Number of submatrices with all 1s

Given a N*N matrix containing only 0s and 1s, the task is to count the number of submatrices containing all 1s.

Examples:

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

Input : {{1, 1, 1},
                   {1, 0, 1},
                   {1, 1, 1}}
Output : 20

For simplicity, we will say a sub-matrix starts from an index (R, C), if that particular index is its top-left corner.

A Simple Solution will be to generate all the possible sub-matrices and then check if all the values inside them are 1. If for a sub-matrix, all the elements are one, we increment the value of the final answer for one. The time complexity of the above approach is O(n6).

Better Approach: To optimize the process, for every index of the matrix, we will try to find the number of submatrices starting from that index having all 1s in it.

Our first step towards solving this problem is creating a matrix ‘p_arr’.

For each index (R, C), if arr[R][C] equals 1, then in p_arr[R][C], we will store the number of 1s to the right of the cell(R, C) along row ‘R’ before we encounter first zero or end of the array plus 1.
If arr[R][C] equals zero, the p_arr[R][C] also equals zero.

For creating this matrix, we will use the following recurrence relation.

IF arr[R][C] is not 0
    p_arr[R][C] = p_arr[R][C+1] + 1
ELSE 
    p_arr[R][C] = 0

arr[][] = {{1, 0, 1, 1},
           {0, 1, 0, 1},
           {1, 1, 1, 0},
           {1, 0, 1, 1}}
p_arr[][] for above will look like
          {{1, 0, 2, 1},
           {0, 1, 0, 1},
           {3, 2, 1, 0},
           {1, 0, 2, 1}}

Once, we have the required matrix p_arr, we will proceed towards the next step. Look at the matrix ‘p_arr’ column-wise. If we are processing jth column of the matrix p_arr, then for each element ‘i’ of this column, we will try to find the number of sub-matrices starting from cell (i, j) with all 1s.

For this, we can use stack data structure.

Algorithm:

  1. Initialize a stack ‘q’ to store the value of the elements getting pushed along with the count(Cij) of the number of elements that were pushed in this stack with a value strictly greater than the value of the current element. We will use a pair to tie up the two data together.
    Initialize a variable ‘to_sum’ with 0. At each step, this variable is updated to store the number of submatrices with all 1s starting from the element being pushed at that step. Thus, using ‘to_sum’, we update the count of number of submatrices with all 1s at each step.
  2. For a column ‘j’, at any step ‘i’, we will prepare to push p_arr[i][j] in the stack. Let Qt represent the topmost element of the stack and Ct represent the number of elements previously pushed in the stack with a value greater than the top-most element of the stack. Before pushing an element ‘p_arr[i][j]’ in the stack, while the stack is not empty or topmost element is greater than the number to be pushed, keep popping the topmost element of the stack and at the same time update to_sum as to_sum += (Ct + 1) * (Qt – p_arr[i][j]). Let Ci, jrepresent the number of elements greater than the current element that were pushed in this stack previously. We also need to keep a track of Ci, j. Thus, before popping an element, we update Ci, j as Ci, j += Ct along with to_sum.
  3. We update the answer as ans += to_sum.
  4. Finally, we push that element in the stack after pairing it with Ci, j.

Create the prefix-array in O(n2) and for each column, we push an element in the stack or pop it out only once. Thus, the time complexity of this algorithm is O(n2).

Below is the implementation of the above approach:

C++

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// C++ program to count number of
// sub-matrices with all 1s
  
#include <iostream>
#include <stack>
  
using namespace std;
  
#define n 3
  
// Function to find required prefix-count for
// each row from right to left
void findPrefixCount(int p_arr[][n], bool arr[][n])
{
    for (int i = 0; i < n; i++) {
        for (int j = n - 1; j >= 0; j--) {
  
            if (!arr[i][j])
                continue;
  
            if (j != n - 1)
                p_arr[i][j] += p_arr[i][j + 1];
  
            p_arr[i][j] += (int)arr[i][j];
        }
    }
}
  
// Function to count the number of
// sub-matrices with all 1s
int matrixAllOne(bool arr[][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, arr);
  
    // variable to store the final answer
    int ans = 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 (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 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;
}
  
// Driver Code
int main()
{
    bool arr[][n] = { { 1, 1, 0 },
                      { 1, 0, 1 },
                      { 0, 1, 1 } };
  
    cout << matrixAllOne(arr);
  
    return 0;
}

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Python3

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# Python3 program to count the number 
# of sub-matrices with all 1s 
  
# Function to find required prefix-count 
# for each row from right to left 
def findPrefixCount(p_arr, arr): 
  
    for i in range(0, n): 
        for j in range(n - 1, -1, -1): 
  
            if not arr[i][j]: 
                continue
  
            if j != n - 1
                p_arr[i][j] += p_arr[i][j + 1
  
            p_arr[i][j] += arr[i][j] 
          
# Function to count the number of 
# sub-matrices with all 1s 
def matrixAllOne(arr): 
  
    # Array to store required prefix count of 
    # 1s 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 final answer 
    ans = 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 1s 
        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[-1][1] + 1
                q.pop() 
              
            to_sum += p_arr[i][j] 
            ans += to_sum 
  
            q.append((p_arr[i][j], c)) 
            i -= 1
          
    return ans 
  
# Driver Code 
if __name__ == "__main__":
  
    arr = [[1, 1, 0], [1, 0, 1], [0, 1, 1]] 
                  
    n = 3
    print(matrixAllOne(arr)) 
  
# This code is contributed by Rituraj Jain

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Output:

10

Time Complexity: O(N2)



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