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📜  在给定数组上执行追加,更新,删除和范围求和查询

📅  最后修改于: 2021-04-17 10:41:01             🧑  作者: Mango

给定大小为N的数组arr [] ,任务是回答以下类型的Q个查询:

  • 1 X 0:在数组后面附加X。
  • 2 XY:设置arr [X] = Y。
  • 3 X 0:删除arr [X]
  • 4 XY:[X,Y]范围内求和。

请注意,删除查询类型3中的arr [X]后,索引X之后的所有元素都将向左移动一个位置。
例子:

天真的方法:解决此问题的天真的方法是直接在给定的数组上执行查询,该数组的时间复杂度为O(Q * N)
高效的方法:

  • 由于存在一些数据结构,例如分段树或BIT(分域树),它们可以为每个查询提供O(logN)中的点更新和范围总和。
  • 那里的帖子使用Fenwick树执行相同的操作,因此强烈建议在Fenwick树中进行点更新。
  • 但是主要的问题是执行删除操作(执行第3类查询)后,需要进行移位,这在最坏的情况下会再次导致O(Q * N)
  • 特技可以使用删除元素之后,只需更新在A [X] = 0的值和索引X之后映射到X之前X +数目的元素被删除
  • 为了找到在X之前删除的元素数量,将使用另一棵fenwick树(在实现中为idx),并在执行删除操作的那个索引处继续加1
  • 意味着在获取或获取特定索引时,可以查询idx树并将索引X更新为X + sum(X,idx)

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// Size of the array (MAX)
const int N = 2e5 + 10;
 
// To store the sum of
// the array elements
vector bit(N, 0);
 
// To keep track of how many type 3
// queries have been performed before
// a particular index
vector idx(N, 0);
 
// Function to perform the point
// update in Fenwick tree
void update(int idx, int val,
            vector& bitt)
{
    while (idx < N) {
        bitt[idx] += val;
        idx += idx & (-idx);
    }
}
 
// Function to return the sum
// of the elements A[1...idx]
// from BIT
int sum(int idx,
        vector& bitt)
{
 
    int res = 0;
    while (idx > 0) {
        res += bitt[idx];
        idx -= idx & (-idx);
    }
 
    return res;
}
 
// Function to perform the queries and
// return the answer vector
vector peformQueries(vector& A,
                          vector >& B)
{
 
    // For 1 bases indexing
    // insert 0 in the vector
    A.insert(A.begin(), 0);
 
    // Updated size of the vector
    int n = (int)A.size();
 
    // Updating the bit tree
    for (int i = 1; i < n; ++i) {
        update(i, A[i], bit);
    }
 
    // Vector to store the answers
    // of range sum
    vector ans;
 
    // Iterating in the query
    // vector
    for (auto i : B) {
 
        int type = i[0];
        int x = i[1], v = i[2];
 
        // If the query is of
        // type 1
        if (type == 1) {
 
            // Updating the tree
            // with x in the last
            update(n, x, bit);
 
            // Pushing back the value
            // in the original vector
            A.push_back(x);
 
            // Incrementing the size
            // of the vector by 1
            n++;
        }
 
        // If the query is of type 2
        else if (type == 2) {
 
            // Getting the updated index
            // in case of any query of
            // type 3 occured before it
            int id = x + sum(x, idx);
 
            // Making the effect to that
            // value to 0 by subtracting
            // same value from the tree
            update(id, -A[id], bit);
 
            // Updating the A[id] to v
            A[id] = v;
 
            // Now update the
            // bit by v at x
            update(id, v, bit);
        }
 
        // If the query is of type 3
        else if (type == 3) {
 
            // Get the current index
            int id = x + sum(x, idx);
 
            // Remove the effect of that
            // index from the bit tree
            update(id, -A[id], bit);
 
            // Update the idx tree
            // because one element has
            // been deleted
            update(x, 1, idx);
 
            // Update the idx val to 0
            A[id] = 0;
        }
 
        // If the query is of type 4
        else {
 
            // Get the updated range
            int xx = x + sum(x, idx);
            int vv = v + sum(v, idx);
 
            // Push_back the value
            ans.push_back(sum(vv, bit)
                          - sum(xx - 1, bit));
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    vector A = { 1, 2, 5, 3, 4 };
 
    // Queries
    vector > B = {
        { 4, 2, 4 },
        { 3, 3, 0 },
        { 1, 6, 0 },
        { 4, 3, 5 },
    };
 
    // Get the answer array
    vector ans = peformQueries(A, B);
 
    // printing the answer
    for (int i : ans)
        cout << i << "\n";
 
    return 0;
}


Java
// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Size of the array (MAX)
    static int N = (int) 2e5 + 10;
 
    // To store the sum of
    // the array elements
    static int[] bit = new int[N];
 
    // To keep track of how many type 3
    // queries have been performed before
    // a particular index
    static int[] idx = new int[N];
 
    // Function to perform the point
    // update in Fenwick tree
    static void update(int idx, int val, int[] bitt) {
        while (idx < N) {
            bitt[idx] += val;
            idx += idx & (-idx);
        }
    }
 
    // Function to return the sum
    // of the elements A[1...idx]
    // from BIT
    static int sum(int idx, int[] bitt) {
 
        int res = 0;
        while (idx > 0) {
            res += bitt[idx];
            idx -= idx & (-idx);
        }
 
        return res;
    }
 
    // Function to perform the queries and
    // return the answer vector
    static Vector peformQueries(Vector A, int[][] B) {
 
        // For 1 bases indexing
        // insert 0 in the vector
        A.insertElementAt(0, 0);
 
        // Updated size of the vector
        int n = (int) A.size();
 
        // Updating the bit tree
        for (int i = 1; i < n; ++i) {
            update(i, A.elementAt(i), bit);
        }
 
        // Vector to store the answers
        // of range sum
        Vector ans = new Vector<>();
 
        // Iterating in the query
        // vector
        for (int[] i : B) {
 
            int type = i[0];
            int x = i[1], v = i[2];
 
            // If the query is of
            // type 1
            if (type == 1) {
 
                // Updating the tree
                // with x in the last
                update(n, x, bit);
 
                // Pushing back the value
                // in the original vector
                A.add(x);
 
                // Incrementing the size
                // of the vector by 1
                n++;
            }
 
            // If the query is of type 2
            else if (type == 2) {
 
                // Getting the updated index
                // in case of any query of
                // type 3 occured before it
                int id = x + sum(x, idx);
 
                // Making the effect to that
                // value to 0 by subtracting
                // same value from the tree
                update(id, -A.elementAt(id), bit);
 
                // Updating the A[id] to v
                A.set(id, v);
 
                // Now update the
                // bit by v at x
                update(id, v, bit);
            }
 
            // If the query is of type 3
            else if (type == 3) {
 
                // Get the current index
                int id = x + sum(x, idx);
 
                // Remove the effect of that
                // index from the bit tree
                update(id, -A.elementAt(id), bit);
 
                // Update the idx tree
                // because one element has
                // been deleted
                update(x, 1, idx);
 
                // Update the idx val to 0
                A.set(id, 0);
            }
 
            // If the query is of type 4
            else {
 
                // Get the updated range
                int xx = x + sum(x, idx);
                int vv = v + sum(v, idx);
 
                // Push_back the value
                ans.add(sum(vv, bit) - sum(xx - 1, bit));
            }
        }
 
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args) {
        Integer[] a = { 1, 2, 5, 3, 4 };
        Vector A = new Vector(Arrays.asList(a));
 
        // Queries
        int[][] B = { { 4, 2, 4 }, { 3, 3, 0 }, { 1, 6, 0 }, { 4, 3, 5 } };
 
        // Get the answer array
        Vector ans = peformQueries(A, B);
 
        // printing the answer
        for (int i : ans)
            System.out.println(i);
    }
}
 
// This code is contributed by
// sanjeev2552


Python3
# Python implementation of the approach
 
# Size of the array (MAX)
N = int(2e5) + 10
 
# To store the sum of
# the array elements
bit = [0] * N
 
# To keep track of how many type 3
# queries have been performed before
# a particular index
idx = [0] * N
 
# Function to perform the point
# update in Fenwick tree
def update(index: int, val: int, bitt: list):
    while index < N:
        bitt[index] += val
        index += index & -index
 
# Function to return the sum
# of the elements A[1...idx]
# from BIT
def summation(index: int, bitt: list) -> int:
    res = 0
    while index > 0:
        res += bitt[index]
        index -= index & -index
    return res
 
# Function to perform the queries and
# return the answer vector
def performQueries(A: list, B: list) -> list:
    global bit, idx
 
    # For 1 bases indexing
    # insert 0 in the vector
    A.insert(0, 0)
 
    # Updated size of the vector
    n = len(A)
 
    # Updating the bit tree
    for i in range(1, n):
        update(i, A[i], bit)
 
    # Vector to store the answers
    # of range sum
    ans = []
 
    # Iterating in the query
    # vector
    for i in B:
        type = i[0]
        x = i[1]
        v = i[2]
 
        # If the query is of
        # type 1
        if type == 1:
 
            # Updating the tree
            # with x in the last
            update(n, x, bit)
 
            # Pushing back the value
            # in the original vector
            A.append(x)
 
            # Incrementing the size
            # of the vector by 1
            n += 1
 
        # If the query is of type 2
        elif type == 2:
 
            # Getting the updated index
            # in case of any query of
            # type 3 occured before it
            id = x + summation(x, idx)
 
            # Making the effect to that
            # value to 0 by subtracting
            # same value from the tree
            update(id, -A[id], bit)
 
            # Updating the A[id] to v
            A[id] = v
 
            # Now update the
            # bit by v at x
            update(id, v, bit)
 
        # If the query is of type 3
        elif type == 3:
 
            # Get the current index
            id = x + summation(x, idx)
 
            # Remove the effect of that
            # index from the bit tree
            update(id, -A[id], bit)
 
            # Update the idx tree
            # because one element has
            # been deleted
            update(x, 1, idx)
 
            # Update the idx val to 0
            A[id] = 0
 
        # If the query is of type 4
        else:
 
            # Get the updated range
            xx = x + summation(x, idx)
            vv = v + summation(v, idx)
 
            # Push_back the value
            ans.append(summation(vv, bit) - summation(xx - 1, bit))
    return ans
 
 
# Driver Code
if __name__ == "__main__":
    A = [1, 2, 5, 3, 4]
 
    # Queries
    B = [[4, 2, 4], [3, 3, 0], [1, 6, 0], [4, 3, 5]]
 
    # Get the answer array
    ans = performQueries(A, B)
 
    # printing the answer
    for i in ans:
        print(i)
 
# This code is contributed by
# sanjeev2552


输出:
10
13

时间复杂度: O(Q * logN + NlogN)