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Two Dimensional Binary Indexed Tree or Fenwick Tree

Prerequisite – Fenwick Tree

We know that to answer range sum queries on a 1-D array efficiently, binary indexed tree (or Fenwick Tree) is the best choice (even better than segment tree due to less memory requirements and a little faster than segment tree).

Can we answer sub-matrix sum queries efficiently using Binary Indexed Tree ?

The answer is yes. This is possible using a 2D BIT which is nothing but an array of 1D BIT.

Algorithm:



We consider the below example. Suppose we have to find the sum of all numbers inside the highlighted area-
fenwick tree

We assume the origin of the matrix at the bottom – O.Then a 2D BIT exploits the fact that-

  
Sum under the marked area = Sum(OB) - Sum(OD) - 
                            Sum(OA) + Sum(OC) 

fenwick tree

In our program, we use the getSum(x, y) function which finds the sum of the matrix from (0, 0) to (x, y).
Hence the below formula :

Sum under the marked area = Sum(OB) - Sum(OD) - 
                            Sum(OA) + Sum(OC) 

The above formula gets reduced to,

Query(x1,y1,x2,y2) = getSum(x2, y2) - 
                     getSum(x2, y1-1) - 
                     getSum(x1-1, y2) + 
                     getSum(x1-1, y1-1) 

where,
x1, y1 = x and y coordinates of C
x2, y2 = x and y coordinates of B

The updateBIT(x, y, val) function updates all the elements under the region – (x, y) to (N, M) where,
N = maximum X co-ordinate of the whole matrix.
M = maximum Y co-ordinate of the whole matrix.

The rest procedure is quite similar to that of 1D Binary Indexed Tree. Below is the C++ implementation of 2D indexed tree

/* C++ program to implement 2D Binary Indexed Tree
  
2D BIT is basically a BIT where each element is another BIT.
Updating by adding v on (x, y) means it's effect will be found
throughout the rectangle [(x, y), (max_x, max_y)],
and query for (x, y) gives you the result of the rectangle
[(0, 0), (x, y)], assuming the total rectangle is
[(0, 0), (max_x, max_y)]. So when you query and update on
this BIT,you have to be careful about how many times you are
subtracting a rectangle and adding it. Simple set union formula
works here.
  
So if you want to get the result of a specific rectangle
[(x1, y1), (x2, y2)], the following steps are necessary:
  
Query(x1,y1,x2,y2) = getSum(x2, y2)-getSum(x2, y1-1) -
                     getSum(x1-1, y2)+getSum(x1-1, y1-1)
  
Here 'Query(x1,y1,x2,y2)' means the sum of elements enclosed
in the rectangle with bottom-left corner's co-ordinates
(x1, y1) and top-right corner's co-ordinates - (x2, y2)
  
Constraints -> x1<=x2 and y1<=y2
  
    /
 y  |
    |           --------(x2,y2)
    |          |       |
    |          |       |
    |          |       |
    |          ---------
    |       (x1,y1)
    |
    |___________________________
   (0, 0)                   x-->
  
In this progrm we have assumed a square matrix. The
program can be easily extended to a rectangular one. */
  
#include<bits/stdc++.h>
using namespace std;
  
#define N 4 // N-->max_x and max_y
  
// A structure to hold the queries
struct Query
{
    int x1, y1; // x and y co-ordinates of bottom left
    int x2, y2; // x and y co-ordinates of top right
};
  
// A function to update the 2D BIT
void updateBIT(int BIT[][N+1], int x, int y, int val)
{
    for (; x <= N; x += (x & -x))
    {
        // This loop update all the 1D BIT inside the
        // array of 1D BIT = BIT[x]
        for (; y <= N; y += (y & -y))
            BIT[x][y] += val;
    }
    return;
}
  
// A function to get sum from (0, 0) to (x, y)
int getSum(int BIT[][N+1], int x, int y)
{
    int sum = 0;
  
    for(; x > 0; x -= x&-x)
    {
        // This loop sum through all the 1D BIT
        // inside the array of 1D BIT = BIT[x]
        for(; y > 0; y -= y&-y)
        {
            sum += BIT[x][y];
        }
    }
    return sum;
}
  
// A function to create an auxiliary matrix
// from the given input matrix
void constructAux(int mat[][N], int aux[][N+1])
{
    // Initialise Auxiliary array to 0
    for (int i=0; i<=N; i++)
        for (int j=0; j<=N; j++)
            aux[i][j] = 0;
  
    // Construct the Auxiliary Matrix
    for (int j=1; j<=N; j++)
        for (int i=1; i<=N; i++)
            aux[i][j] = mat[N-j][i-1];
  
    return;
}
  
// A function to construct a 2D BIT
void construct2DBIT(int mat[][N], int BIT[][N+1])
{
    // Create an auxiliary matrix
    int aux[N+1][N+1];
    constructAux(mat, aux);
  
    // Initialise the BIT to 0
    for (int i=1; i<=N; i++)
        for (int j=1; j<=N; j++)
            BIT[i][j] = 0;
  
    for (int j=1; j<=N; j++)
    {
        for (int i=1; i<=N; i++)
        {
            // Creating a 2D-BIT using update function
            // everytime we/ encounter a value in the
            // input 2D-array
            int v1 = getSum(BIT, i, j);
            int v2 = getSum(BIT, i, j-1);
            int v3 = getSum(BIT, i-1, j-1);
            int v4 = getSum(BIT, i-1, j);
  
            // Assigning a value to a particular element
            // of 2D BIT
            updateBIT(BIT, i, j, aux[i][j]-(v1-v2-v4+v3));
        }
    }
  
    return;
}
  
// A function to answer the queries
void answerQueries(Query q[], int m, int BIT[][N+1])
{
    for (int i=0; i<m; i++)
    {
        int x1 = q[i].x1 + 1;
        int y1 = q[i].y1 + 1;
        int x2 = q[i].x2 + 1;
        int y2 = q[i].y2 + 1;
  
        int ans = getSum(BIT, x2, y2)-getSum(BIT, x2, y1-1)-
                  getSum(BIT, x1-1, y2)+getSum(BIT, x1-1, y1-1);
  
        printf ("Query(%d, %d, %d, %d) = %d ",
                q[i].x1, q[i].y1, q[i].x2, q[i].y2, ans);
    }
    return;
}
  
// Driver program
int main()
{
    int mat[N][N] = {{1, 2, 3, 4},
                    {5, 3, 8, 1},
                    {4, 6, 7, 5},
                    {2, 4, 8, 9}};
  
    // Create a 2D Binary Indexed Tree
    int BIT[N+1][N+1];
    construct2DBIT(mat, BIT);
  
    /* Queries of the form - x1, y1, x2, y2
       For example the query- {1, 1, 3, 2} means the sub-matrix-
    y
    /
 3  |       1 2 3 4      Sub-matrix
 2  |       5 3 8 1      {1,1,3,2}      --->     3 8 1
 1  |       4 6 7 5                                 6 7 5
 0  |       2 4 8 9
    |
  --|------ 0 1 2 3 ----> x
    |
  
    Hence sum of the sub-matrix = 3+8+1+6+7+5 = 30
  
    */
  
    Query q[] = {{1, 1, 3, 2}, {2, 3, 3, 3}, {1, 1, 1, 1}};
    int m = sizeof(q)/sizeof(q[0]);
  
    answerQueries(q, m, BIT);
  
    return(0);
}

Output:

Query(1, 1, 3, 2) = 30
Query(2, 3, 3, 3) = 7
Query(1, 1, 1, 1) = 6

Time Complexity:

  • Both updateBIT(x, y, val) function and getSum(x, y) function takes O(log(NM)) time.
  • Building the 2D BIT takes O(NM log(NM)).
  • Since in each of the queries we are calling getSum(x, y) function so answering all the Q queries takes O(Q.log(NM)) time.

Hence the overall time complexity of the program is O((NM+Q).log(NM)) where,
N = maximum X co-ordinate of the whole matrix.
M = maximum Y co-ordinate of the whole matrix.
Q = Number of queries.

Auxiliary Space: O(NM) to store the BIT and the auxiliary array

References: https://www.topcoder.com/community/data-science/data-science-tutorials/binary-indexed-trees/

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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