 6]
    K = [6, 8]
 
    # Number of queries
    q = len(L)
 
     performQueries(L, R, K, n, q, tree)
      
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG {
  
    // Merge procedure to merge two
    // vectors into a single vector
    static List merge(List v1,
                               List v2)
    {
        int i = 0, j = 0;
  
        // Final vector to return
        // after merging
        List v = new List();
  
        // Loop continues until it reaches
        // the end of one of the vectors
        while (i < v1.Count && j < v2.Count)
        {
            if (v1[i] <= v2[j])
            {
                v.Add(v1[i]);
                i++;
            }
            else
            {
                v.Add(v2[j]);
                j++;
            }
        }
  
        // Here, simply add the remaining
        // elements to the vector v
        for (int k = i; k < v1.Count; k++)
            v.Add(v1[k]);
        for (int k = j; k < v2.Count; k++)
            v.Add(v2[k]);
        return v;
    }
  
    // Procedure to build the segment tree
    static void buildTree(List[] tree, int[] arr,
                        int index, int s, int e)
    {
  
        // Reached the leaf node
        // of the segment tree
        if (s == e)
        {
            tree[index].Add(arr[s]);
            return;
        }
  
        // Recursively call the buildTree
        // on both the nodes of the tree
        int mid = (s + e) / 2;
        buildTree(tree, arr, 2 * index, s, mid);
        buildTree(tree, arr, 2 * index + 1, mid + 1, e);
  
        // Storing the readonly vector after merging
        // the two of its sorted child vector
        tree[index] = merge(tree[2 * index], tree[2 * index + 1]);
    }
  
    // Query procedure to get the answer
    // for each query l and r are query range
    static int query(List[] tree, int index, int s,
                    int e, int l, int r, int k)
    {
  
        // out of bound or no overlap
        if (r < s || l > e)
            return 0;
  
        // Complete overlap
        // Query range completely lies in
        // the segment tree node range
        if (s >= l && e <= r)
        {
              
            // binary search to find index of k
            return (tree[index].Count - lowerBound(tree[index],
                    tree[index].Count, k));
        }
  
        // Partially overlap
        // Query range partially lies in
        // the segment tree node range
        int mid = (s + e) / 2;
        return (query(tree, 2 * index, s, mid, l, r, k) +
                query(tree, 2 * index + 1, mid + 1, e, l, r, k));
    }
  
    // Function to perform the queries
    static void performQueries(int []L, int []R, int []K,
                        int n, int q, List []tree)
    {
        for (int i = 0; i < q; i++)
        {
            Console.WriteLine(query(tree, 1, 0, n - 1,
                                    L[i] - 1, R[i] - 1, K[i]));
        }
    }
  
    static int lowerBound(List array,
                        int length, int value)
    {
        int low = 0;
        int high = length;
        while (low < high)
        {
            int mid = (low + high) / 2;
            if (value <= array[mid])
            {
                high = mid;
            }
            else
            {
                low = mid + 1;
            }
        }
        return low;
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        int []arr = { 7, 3, 9, 13, 5, 4 };
        int n = arr.Length;
        List[] tree = new List[4 * n + 1];
        for (int i = 0; i < (4 * n + 1); i++)
        {
            tree[i] = new List();
        }
  
        buildTree(tree, arr, 1, 0, n - 1);
  
        // 1-based indexing
        int []L = { 1, 2 };
        int []R = { 4, 6 };
        int []K = { 6, 8 };
  
        // Number of queries
        int q = L.Length;
  
        performQueries(L, R, K, n, q, tree);
    }
}
 
// This code is contributed by PrinciRaj1992

输出：

3
2

另一种方法：

使用段树的另一种方法是在该范围内的每个节点(如果存在)中存储大于K的第一个元素，否则存储0 。

在这里，我们需要考虑3种情况来构建树。

1. 如果左和右子级都包含非0的数字，则答案始终是左子级。 (我们需要考虑大于K的数字的第一次出现。)
2. 如果左或右子级中的任何一个包含0 ，则答案始终是非0的数字。
3. 如果左右两个孩子都包含0 ，则答案始终为0 (表示该范围内不存在大于K的数字)。

查询函数始终保持不变。

请考虑以下示例： arr [] = {7，3，9，13，13，5，4}，K = 6

在这种情况下，树将如下所示：

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
vector arr(1000000), tree(4 * arr.size());
 
// combine function to make parent node
int combine(int a, int b)
{
    if (a != 0 && b != 0) {
        return a;
    }
    if (a >= b) {
        return a;
    }
    return b;
}
 
// building the tree
void buildTree(int ind, int low, int high, int x)
{
    // leaf node
    if (low == high) {
        if (arr[low] > x) {
            tree[ind] = arr[low];
        }
        else {
            tree[ind] = 0;
        }
        return;
    }
    int mid = (low + high) / 2;
    buildTree(2 * ind + 1, low, mid, x);
    buildTree(2 * ind + 2, mid + 1, high, x);
 
    // merging the nodes while backtracking.
    tree[ind]
        = combine(tree[2 * ind + 1], tree[2 * ind + 2]);
}
// performing query
int query(int ind, int low, int high, int l, int r)
{
    int mid = (low + high) / 2;
    // Out of Bounds
    if (low > r || high < l) {
        return 0;
    }
    // completely overlaps
    if (l <= low && r >= high) {
        return tree[ind];
    }
    // partially overlaps
    return combine(query(2 * ind + 1, low, mid, l, r),
                   query(2 * ind + 2, mid + 1, high, l, r));
}
 
// Driver Code
int main()
{
    arr = { 7, 3, 9, 13, 5, 4 };
    int n = 6;
    int k = 6;
 
    // 1-based indexing
    int l = 1, r = 4;
    buildTree(0, 0, n - 1, k);
    cout << query(0, 0, n - 1, l - 1, r - 1);
    return 0;
}
// This code is contributed by yashbeersingh42

输出：

7

时间复杂度： O(N * log N)来构建树，O(log N)用于每个查询。

空间复杂度： O(N)