Maximum size rectangle binary sub-matrix with all 1s

4.3

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

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

We have discussed a dynamic programming based solution for finding largest square with 1s.

In this post an interesting method is discussed that uses largest rectangle under histogram as a subroutine. Below are steps. The idea is to update each column of a given row with corresponding column of previous row and find largest histogram area for for that row.

Step 1: Find maximum area for row[0]
Step 2:
    for each row in 1 to N - 1
        for each column in that row
            if A[row][column] == 1
              update A[row][column] with
                A[row][column] += A[row - 1][column]
    find area for that row
    and update maximum area so far 

Illustration :

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

Below is the implementation. It is strongly recommended to refer this post first as most of the code taken from there.

C++

// 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.
// http://www.geeksforgeeks.org/largest-rectangle-under-histogram/
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, 1, 1},
                   {1, 1, 0, 0},
                 };

    cout << "Area of maximum rectangle is "
         << maxRectangle(A);

    return 0;
}

Java

// Java program to find largest rectangle with all 1s
// in a binary matrix
import java.io.*;
import java.util.*;
 
class GFG 
{
    // Finds the maximum area under the histogram represented
    // by histogram.  See below article for details.
    // http://www.geeksforgeeks.org/largest-rectangle-under-histogram/
    static int maxHist(int R,int C,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<Integer> result = new Stack<Integer>();
     
        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.peek()] <= 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.peek()];
                result.pop();
                area = top_val * i;
     
                if (!result.empty())
                    area = top_val * (i - result.peek() - 1 );
                max_area = Math.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.peek()];
            result.pop();
            area = top_val * i;
            if (!result.empty())
                area = top_val * (i - result.peek() - 1 );
     
            max_area = Math.max(area, max_area);
        }
        return max_area;
    }

    // Returns area of the largest rectangle with all 1s in 
    // A[][]
    static int maxRectangle(int R,int C,int A[][])
    {
        // Calculate area for first row and initialize it as
        // result
        int result = maxHist(R,C,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] == 1) A[i][j] += A[i - 1][j];
     
     
            // Update result if area with current row (as last
            // row of rectangle) is more
            result = Math.max(result, maxHist(R,C,A[i]));
        }
     
        return result;
    }
     
    // Driver code
    public static void main (String[] args) 
    {
        int R = 4;
        int C = 4;

        int A[][] = { {0, 1, 1, 0},
                      {1, 1, 1, 1},
                      {1, 1, 1, 1},
                      {1, 1, 0, 0},
                    };
        System.out.print("Area of maximum rectangle is " + 
                                  maxRectangle(R,C,A));
    }
}
 
// Contributed by Prakriti Gupta


Output :
Area of maximum rectangle is 8

Time Complexity : O(R x X)

Asked in: AmazonMakeMyTripMicrosoft

This article is contributed by Sanjiv Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



4.3 Average Difficulty : 4.3/5.0
Based on 91 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.