Maximum size rectangle binary sub-matrix with all 1s

brahmajit-mohapatra-f8fe5582 · 25 May 2021 20:16

Hello Everyone,

Given a binary matrix, find the maximum size rectangle binary-sub-matrix with all 1’s.

Example:

Input: 0 1 1 0 1 1 1 1 1 1 1 1 1 1 0 0 Output : 1 1 1 1 1 1 1 1 Explanation : The largest rectangle with only 1’s is from (1, 0) to (2, 3) which is 1 1 1 1 1 1 1 1 Input: 0 1 1 1 1 1 0 1 1 Output: 1 1 1 1 1 1 Explanation : The largest rectangle with only 1’s is from (0, 1) to (2, 2) which is 1 1 1 1 1 1

If the height of bars of the histogram is given then the largest area of the histogram can be found. This way in each row, the largest area of bars of the histogram can be found. To get the largest rectangle full of 1’s, update the next row with the previous row and find the largest area under the histogram, i.e. consider each 1’s as filled squares and 0’s with an empty square and consider each row as the base.

Illustration:

Input : 0 1 1 0 1 1 1 1 1 1 1 1 1 1 0 0 Step 1: 0 1 1 0 maximum area = 2 Step 2: row 1 1 2 2 1 area = 4, maximum area becomes 4 row 2 2 3 3 2 area = 8, maximum area becomes 8 row 3 3 4 0 0 area = 6, maximum area remains 8

Algorithm:

Run a loop to traverse through the rows.
Now If the current row is not the first row then update the row as follows, if matrix[i][j] is not zero then matrix[i][j] = matrix[i-1][j] + matrix[i][j].
Find the maximum rectangular area under the histogram, consider the ith row as heights of bars of a histogram.
Do the previous two steps for all rows and print the maximum area of all the rows.

Implementation

// C++ program to find largest

// rectangle with all 1s

// in a binary matrix

#include <bits/stdc++.h>

using namespace std;

// Rows and columns in input matrix

#define R 4

#define C 4

// Finds the maximum area under

// the histogram represented

// by histogram. See below article for details.

int maxHist( int row[])

{

// Create an empty stack.

// The stack holds indexes of

// hist[] array/ The bars stored

// in stack are always

// in increasing order of their heights.

stack< int > result;

int top_val; // Top of stack

int max_area = 0; // Initialize max area in current

// row (or histogram)

int area = 0; // Initialize area with current top

// Run through all bars of given histogram (or row)

int i = 0;

while (i < C) {

// If this bar is higher than the bar on top stack,

// push it to stack

if (result.empty() || row[result.top()] <= row[i])

result.push(i++);

else {

// If this bar is lower than top of stack, then

// calculate area of rectangle with stack top as

// the smallest (or minimum height) bar. 'i' is

// 'right index' for the top and element before

// top in stack is 'left index'

top_val = row[result.top()];

result.pop();

area = top_val * i;

if (!result.empty())

area = top_val * (i - result.top() - 1);

max_area = max(area, max_area);

}

// Now pop the remaining bars from stack and calculate

// area with every popped bar as the smallest bar

while (!result.empty()) {

top_val = row[result.top()];

result.pop();

area = top_val * i;

if (!result.empty())

area = top_val * (i - result.top() - 1);

max_area = max(area, max_area);

}

return max_area;

}

// Returns area of the largest rectangle with all 1s in

// A[][]

int maxRectangle( int A[][C])

{

// Calculate area for first row and initialize it as

// result

int result = maxHist(A[0]);

// iterate over row to find maximum rectangular area

// considering each row as histogram

for ( int i = 1; i < R; i++) {

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

// if A[i][j] is 1 then add A[i -1][j]

if (A[i][j])

A[i][j] += A[i - 1][j];

// Update result if area with current row (as last

// row) of rectangle) is more

result = max(result, maxHist(A[i]));

}

return result;

}

// Driver code

int main()

{

int A[][C] = {

{ 0, 1, 1, 0 },

{ 1, 1, 1, 1 },

{ 1, 1, 0, 0 },

};

cout << "Area of maximum rectangle is "

<< maxRectangle(A);

return 0;

}

Output

Area of maximum rectangle is 8

Complexity Analysis:

Time Complexity: O(R x C).
Only one traversal of the matrix is required, so the time complexity is O(R X C)