Maximum size rectangle binary sub-matrix with all 1s

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

There is already an algorithm discussed a dynamic programming based solution for finding largest square with 1s.

Approach: In this post an interesting method is discussed that uses largest rectangle under histogram as a subroutine.
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:

  1. Run a loop to traverse through the rows.
  2. 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].
  3. Find the maximum rectangular area under the histogram, consider the ith row as heights of bars of a histogram. This can be calculated as given in this article Largest Rectangular Area in a Histogram
  4. Do the previous two steps for all rows and print the maximum area of all the rows.

Note: It is strongly recommended to refer this post first as most of the code taken from there.
Implementation

C++

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// 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.
// https:// 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;
}

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Java

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// 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.
    // https:// 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

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Python3

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# Python3 program to find largest rectangle 
# with all 1s in a binary matrix 
  
# Rows and columns in input matrix 
R = 4
C = 4
  
# Finds the maximum area under the histogram represented 
# by histogram. See below article for details. 
# https:# www.geeksforgeeks.org / largest-rectangle-under-histogram / def maxHist(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. 
    result = []
  
    top_val = 0     # Top of stack 
  
    max_area = 0 # Initialize max area in current 
                 # row (or histogram) 
  
    area = 0 # Initialize area with current top 
  
    # Run through all bars of given
    # histogram (or row) 
    i = 0
    while (i < C): 
      
        # If this bar is higher than the 
        # bar on top stack, push it to stack 
        if (len(result) == 0) or (row[result[0]] <= row[i]):
            result.append(i)
            i += 1
        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[0]] 
            result.pop(0
            area = top_val *
  
            if (len(result)):
                area = top_val * (i - result[0] - 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 (len(result)):
        top_val = row[result[0]] 
        result.pop(0
        area = top_val *
        if (len(result)): 
            area = top_val * (i - result[0] - 1
  
        max_area = max(area, max_area) 
      
    return max_area 
  
# Returns area of the largest rectangle 
# with all 1s in A 
def maxRectangle(A):
      
    # Calculate area for first row and 
    # initialize it as result 
    result = maxHist(A[0]) 
  
    # iterate over row to find maximum rectangular 
    # area considering each row as histogram 
    for i in range(1, R):
      
        for j in range(C):
  
            # 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 
if __name__ == '__main__':
    A = [[0, 1, 1, 0],
         [1, 1, 1, 1], 
         [1, 1, 1, 1], 
         [1, 1, 0, 0]] 
  
    print("Area of maximum rectangle is"
                         maxRectangle(A))
      
# This code is contributed 
# by SHUBHAMSINGH10

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C#

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// C# program to find largest rectangle
// with all 1s in a binary matrix
using System;
using System.Collections.Generic;
  
class GFG {
    // Finds the maximum area under the
    // histogram represented by histogram.
    // See below article for details.
    // https:// www.geeksforgeeks.org/largest-rectangle-under-histogram/
    public 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<int> result = new Stack<int>();
  
        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.Count == 0 || 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.Count > 0) {
                    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.Count > 0) {
            top_val = row[result.Peek()];
            result.Pop();
            area = top_val * i;
            if (result.Count > 0) {
                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[][]
    public 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 = new int[][] {
            new int[] { 0, 1, 1, 0 },
            new int[] { 1, 1, 1, 1 },
            new int[] { 1, 1, 1, 1 },
            new int[] { 1, 1, 0, 0 }
        };
        Console.Write("Area of maximum rectangle is " + maxRectangle(R, C, A));
    }
}
  
// This code is contributed by Shrikant13

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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)
  • Space Complexity: O(C).
    Stack is required to store the columns, so so space complexity is O(C)

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.

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