查询具有更新的给定范围内的值的数组元素的计数

给定大小为N的数组arr[]和由以下两种类型的查询组成的矩阵Q ：

• 1 LR ：打印位于 [L, R] 范围内的元素数量。
• 2 ix :设置 arr[i] = x

例子：

Input: arr[] = {1, 2, 2, 3, 4, 4, 5, 6}, Q = {{1, {3, 5}}, {1, {2, 4}}, {1, {1, 2}}, {2, {1, 7}}, {1, {1, 2}}} 
Output: 4 5 3 2 
Explanation: 
Array elements from the range [3, 5] are {3, 4, 4, 5} 
Array elements from the range [2, 4] are {2, 2, 3, 4, 4} 
Array elements from the range [1, 2] are {1, 2, 2} 
Replacing arr[1] by 7 modifies the array to {1, 7, 2, 3, 4, 4, 5, 6} 
Elements that lie in range [1, 2] are {1, 2}

Input: arr = {5, 5, 1, 3, 4, 4, 2, 3}, Q = {{1, {3, 6}}, {1, {2, 4}}, {1, {10, 20}}} 
Output: 6 5 0 
Explanation: 
Array elements from the range [3, 6] are {3, 3, 4, 4, 5, 5} 
Array elements from the range [2, 4] are {2, 3, 3, 4, 4} 
No element from the range [10, 20] exists in the array. 

天真的方法：
解决这个问题最简单的方法如下：

• 对于(1 LR)类型的查询，遍历整个数组并计算数组中元素的数量，使得L ≤ arr[i] ≤ R 。最后，打印计数。
• 对于类型(2 ix)的查询，将arr[i]替换为x 。

下面是上述方法的实现：

C++

// C++ code for queries for number
// of elements that lie in range
// [l, r] (with updates)
#include 
using namespace std;
 
// Function to set arr[index] = x
void setElement(int* arr, int n,
                int index, int x)
{
    arr[index] = x;
}
 
// Function to get count of elements
// that lie in range [l, r]
int getCount(int* arr, int n,
            int l, int r)
{
    int count = 0;
 
    // Traverse array
    for (int i = 0; i < n; i++) {
 
        // If element lies in the
        // range [L, R]
        if (arr[i] >= l
            && arr[i] <= r) {
 
            // Increase count
            count++;
        }
    }
    return count;
}
 
// Function to solve each query
void SolveQuery(int arr[], int n,
                vector > >
                    Q)
{
    int x;
 
    for (int i = 0; i < Q.size(); i++) {
        if (Q[i].first == 1) {
            x = getCount(arr, n,
                        Q[i].second.first,
                        Q[i].second.second);
 
            cout << x << " ";
        }
        else {
            setElement(arr, n,
                    Q[i].second.first,
                    Q[i].second.second);
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 2, 3, 4, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    vector > > Q
        = { { 1, { 3, 5 } },
            { 1, { 2, 4 } },
            { 1, { 1, 2 } },
            { 2, { 1, 7 } },
            { 1, { 1, 2 } } };
    SolveQuery(arr, n, Q);
 
    return 0;
}

Java

// Java code for queries for number
// of elements that lie in range
// [l, r] (with updates)
import java.util.*;
import java.lang.*;
 
class GFG{
     
// Function to set arr[index] = x
static void setElement(int[] arr, int n,
                       int index, int x)
{
    arr[index] = x;
}
 
// Function to get count of elements
// that lie in range [l, r]
static int getCount(int[] arr, int n,
                    int l, int r)
{
    int count = 0;
     
    // Traverse array
    for(int i = 0; i < n; i++)
    {
         
        // If element lies in the
        // range [L, R]
        if (arr[i] >= l && arr[i] <= r)
        {
            // Increase count
            count++;
        }
    }
    return count;
}
 
// Function to solve each query
static void SolveQuery(int arr[], int n,
                       ArrayList> Q)
{
    int x;
  
    for(int i = 0; i < Q.size(); i++)
    {
        if (Q.get(i).get(0) == 1)
        {
            x = getCount(arr, n,
                         Q.get(i).get(1),
                         Q.get(i).get(2));
  
            System.out.print(x + " ");
        }
        else
        {
            setElement(arr, n,
                       Q.get(i).get(1),
                       Q.get(i).get(2));
        }
    }
}
  
// Driver code
public static void main (String[] args)
{
    int arr[] = { 1, 2, 2, 3, 4, 4, 5, 6 };
    int n = arr.length;
     
    ArrayList> Q = new ArrayList<>();
    Q.add(Arrays.asList(1, 3, 5));
    Q.add(Arrays.asList(1, 2, 4)); 
    Q.add(Arrays.asList(1, 1, 2));
    Q.add(Arrays.asList(2, 1, 7));
    Q.add(Arrays.asList(1, 1, 2));
     
    SolveQuery(arr, n, Q);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 code for queries for number
# of elements that lie in range
# [l, r] (with updates)
from typing import Generic, List, TypeVar
 
T = TypeVar('T')
V = TypeVar('V')
 
class Pair(Generic[V, T]):
     
    def __init__(self, first: V, second: T) -> None:
         
        self.first = first
        self.second = second
 
# Function to set arr[index] = x
def setElement(arr: List[int], n: int,
                   index: int, x: int) -> None:
                        
    arr[index] = x
 
# Function to get count of elements
# that lie in range [l, r]
def getCount(arr: List[int], n: int,
                     l: int, r: int) -> int:
 
    count = 0
 
    # Traverse array
    for i in range(n):
 
        # If element lies in the
        # range [L, R]
        if (arr[i] >= l and arr[i] <= r):
 
            # Increase count
            count += 1
 
    return count
 
# Function to solve each query
def SolveQuery(arr: List[int], n: int,
            Q: List[Pair[int, Pair[int, int]]]):
 
    x = 0
 
    for i in range(len(Q)):
        if (Q[i].first == 1):
            x = getCount(arr, n, Q[i].second.first,
                                 Q[i].second.second)
            print(x, end = " ")
        else:
            setElement(arr, n, Q[i].second.first,
                               Q[i].second.second)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ 1, 2, 2, 3, 4, 4, 5, 6 ]
    n = len(arr)
 
    Q = [ Pair(1, Pair(3, 5)),
          Pair(1, Pair(2, 4)),
          Pair(1, Pair(1, 2)),
          Pair(2, Pair(1, 7)),
          Pair(1, Pair(1, 2)) ]
           
    SolveQuery(arr, n, Q)
 
# This code is contributed by sanjeev2552

C#

// C# code for queries for number
// of elements that lie in range
// [l, r] (with updates)
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to set arr[index] = x
static void setElement(int[] arr, int n,
                       int index, int x)
{
    arr[index] = x;
}
 
// Function to get count of elements
// that lie in range [l, r]
static int getCount(int[] arr, int n,
                    int l, int r)
{
    int count = 0;
     
    // Traverse array
    for(int i = 0; i < n; i++)
    {
         
        // If element lies in the
        // range [L, R]
        if (arr[i] >= l && arr[i] <= r)
         
            // Increase count
            count += 1;
    }
    return count;
}
 
// Function to solve each query
static void SolveQuery(int[] arr, int n,
             List> Q)
{
    int x;
 
    for(int i = 0; i < Q.Count; i++)
    {
        if (Q[i][0] == 1)
        {
            x = getCount(arr, n,
                         Q[i][1],
                         Q[i][2]);
 
            Console.Write(x + " ");
        }
        else
        {
            setElement(arr, n,
                       Q[i][1],
                       Q[i][2]);
        }
    }
}
 
// Driver code
public static void Main(string[] args)
{
    int[] arr = { 1, 2, 2, 3, 4, 4, 5, 6 };
    int n = arr.Length;
 
    List> myList = new List>();
    myList.Add(new List{ 1, 3, 5 });
    myList.Add(new List{ 1, 2, 4 });
    myList.Add(new List{ 1, 1, 2 });
    myList.Add(new List{ 2, 1, 7 });
    myList.Add(new List{ 1, 1, 2 });
 
    SolveQuery(arr, n, myList);
}
}
 
// This code is contributed by grand_master

输出：

4 5 3 2

时间复杂度： O(Q * N)
辅助空间： O(1)

有效的方法：
上述方法可以使用 Fenwick Tree 进行优化。请按照以下步骤解决问题：

• 从给定的数组构造一个 Fenwick 树。
• Fenwick树将表示为一个大小等于数组中最大元素的数组，这样数组元素就可以作为索引(思路和这个方法类似)。
• 通过调用Fenwick树的update方法遍历数组，增加每个元素出现的频率。
• 对于每个类型为(1 LR) 的查询，调用 Fenwick 树的getSum方法。类型 1 查询的答案将是：

getSum(R) – getSum(L – 1)

• 对于每个类型为(2 ix) 的查询，调用 Fenwick 树的update方法来增加添加元素的频率并减少要替换的元素的数量。

下面是上述方法的实现：

C++

// C++ code for queries for number
// of elements that lie in range
// [l, r] (with updates)
#include 
using namespace std;
 
class FenwickTree {
public:
    int* BIT;
    int N;
 
    FenwickTree(int N)
    {
        this->N = N;
        BIT = new int[N];
        for (int i = 0; i < N; i++) {
            BIT[i] = 0;
        }
    }
 
    // Traverse all ancestors and
    // increase frequency of index
    void update(int index, int increment)
    {
        while (index < N) {
 
            // Increase count of the current
            // node of BIT Tree
            BIT[index] += increment;
 
            // Update index to that of parent
            // in update View
            index += (index & -index);
        }
    }
    // Function to return the
    // sum of arr[0..index]
    int getSum(int index)
    {
        int sum = 0;
 
        // Traverse ancestors of
        // BITree[index]
        while (index > 0) {
 
            // Add current element of
            // BITree to sum
            sum += BIT[index];
 
            // Move index to parent node in
            // getSum View
            index -= (index & -index);
        }
        return sum;
    }
};
 
// Function to set arr[index] = x
void setElement(int* arr, int n,
                int index, int x,
                FenwickTree* fenTree)
{
    int removedElement = arr[index];
    fenTree->update(removedElement, -1);
    arr[index] = x;
    fenTree->update(x, 1);
}
 
// Function to get count of
// elements that lie in
// range [l, r]
int getCount(int* arr, int n,
            int l, int r,
            FenwickTree* fenTree)
{
    int count = fenTree->getSum(r)
                - fenTree->getSum(l - 1);
    return count;
}
 
// Function to solve each query
void SolveQuery(int arr[], int n,
                vector > >
                    Q)
{
    int N = 100001;
 
    FenwickTree* fenTree = new FenwickTree(N);
 
    for (int i = 0; i < n; i++) {
        fenTree->update(arr[i], 1);
    }
 
    int x;
 
    for (int i = 0; i < Q.size(); i++) {
        if (Q[i].first == 1) {
            x = getCount(arr, n,
                        Q[i].second.first,
                        Q[i].second.second,
                        fenTree);
 
            cout << x << " ";
        }
        else {
            setElement(arr, n,
                    Q[i].second.first,
                    Q[i].second.second,
                    fenTree);
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 2, 3,
                4, 4, 5, 6 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
    vector > > Q
        = { { 1, { 3, 5 } },
            { 1, { 2, 4 } },
            { 1, { 1, 2 } },
            { 2, { 1, 7 } },
            { 1, { 1, 2 } } };
 
    SolveQuery(arr, n, Q);
 
    return 0;
}

输出：

4 5 3 2

时间复杂度： O(n*log(N) + Q*log(N))
辅助空间： O(maxm)，其中 maxm 是数组中存在的最大元素。

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live