Sum of bitwise AND of all submatrices

Given a NxN matrix, the task is to find the sum of bit-wise AND of all of its rectangular sub-matrices.

Examples:

Input : arr[][] = {{1, 1, 1},
                   {1, 1, 1},
                   {1, 1, 1}}
Output : 36
Explanation: All the possible submatrices will have AND value 1.
Since, there are 36 submatrices in total, ans = 36

Input : arr[][] = {{9, 7, 4},
                   {8, 9, 2},
                   {11, 11, 5}}
Output : 135

Prerequisite: Number of rectangular submatrices of a binary matrix with all 1s.

Naive Solution: A simple solution is to generate all of the sub-matrices and find the required AND for each of them. The time complexity of this approach will be O(N6).

Efficient Approach: For the sake of better understanding, let’s assume that any bit of an element is represented by the variable ‘i’ and the variable ‘sum’ is used to store the final sum.

The idea here is, we will try to find the number of AND values(sub-matrices with bit-wise and(&)) with ith bit set. Let us suppose, there are ‘Si‘ number of sub-matrices with ith bit set. For, ith bit, sum can be updated as sum += (2i * Si).

For each bit ‘i’, create a boolean matrix set_bit which stores ‘1’ at an index (R, C) if ith bit of arr[R][C] is set. Otherwise, it stores ‘0’. Then, for this boolean array, we try to find the number of rectangular submatrices with all 1s(Si). For, ith bit, the final sum will be updated as:

sum += 2i * Si

Below is the implementation of the above approach:

C++

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// C++ program to find sum of Bit-wise AND
// of all submatrices
  
#include <iostream>
#include <stack>
  
using namespace std;
  
#define n 3
  
// Function to find prefix-count for each row
// from right to left
void findPrefixCount(int p_arr[][n], bool set_bit[][n])
{
    for (int i = 0; i < n; i++) {
        for (int j = n - 1; j >= 0; j--) {
            if (!set_bit[i][j])
                continue;
  
            if (j != n - 1)
                p_arr[i][j] += p_arr[i][j + 1];
  
            p_arr[i][j] += (int)set_bit[i][j];
        }
    }
}
  
// Function to find the number of submatrices
// with all 1s
int matrixAllOne(bool set_bit[][n])
{
    // Array to store required prefix count of 1s from
    // right to left for boolean array
    int p_arr[n][n] = { 0 };
  
    findPrefixCount(p_arr, set_bit);
  
    // Variable to store the final answer
    int ans = 0;
  
    // For each index of a column, determine the number
    // of sub-matrices starting from that index
    // and has all 1s
    for (int j = 0; j < n; j++) {
        int i = n - 1;
  
        // Stack to store elements and the count
        // of the numbers they popped
        // First part of pair is value of inserted element
        // Second part is count of the number of elements
        // pushed before with a greater value
        stack<pair<int, int> > q;
  
        // variable to store the number of submatrices
        // with all 1s
        int to_sum = 0;
  
        while (i >= 0) {
            int c = 0;
            while (q.size() != 0 and q.top().first > p_arr[i][j]) {
                to_sum -= (q.top().second + 1) * (q.top().first - p_arr[i][j]);
                c += q.top().second + 1;
                q.pop();
            }
  
            to_sum += p_arr[i][j];
            ans += to_sum;
  
            q.push({ p_arr[i][j], c });
            i--;
        }
    }
  
    return ans;
}
  
// Function to find the sum of Bitwise-AND
// of all submatrices
int sumAndMatrix(int arr[][n])
{
    int sum = 0;
  
    int mul = 1;
  
    for (int i = 0; i < 30; i++) {
        // matrix to store the status
        // of ith bit of each element
        // of matrix arr
        bool set_bit[n][n];
  
        for (int R = 0; R < n; R++)
            for (int C = 0; C < n; C++)
                set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0);
  
        sum += (mul * matrixAllOne(set_bit));
  
        mul *= 2;
    }
  
    return sum;
}
  
// Driver Code
int main()
{
    int arr[][n] = { { 9, 7, 4 },
                     { 8, 9, 2 },
                     { 11, 11, 5 } };
  
    cout << sumAndMatrix(arr);
  
    return 0;
}

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Java

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// Java program to find sum of Bit-wise AND 
// of all submatrices 
import java.util.*;
  
class GFG
{
  
static int n = 3
  
// Function to find prefix-count for  
// each row from right to left 
static void findPrefixCount(int p_arr[][], 
                            boolean set_bit[][]) 
    for (int i = 0; i < n; i++) 
    
        for (int j = n - 1; j >= 0; j--) 
        
            if (!set_bit[i][j]) 
                continue
  
            if (j != n - 1
                p_arr[i][j] += p_arr[i][j + 1]; 
  
            p_arr[i][j] += (set_bit[i][j]) ? 1 : 0
        
    
static class pair
{
    int first,second;
    pair(){}
      
    pair(int a, int b)
    {
        first = a;
        second = b;
    }
}
  
// Function to find the number of 
// submatrices with all 1s 
static int matrixAllOne(boolean set_bit[][]) 
    // Array to store required prefix count of 1s from 
    // right to left for boolean array 
    int p_arr[][] = new int[n][n]; 
      
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n;j++)
            p_arr[i][j] = 0;
      
    findPrefixCount(p_arr, set_bit); 
  
    // Variable to store the final answer 
    int ans = 0
  
    // For each index of a column, determine the number 
    // of sub-matrices starting from that index 
    // and has all 1s 
    for (int j = 0; j < n; j++) 
    
        int i = n - 1
  
        // Stack to store elements and the count 
        // of the numbers they popped 
        // First part of pair is value of inserted element 
        // Second part is count of the number of elements 
        // pushed before with a greater value 
        Stack<pair > q = new Stack<pair >(); 
  
        // variable to store the number of submatrices 
        // with all 1s 
        int to_sum = 0
  
        while (i >= 0)
        
            int c = 0
            while (q.size() != 0 && 
                    q.peek().first > p_arr[i][j]) 
            
                to_sum -= (q.peek().second + 1) *
                            (q.peek().first - p_arr[i][j]); 
                c += q.peek().second + 1
                q.pop(); 
            
  
            to_sum += p_arr[i][j]; 
            ans += to_sum; 
  
            q.push(new pair( p_arr[i][j], c )); 
            i--; 
        
    
    return ans; 
  
// Function to find sum of Bitwise-OR of 
// all submatrices 
static int sumAndMatrix(int arr[][]) 
    int sum = 0
  
    int mul = 1
  
    for (int i = 0; i < 30; i++) 
    
        // matrix to store the status 
        // of ith bit of each element 
        // of matrix arr 
        boolean set_bit[][] = new boolean[n][n]; 
  
        for (int R = 0; R < n; R++) 
            for (int C = 0; C < n; C++) 
                set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0); 
  
        sum += (mul * matrixAllOne(set_bit)); 
  
        mul *= 2
    
    return sum; 
  
// Driver Code 
public static void main(String args[])
    int arr[][] = { { 9, 7, 4 }, 
                    { 8, 9, 2 }, 
                    { 11, 11, 5 } }; 
  
    System.out.println( sumAndMatrix(arr)); 
}
  
// This code is contributed by Arnab Kundu 

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Python3

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# Python3 program to find sum of 
# Bitwise-AND of all submatrices 
  
# Function to find prefix-count for 
# each row from right to left 
def findPrefixCount(p_arr, set_bit): 
  
    for i in range(0, n):
        for j in range(n - 1, -1, -1): 
  
            if not set_bit[i][j]: 
                continue
            if j != n - 1
                p_arr[i][j] += p_arr[i][j + 1
  
            p_arr[i][j] += int(set_bit[i][j]) 
  
# Function to create a boolean matrix 
# set_bit which stores ‘1’ at an index 
# (R, C) if ith bit of arr[R][C] is set. 
def matrixAllOne(set_bit): 
  
    # Array to store prefix count of zeros 
    # from right to left for boolean array 
    p_arr = [[0 for i in range(n)] 
                for j in range(n)] 
  
    findPrefixCount(p_arr, set_bit) 
  
    # Variable to store the final answer 
    ans = 0
  
    # For each index of a column we 
    # will try to determine the number 
    # of sub-matrices starting from 
    # that index and has all 1s 
    for j in range(0, n): 
  
        i = n - 1
          
        # stack to store elements and the 
        # count of the numbers they popped 
  
        # First part of pair will be the 
        # value of inserted element. 
        # Second part will be the count 
        # of the number of elements pushed 
        # before with a greater value 
        q = [] 
  
        # Variable to store the number 
        # of submatrices with all 0s 
        to_sum = 0
          
        while i >= 0
  
            c = 0
            while (len(q) != 0 and
                   q[-1][0] > p_arr[i][j]): 
  
                to_sum -= ((q[-1][1] + 1) * 
                           (q[-1][0] - p_arr[i][j])) 
  
                c += q.pop()[1] + 1
  
            to_sum += p_arr[i][j] 
            ans += to_sum 
  
            q.append((p_arr[i][j], c)) 
            i -= 1
  
    # Return the final answer 
    return ans 
  
# Function to find sum of 
# Bitwise-AND of all submatrices 
def sumAndMatrix(arr): 
  
    Sum, mul = 0, 1
    for i in range(0, 30): 
      
        # matrix to store the status 
        # of ith bit of each element 
        # of matrix arr 
        set_bit = [[False for i in range(n)] 
                          for j in range(n)]
  
        for R in range(0, n): 
            for C in range(0, n): 
                set_bit[R][C] = ((arr[R][C] & 
                                 (1 << i)) != 0
  
        Sum += (mul * matrixAllOne(set_bit)) 
        mul *= 2
  
    return Sum
  
# Driver Code 
if __name__ == "__main__":
      
    n = 3
    arr = [[9, 7, 4], 
        [8, 9, 2], 
        [11, 11, 5]] 
  
    print(sumAndMatrix(arr)) 
  
# This code is contributed by Rituraj Jain

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C#

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// C# program to find sum of Bit-wise AND 
// of all submatrices
using System;
using System.Collections.Generic;
  
class GFG
{
  
static int n = 3; 
  
// Function to find prefix-count for 
// each row from right to left 
static void findPrefixCount(int [,]p_arr, 
                            bool [,]set_bit) 
    for (int i = 0; i < n; i++) 
    
        for (int j = n - 1; j >= 0; j--) 
        
            if (!set_bit[i, j]) 
                continue
  
            if (j != n - 1) 
                p_arr[i, j] += p_arr[i, j + 1]; 
  
            p_arr[i, j] += (set_bit[i, j]) ? 1 : 0; 
        
    
public class pair
{
    public int first,second;
    public pair(){}
      
    public pair(int a, int b)
    {
        first = a;
        second = b;
    }
}
  
// Function to find the number of 
// submatrices with all 1s 
static int matrixAllOne(bool [,]set_bit) 
    // Array to store required prefix count of 1s from 
    // right to left for boolean array 
    int [,]p_arr = new int[n, n]; 
      
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n;j++)
            p_arr[i, j] = 0;
      
    findPrefixCount(p_arr, set_bit); 
  
    // Variable to store the final answer 
    int ans = 0; 
  
    // For each index of a column, determine the number 
    // of sub-matrices starting from that index 
    // and has all 1s 
    for (int j = 0; j < n; j++) 
    
        int i = n - 1; 
  
        // Stack to store elements and the count 
        // of the numbers they popped 
        // First part of pair is value of inserted element 
        // Second part is count of the number of elements 
        // pushed before with a greater value 
        Stack<pair > q = new Stack<pair >(); 
  
        // variable to store the number of submatrices 
        // with all 1s 
        int to_sum = 0; 
  
        while (i >= 0)
        
            int c = 0; 
            while (q.Count != 0 && 
                    q.Peek().first > p_arr[i,j]) 
            
                to_sum -= (q.Peek().second + 1) *
                            (q.Peek().first - p_arr[i,j]); 
                c += q.Peek().second + 1; 
                q.Pop(); 
            
  
            to_sum += p_arr[i,j]; 
            ans += to_sum; 
  
            q.Push(new pair( p_arr[i,j], c )); 
            i--; 
        
    
    return ans; 
  
// Function to find sum of Bitwise-OR of 
// all submatrices 
static int sumAndMatrix(int [,]arr) 
    int sum = 0; 
  
    int mul = 1; 
  
    for (int i = 0; i < 30; i++) 
    
        // matrix to store the status 
        // of ith bit of each element 
        // of matrix arr 
        bool [,]set_bit = new bool[n,n]; 
  
        for (int R = 0; R < n; R++) 
            for (int C = 0; C < n; C++) 
                set_bit[R, C] = ((arr[R, C] & (1 << i)) != 0); 
  
        sum += (mul * matrixAllOne(set_bit)); 
  
        mul *= 2; 
    
    return sum; 
  
// Driver Code 
public static void Main(String []args)
    int [,]arr = { { 9, 7, 4 }, 
                    { 8, 9, 2 }, 
                    { 11, 11, 5 } }; 
  
    Console.WriteLine(sumAndMatrix(arr)); 
}
  
// This code contributed by Rajput-Ji

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Output:

135

Time Complexity: O(N2).



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