Queries for bitwise AND in the given matrix

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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 i^th bit of our answer to be set is that i^th 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 i^th bit set. If it is equal to the total elements of the sub-matrix then the i^th 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 i^th bit set. Refer
this article to understand prefix_count in case of matrix.

Below is the implementation of the above approach:

C++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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 */

Javascript

<script>
 
// Javascript implementation of the approach 
let bitscount = 32;
let n = 3 ;
 
// Array to store bit-wise 
// prefix count 
let prefix_count = new Array(bitscount); 
for(let i = 0; i < bitscount; i++)
{
    prefix_count[i] = new Array(n);
    for(let j = 0; j < n; j++)
    {
        prefix_count[i][j] = new Array(n);
        for(let k = 0; k < n; k++)
        {
            prefix_count[i][j][k] = 0;
        }
    }
}
 
// Function to find the prefix sum 
function findPrefixCount(arr) 
{ 
     
    // Loop for each bit 
    for(let i = 0; i < bitscount; i++) 
    { 
         
        // Loop to find prefix-count 
        // for each row 
        for(let j = 0; j < n; j++)
        { 
            prefix_count[i][j][0] = ((arr[j][0] >> i) & 1); 
            for(let 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(let i = 0; i < bitscount; i++) 
        for(let j = 1; j < n; j++) 
            for(let k = 0; k < n; k++) 
                prefix_count[i][j][k] += 
                prefix_count[i][j - 1][k]; 
} 
 
// Function to return the result for a query 
function rangeAnd(x1, y1, x2, y2) 
{ 
     
    // To store the answer 
    let ans = 0; 
 
    // Loop for each bit 
    for(let i = 0; i < bitscount; i++)
    { 
         
        // To store the number of variables 
        // with ith bit set 
        let 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
let arr = [ [ 1, 2, 3 ], 
            [ 4, 5, 6 ], 
            [ 7, 8, 9 ] ]; 
 
findPrefixCount(arr); 
 
let queries = [ [ 1, 1, 1, 1 ], 
                [ 1, 2, 2, 2 ] ]; 
let q = queries.length; 
 
for(let i = 0; i < q; i++) 
    document.write(rangeAnd(queries[i][0], 
                            queries[i][1], 
                            queries[i][2], 
                            queries[i][3]) + "</br>");
                             
// This code is contributed by divyeshrabadiya07
 
</script>

Output:

5
0

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

Auxiliary Space: O(n²)

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