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Perform append, update, delete and range sum queries on the given array
  • Last Updated : 20 Mar, 2020

Given an array arr[] of size N and the task is to answer Q queries of the following types:

  • 1 X 0: Append X at the back of array.
  • 2 X Y: Set arr[X] = Y.
  • 3 X 0: Delete arr[X].
  • 4 X Y: Find the sum in the range [X, Y].

Note that after deleting arr[X] in query type 3, all the elements after index X will be shifted left by one position.

Examples:

Input: arr[] = {1, 2, 5, 3, 4}, Q[][] = {
{4, 2, 4},
{3, 3, 0},
{1, 6, 0},
{4, 3, 5}}
Output:
10
13

First Query -> sum(arr[1], arr[2], arr[3])
Second Query -> arr[] = { 1, 2, 3, 4 }
Third Query -> arr[] = { 1, 2, 3, 4, 6 }
Fourth Query -> sum(arr[2], arr[3], arr[4])



Input: arr[] = {1, 2, 3, 4, 5}, Q[][] = {
{4, 1, 5},
{3, 3, 0},
{1, 6, 0},
{4, 3, 5},
{2, 4, 10}.
{4, 1, 5}}
Output:
15
15
23

Naive approach: The naive way to solve this problem is to execute the queries directly on the given array which will have a time complexity of O(Q * N).

Efficient approach:

  • As there are some data structures like Segment tree or BIT(Fenwick Tree) that provide the point update and range sum in O(logN) per queries.
  • There post uses a Fenwick Tree to do the same, so it is highly recommended to go through point update in Fenwick Tree.
  • But the main problem is to perform the delete operation ( type-3 queries ), after performing there is a need of shifting, that will again lead to O(Q * N) in the worst case.
  • A trick can be used that after deleting the element, just update the value at A[X] = 0 and map the index after X to X + number of elements deleted before X.
  • To find the number of elements deleted before X, there will be another fenwick tree ( as idx in the implementation ) used and keep adding 1 at that index where the delete operation is being performed.
  • Means at the time of fetching or getting a particular index, a query can be made to the idx tree and update index X to X + sum(X, idx).

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Size of the array (MAX)
const int N = 2e5 + 10;
  
// To store the sum of
// the array elements
vector<int> bit(N, 0);
  
// To keep track of how many type 3
// queries have been performed before
// a particular index
vector<int> idx(N, 0);
  
// Function to perform the point
// update in Fenwick tree
void update(int idx, int val,
            vector<int>& 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<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
vector<int> peformQueries(vector<int>& A,
                          vector<vector<int> >& 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<int> 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 vaule 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<int> A = { 1, 2, 5, 3, 4 };
  
    // Queries
    vector<vector<int> > B = {
        { 4, 2, 4 },
        { 3, 3, 0 },
        { 1, 6, 0 },
        { 4, 3, 5 },
    };
  
    // Get the answer array
    vector<int> 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<Integer> peformQueries(Vector<Integer> 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<Integer> 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 vaule 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<Integer> A = new Vector<Integer>(Arrays.asList(a));
  
        // Queries
        int[][] B = { { 4, 2, 4 }, { 3, 3, 0 }, { 1, 6, 0 }, { 4, 3, 5 } };
  
        // Get the answer array
        Vector<Integer> 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 vaule 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


Output:

10
13

Time Complexity: O(Q * logN + NlogN)

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