Queries for bitwise AND in the given matrix

Given an N * N matrix mat[][] consisting of non-negative integers and some queries consisting of top-left and bottom-right corner of the sub-matrix, the task is to find the bit-wise AND of all the elements of the sub-matrix given in each query.

Examples:

Input: mat[][] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}},
q[] = {{1, 1, 1, 1}, {1, 2, 2, 2}}
Output:
5
0
Query 1: Only element in the sub-matrix is 5.
Query 2: 6 AND 9 = 0

Input: mat[][] = {
{12, 23, 13},
{41, 15, 46},
{75, 82, 123}},
q[] = {{0, 0, 2, 2}, {1, 1, 2, 2}}
Output:
0
2

Naive approach: Iterate through the sub-matrix and find the bit-wise AND of all the numbers in that range. This will take O(n2) time for each query in the worst case.

Efficient approach: If we look at the integers as binary number, we can easily see that condition for ith bit of our answer to be set is that ith bit of all the integers in the sub-matrix should be set.
So, we will calculate prefix-count for each bit. We will use this to find the number of integers in the sub-matrix with ith bit set. If it is equal to the total elements of the sub-matrix then the ith bit of our answer will also be set.
For this, we will create a 3d-array, prefix_count[][][] where prefix_count[i][x][y] will store the count of all the elements of the sub-matrix with top left corner at {0, 0} and bottom right corner at {x, y} and ith bit set. Refer
this article to understand prefix_count in case of matrix.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
#define bitscount 32
#define n 3
using namespace std;
  
// Array to store bit-wise
// prefix count
int prefix_count[bitscount][n][n];
  
// Function to find the prefix sum
void findPrefixCount(int arr[][n])
{
  
    // Loop for each bit
    for (int i = 0; i < bitscount; i++) {
  
        // Loop to find prefix-count
        // for each row
        for (int j = 0; j < n; j++) {
            prefix_count[i][j][0] = ((arr[j][0] >> i) & 1);
            for (int k = 1; k < n; k++) {
                prefix_count[i][j][k] = ((arr[j][k] >> i) & 1);
                prefix_count[i][j][k] += prefix_count[i][j][k - 1];
            }
        }
    }
  
    // Finding column-wise prefix
    // count
    for (int i = 0; i < bitscount; i++)
        for (int j = 1; j < n; j++)
            for (int k = 0; k < n; k++)
                prefix_count[i][j][k] += prefix_count[i][j - 1][k];
}
  
// Function to return the result for a query
int rangeAnd(int x1, int y1, int x2, int y2)
{
  
    // To store the answer
    int ans = 0;
  
    // Loop for each bit
    for (int i = 0; i < bitscount; i++) {
  
        // To store the number of variables
        // with ith bit set
        int p;
        if (x1 == 0 and y1 == 0)
            p = prefix_count[i][x2][y2];
        else if (x1 == 0)
            p = prefix_count[i][x2][y2]
                - prefix_count[i][x2][y1 - 1];
        else if (y1 == 0)
            p = prefix_count[i][x2][y2]
                - prefix_count[i][x1 - 1][y2];
        else
            p = prefix_count[i][x2][y2]
                - prefix_count[i][x1 - 1][y2]
                - prefix_count[i][x2][y1 - 1]
                + prefix_count[i][x1 - 1][y1 - 1];
  
        // If count of variables whose ith bit
        // is set equals to the total
        // elements in the sub-matrix
        if (p == (x2 - x1 + 1) * (y2 - y1 + 1))
            ans = (ans | (1 << i));
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int arr[][n] = { { 1, 2, 3 },
                     { 4, 5, 6 },
                     { 7, 8, 9 } };
  
    findPrefixCount(arr);
  
    int queries[][4] = { { 1, 1, 1, 1 }, { 1, 2, 2, 2 } };
    int q = sizeof(queries) / sizeof(queries[0]);
  
    for (int i = 0; i < q; i++)
        cout << rangeAnd(queries[i][0],
                         queries[i][1],
                         queries[i][2],
                         queries[i][3])
             << endl;
  
    return 0;
}

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Java

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// Java implementation of the approach 
  
class GFG
{
  
    final static int bitscount = 32 ;
    final static int n = 3 ;
  
    // Array to store bit-wise 
    // prefix count 
    static int prefix_count[][][] = new int [bitscount][n][n]; 
      
    // Function to find the prefix sum 
    static void findPrefixCount(int arr[][]) 
    
      
        // Loop for each bit 
        for (int i = 0; i < bitscount; i++) 
        
      
            // Loop to find prefix-count 
            // for each row 
            for (int j = 0; j < n; j++)
            
                prefix_count[i][j][0] = ((arr[j][0] >> i) & 1); 
                for (int k = 1; k < n; k++) 
                
                    prefix_count[i][j][k] = ((arr[j][k] >> i) & 1); 
                    prefix_count[i][j][k] += prefix_count[i][j][k - 1]; 
                
            
        
      
        // Finding column-wise prefix 
        // count 
        for (int i = 0; i < bitscount; i++) 
            for (int j = 1; j < n; j++) 
                for (int k = 0; k < n; k++) 
                    prefix_count[i][j][k] += prefix_count[i][j - 1][k]; 
    
      
    // Function to return the result for a query 
    static int rangeAnd(int x1, int y1, int x2, int y2) 
    
      
        // To store the answer 
        int ans = 0
      
        // Loop for each bit 
        for (int i = 0; i < bitscount; i++)
        
      
            // To store the number of variables 
            // with ith bit set 
            int p; 
            if (x1 == 0 && y1 == 0
                p = prefix_count[i][x2][y2]; 
            else if (x1 == 0
                p = prefix_count[i][x2][y2] 
                    - prefix_count[i][x2][y1 - 1]; 
            else if (y1 == 0
                p = prefix_count[i][x2][y2] 
                    - prefix_count[i][x1 - 1][y2]; 
            else
                p = prefix_count[i][x2][y2] 
                    - prefix_count[i][x1 - 1][y2] 
                    - prefix_count[i][x2][y1 - 1
                    + prefix_count[i][x1 - 1][y1 - 1]; 
      
        // If count of variables whose ith bit 
        // is set equals to the total 
        // elements in the sub-matrix 
        if (p == (x2 - x1 + 1) * (y2 - y1 + 1)) 
            ans = (ans | (1 << i)); 
        
      
        return ans; 
    
      
    // Driver code 
    public static void main (String[] args)
    {
        int arr[][] = { { 1, 2, 3 }, 
                        { 4, 5, 6 }, 
                        { 7, 8, 9 } }; 
      
        findPrefixCount(arr); 
      
        int queries[][] = { { 1, 1, 1, 1 }, { 1, 2, 2, 2 } }; 
        int q = queries.length; 
      
        for (int i = 0; i < q; i++) 
            System.out.println( rangeAnd(queries[i][0], 
                            queries[i][1], 
                            queries[i][2], 
                            queries[i][3]) );
    }
}
  
// This code is contributed by AnkitRai

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Python3

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# Python 3 implementation of the approach
bitscount = 32
n = 3
  
# Array to store bit-wise
# prefix count
prefix_count = [[[0 for i in range(n)] for j in range(n)] for k in range(bitscount)]
  
# Function to find the prefix sum
def findPrefixCount(arr):
      
    # Loop for each bit
    for i in range(bitscount):
          
        # Loop to find prefix-count
        # for each row
        for j in range(n):
            prefix_count[i][j][0] = ((arr[j][0] >> i) & 1)
            for k in range(1,n):
                prefix_count[i][j][k] = ((arr[j][k] >> i) & 1)
                prefix_count[i][j][k] += prefix_count[i][j][k - 1]
  
    # Finding column-wise prefix
    # count
    for i in range(bitscount):
        for j in range(1,n):
            for k in range(n):
                prefix_count[i][j][k] += prefix_count[i][j - 1][k]
  
# Function to return the result for a query
def rangeOr(x1, y1, x2, y2):
      
    # To store the answer
    ans = 0
  
    # Loop for each bit
    for i in range(bitscount):
          
        # To store the number of variables
        # with ith bit set
        if (x1 == 0 and y1 == 0):
            p = prefix_count[i][x2][y2]
        elif (x1 == 0):
            p = prefix_count[i][x2][y2] - prefix_count[i][x2][y1 - 1]
        elif (y1 == 0):
            p = prefix_count[i][x2][y2] - prefix_count[i][x1 - 1][y2]
        else:
            p = prefix_count[i][x2][y2] - prefix_count[i][x1 - 1][y2] - prefix_count[i][x2][y1 - 1] + prefix_count[i][x1 - 1][y1 - 1];
  
        # If count of variables with ith bit
        # set is greater than 0
  
              
        if (p == (x2 - x1 + 1) * (y2 - y1 + 1)):
            ans = (ans | (1 << i))
  
    return ans
  
# Driver code
if __name__ == '__main__':
    arr = [[1, 2, 3],
            [4, 5, 6],
            [7, 8, 9]]
  
    findPrefixCount(arr)
    queries = [[1, 1, 1, 1],
                        [1, 2, 2, 2]]
    q = len(queries)
  
    for i in range(q):
        print(rangeOr(queries[i][0],queries[i][1],queries[i][2],queries[i][3]))
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# implementation of the approach
using System;
      
class GFG
{
  
    static int bitscount = 32 ;
    static int n = 3 ;
  
    // Array to store bit-wise 
    // prefix count 
    static int [,,]prefix_count = new int [bitscount,n,n]; 
      
    // Function to find the prefix sum 
    static void findPrefixCount(int [,]arr) 
    
      
        // Loop for each bit 
        for (int i = 0; i < bitscount; i++) 
        
      
            // Loop to find prefix-count 
            // for each row 
            for (int j = 0; j < n; j++)
            
                prefix_count[i,j,0] = ((arr[j,0] >> i) & 1); 
                for (int k = 1; k < n; k++) 
                
                    prefix_count[i, j, k] = ((arr[j,k] >> i) & 1); 
                    prefix_count[i, j, k] += prefix_count[i, j, k - 1]; 
                }  
            
        
      
        // Finding column-wise prefix 
        // count 
        for (int i = 0; i < bitscount; i++) 
            for (int j = 1; j < n; j++) 
                for (int k = 0; k < n; k++) 
                    prefix_count[i, j, k] += prefix_count[i, j - 1, k]; 
    
      
    // Function to return the result for a query 
    static int rangeAnd(int x1, int y1, int x2, int y2) 
    
      
        // To store the answer 
        int ans = 0; 
      
        // Loop for each bit 
        for (int i = 0; i < bitscount; i++)
        
      
            // To store the number of variables 
            // with ith bit set 
            int p; 
            if (x1 == 0 && y1 == 0) 
                p = prefix_count[i, x2, y2]; 
            else if (x1 == 0) 
                p = prefix_count[i, x2, y2] 
                    - prefix_count[i, x2, y1 - 1]; 
            else if (y1 == 0) 
                p = prefix_count[i, x2, y2] 
                    - prefix_count[i, x1 - 1, y2]; 
            else
                p = prefix_count[i, x2, y2] 
                    - prefix_count[i, x1 - 1, y2] 
                    - prefix_count[i, x2, y1 - 1] 
                    + prefix_count[i, x1 - 1, y1 - 1]; 
      
        // If count of variables whose ith bit 
        // is set equals to the total 
        // elements in the sub-matrix 
        if (p == (x2 - x1 + 1) * (y2 - y1 + 1)) 
            ans = (ans | (1 << i)); 
        
      
        return ans; 
    
      
    // Driver code 
    public static void Main (String[] args)
    {
        int [,]arr = { { 1, 2, 3 }, 
                        { 4, 5, 6 }, 
                        { 7, 8, 9 } }; 
      
        findPrefixCount(arr); 
      
        int [,]queries = { { 1, 1, 1, 1 }, { 1, 2, 2, 2 } }; 
        int q = queries.GetLength(0); 
      
        for (int i = 0; i < q; i++) 
            Console.WriteLine( rangeAnd(queries[i,0], 
                            queries[i,1], 
                            queries[i,2], 
                            queries[i,3]) );
    }
}
  
/* This code contributed by PrinciRaj1992 */

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Output:

5
0

Time complexity for pre-computation is O(n2) and each query can be answered in O(1)



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