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Sum of Bitwise-OR of all Submatrices

  • Difficulty Level : Hard
  • Last Updated : 13 Jul, 2021

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

Input : arr[][] = {{1, 0, 0},
                   {0, 0, 0},
                   {0, 0, 0}}
Output : 9
Explanation: All the submatrices starting from 
the index (0, 0) will have OR value as 1.
Thus, ans = 9

Input : arr[][] = {{9, 7, 4},
                    {8, 9, 2},
                    {11, 11, 5}}
Output : 398

 

Prerequisite: Number of submatrices with OR value 1
Simple Solution: A simple solution is to generate all the sub-matrices and find the required OR value for each of them. The time complexity of this approach will be O(N6).
Efficient Solution: 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 OR values(sub-matrices with bit-wise or( | )) 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 OR value 1(Si). For, ith bit, the final sum will be updated as: 
 

sum += 2i * Si

Below is the implementation of the above approach: 
 

C++




// C++ program to find sum of Bitwise-OR 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 create a boolean matrix set_bit which
// stores ‘1’ at an index (R, C) if ith bit
// of arr[R][C] is set.
int matrixOrValueOne(bool set_bit[][n])
{
    // array to store prefix count of zeros from
    // right to left for boolean array
    int p_arr[n][n] = { 0 };
 
    findPrefixCount(p_arr, set_bit);
 
    // variable to store the count of
    // submatrices with OR value 0
    int count_zero_submatrices = 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 (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 will be the
        // value of inserted element.
        // Second part will be the 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 0s
        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];
            count_zero_submatrices += to_sum;
            q.push({ p_arr[i][j], c });
 
            i--;
        }
    }
 
    return (n * (n + 1) * n * (n + 1)) /
                    4 - count_zero_submatrices;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
int sumOrMatrix(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 * matrixOrValueOne(set_bit));
 
        mul *= 2;
    }
 
    return sum;
}
 
// Driver Code
int main()
{
    int arr[][n] = { { 9, 7, 4 },
                     { 8, 9, 2 },
                     { 11, 11, 5 } };
 
    cout << sumOrMatrix(arr);
 
    return 0;
}

Java




// Java program to find sum of Bitwise-OR 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 create a booleanean
// matrix set_bit which stores '1'
// at an index (R, C) if ith bit
// of arr[R][C] is set.
static int matrixOrValueOne(boolean set_bit[][])
{
    // array to store prefix count of zeros from
    // right to left for booleanean 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 count of
    // submatrices with OR value 0
    int count_zero_submatrices = 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 (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 will be the
        // value of inserted element.
        // Second part will be the 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 0s
        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];
            count_zero_submatrices += to_sum;
            q.push(new pair( p_arr[i][j], c ));
 
            i--;
        }
    }
 
    return (n * (n + 1) * n * (n + 1)) /
                4 - count_zero_submatrices;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
static int sumOrMatrix(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 * matrixOrValueOne(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( sumOrMatrix(arr));
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python3 program to find sum of
# Bitwise-OR 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 set_bit[i][j]:
                continue
            if j != n - 1:
                p_arr[i][j] += p_arr[i][j + 1]
 
            p_arr[i][j] += int(not 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 matrixOrValueOne(arr):
 
    # 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, arr)
 
    # Variable to store the count of
    # submatrices with OR value 0
    count_zero_submatrices = 0
 
    # Loop to evaluate each column of
    # the prefix matrix uniquely.
    # 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]
            count_zero_submatrices += to_sum
 
            q.append((p_arr[i][j], c))
            i -= 1
 
    # Return the final answer
    return ((n * (n + 1) * n * (n + 1)) //
             4 - count_zero_submatrices)
 
# Function to find sum of
# Bitwise-OR of all submatrices
def sumOrMatrix(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 * matrixOrValueOne(set_bit))
        mul *= 2
 
    return Sum
 
# Driver Code
if __name__ == "__main__":
     
    n = 3
    arr = [[9, 7, 4],
        [8, 9, 2],
        [11, 11, 5]]
 
    print(sumOrMatrix(arr))
 
# This code is contributed by Rituraj Jain

C#




// C# program to find sum of Bitwise-OR 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,
                            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;
        }
    }
}
 
public class pair
{
    public int first,second;
    public pair(){}
     
    public pair(int a, int b)
    {
        first = a;
        second = b;
    }
}
 
// Function to create a booleanean
// matrix set_bit which stores '1'
// at an index (R, C) if ith bit
// of arr[R,C] is set.
static int matrixOrValueOne(Boolean [,]set_bit)
{
    // array to store prefix count of zeros from
    // right to left for booleanean 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 count of
    // submatrices with OR value 0
    int count_zero_submatrices = 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 (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 will be the
        // value of inserted element.
        // Second part will be the 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 0s
        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];
            count_zero_submatrices += to_sum;
            q.Push(new pair(p_arr[i, j], c));
 
            i--;
        }
    }
 
    return (n * (n + 1) * n * (n + 1)) /
            4 - count_zero_submatrices;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
static int sumOrMatrix(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 * matrixOrValueOne(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( sumOrMatrix(arr));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
// Javascript program to find sum of Bitwise-OR of
// all submatrices
 
let n = 3;
// Function to find prefix-count for 
// each row from right to left
function findPrefixCount(p_arr,set_bit)
{
    for (let i = 0; i < n; i++)
    {
        for (let 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;
        }
    }
}
 
class pair
{
    constructor(a,b)
    {
        this.first=a;
        this.second=b;
    }
}
 
// Function to create a booleanean
// matrix set_bit which stores '1'
// at an index (R, C) if ith bit
// of arr[R][C] is set.
function matrixOrValueOne(set_bit)
{
    // array to store prefix count of zeros from
    // right to left for booleanean array
    let p_arr = new Array(n);
     
       
    for(let i = 0; i < n; i++)
    {
        p_arr[i]=new Array(n);
        for(let j = 0; j < n; j++)
            p_arr[i][j] = 0;
      }
    findPrefixCount(p_arr, set_bit);
   
    // variable to store the count of
    // submatrices with OR value 0
    let count_zero_submatrices = 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 (let j = 0; j < n; j++)
    {
        let 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
        let q = [];
   
        // variable to store the number of submatrices
        // with all 0s
        let to_sum = 0;
        while (i >= 0)
        {
            let c = 0;
   
            while (q.length != 0 && q[q.length-1].first > p_arr[i][j])
            {
                to_sum -= (q[q.length-1].second + 1) *
                            (q[q.length-1].first - p_arr[i][j]);
                c += q[q.length-1].second + 1;
                q.pop();
            }
   
            to_sum += p_arr[i][j];
            count_zero_submatrices += to_sum;
            q.push(new pair( p_arr[i][j], c ));
   
            i--;
        }
    }
   
    return (n * (n + 1) * n * (n + 1)) /
                4 - count_zero_submatrices;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
function sumOrMatrix(arr)
{
    let sum = 0;
   
    let mul = 1;
   
    for (let i = 0; i < 30; i++)
    {
        // matrix to store the status
        // of ith bit of each element
        // of matrix arr
        let set_bit = new Array(n);
        for(let i=0;i<n;i++)
            set_bit[i]=new Array(n);
   
        for (let R = 0; R < n; R++)
            for (let C = 0; C < n; C++)
                set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0);
   
        sum += (mul * matrixOrValueOne(set_bit));
   
        mul *= 2;
    }
   
    return sum;
}
 
// Driver Code
let arr=[[ 9, 7, 4 ],
                    [ 8, 9, 2 ],
                    [ 11, 11, 5 ]];
document.write( sumOrMatrix(arr));
 
// This code is contributed by avanitrachhadiya2155
</script>
Output: 
398

 

Time Complexity: O(N2).
 


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