Number of submatrices with all 1s - GeeksforGeeks (2022)

Table of Contents
C++ Java Python3 C# Javascript

Given an 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 : 36Explanation: All the possible submatrices will have only 1s.Since, there are 36 submatrices in total, ans = 36Input : {{1, 1, 1}, {1, 0, 1}, {1, 1, 1}}Output : 20 

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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] + 1ELSE p_arr[R][C] = 0arr[][] = {{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 the 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 the cell (i, j) with all 1s.
For this, we can use the 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 the 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, represent the number of elements greater than the current element that was 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++

#include <iostream>

#include <stack>

using namespace std;

#define n 3

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];

}

}

}

int matrixAllOne(bool arr[][n])

{

int p_arr[n][n] = { 0 };

findPrefixCount(p_arr, arr);

int ans = 0;

for (int j = 0; j < n; j++) {

int i = n - 1;

stack<pair<int, int> > q;

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;

}

int main()

{

bool arr[][n] = { { 1, 1, 0 },

{ 1, 0, 1 },

{ 0, 1, 1 } };

cout << matrixAllOne(arr);

return 0;

}

Java

import java.util.*;

class GFG

{

static int n = 3;

static class pair

{

int first, second;

public pair(int first, int second)

{

this.first = first;

this.second = second;

}

}

static void findPrefixCount(int p_arr[][],

boolean arr[][])

{

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] += arr[i][j] == true ? 1 : 0;

}

}

}

static int matrixAllOne(boolean arr[][])

{

int [][]p_arr = new int[n][n];

findPrefixCount(p_arr, arr);

int ans = 0;

for (int j = 0; j < n; j++)

{

int i = n - 1;

Stack<pair> q = new Stack<pair>();

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.add(new pair(p_arr[i][j], c));

i--;

}

}

return ans;

}

public static void main(String[] args)

{

boolean arr[][] = { { true, true, false },

{ true, false, true },

{ false, true, true } };

System.out.println(matrixAllOne(arr));

}

}

Python3

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]

def matrixAllOne(arr):

p_arr = [[0 for i in range(n)] for j in range(n)]

findPrefixCount(p_arr, arr)

ans = 0

for j in range(0, n):

i = n - 1

q = []

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

if __name__ == "__main__":

arr = [[1, 1, 0], [1, 0, 1], [0, 1, 1]]

n = 3

print(matrixAllOne(arr))

C#

using System;

using System.Collections.Generic;

class GFG

{

static int n = 3;

class pair

{

public int first, second;

public pair(int first, int second)

{

this.first = first;

this.second = second;

}

}

static void findPrefixCount(int [,]p_arr,

Boolean [,]arr)

{

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] += arr[i, j] == true ? 1 : 0;

}

}

}

static int matrixAllOne(Boolean [,]arr)

{

int [,]p_arr = new int[n, n];

findPrefixCount(p_arr, arr);

int ans = 0;

for (int j = 0; j < n; j++)

{

int i = n - 1;

Stack<pair> q = new Stack<pair>();

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;

}

public static void Main(String[] args)

{

Boolean [,]arr = {{ true, true, false },

{ true, false, true },

{ false, true, true }};

Console.WriteLine(matrixAllOne(arr));

}

}

Javascript

<script>

var n = 3;

function findPrefixCount(p_arr, arr)

{

for (var i = 0; i < n; i++) {

for (var 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] += arr[i][j];

}

}

}

function matrixAllOne(arr)

{

var p_arr = Array.from(Array(n), ()=>Array(n).fill(0));

findPrefixCount(p_arr, arr);

var ans = 0;

for (var j = 0; j < n; j++) {

var i = n - 1;

var q = [];

var to_sum = 0;

while (i >= 0) {

var c = 0;

while (q.length != 0 && q[q.length-1][0] > p_arr[i][j]) {

to_sum -= (q[q.length-1][1] + 1) *

(q[q.length-1][0] - p_arr[i][j]);

c += q[q.length-1][1] + 1;

q.pop();

}

to_sum += p_arr[i][j];

ans += to_sum;

q.push([ p_arr[i][j], c ]);

i--;

}

}

return ans;

}

var arr = [ [ 1, 1, 0 ],

[ 1, 0, 1 ],

[ 0, 1, 1 ] ];

document.write( matrixAllOne(arr));

</script>

Output:

10

Time Complexity: O(N2)
Auxiliary Space: O(N2), since N2 extra space has been taken.


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