Find the largest rectangle of 1’s with swapping of columns allowed
Given a matrix with 0 and 1’s, find the largest rectangle of all 1’s in the matrix. The rectangle can be formed by swapping any pair of columns of given matrix.
Example:
Input: bool mat[][] = { {0, 1, 0, 1, 0},
{0, 1, 0, 1, 1},
{1, 1, 0, 1, 0}
};
Output: 6
The largest rectangle's area is 6. The rectangle
can be formed by swapping column 2 with 3
The matrix after swapping will be
0 0 1 1 0
0 0 1 1 1
1 0 1 1 0
Input: bool mat[R][C] = { {0, 1, 0, 1, 0},
{0, 1, 1, 1, 1},
{1, 1, 1, 0, 1},
{1, 1, 1, 1, 1}
};
Output: 9
The idea is to use an auxiliary matrix to store count of consecutive 1’s in every column. Once we have these counts, we sort all rows of auxiliary matrix in non-increasing order of counts. Finally traverse the sorted rows to find the maximum area.
Note : After forming the auxiliary matrix each row becomes independent, hence we can swap or sort each row independently.It is because we can only swap columns, so we have made each row independent and find the max area of rectangle possible with row and column.
Below are detailed steps for first example mentioned above.
Step 1: First of all, calculate no. of consecutive 1’s in every column. An auxiliary array hist[][] is used to store the counts of consecutive 1’s. So for the above first example, contents of hist[R][C] would be
0 1 0 1 0
0 2 0 2 1
1 3 0 3 0
Time complexity of this step is O(R*C)
Step 2: Sort the rows in non-increasing fashion. After sorting step the matrix hist[][] would be
1 1 0 0 0
2 2 1 0 0
3 3 1 0 0
This step can be done in O(R * (R + C)). Since we know that the values are in range from 0 to R, we can use counting sort for every row.
The sorting is actually the swapping of columns. If we look at the 3rd row under step 2:
3 3 1 0 0
The sorted row corresponds to swapping the columns so that the column with the highest possible rectangle is placed first, after that comes the column that allows the second highest rectangle and so on. So, in the example there are 2 columns that can form a rectangle of height 3. That makes an area of 3*2=6. If we try to make the rectangle wider the height drops to 1, because there are no columns left that allow a higher rectangle on the 3rd row.
Step 3: Traverse each row of hist[][] and check for the max area. Since every row is sorted by count of 1’s, current area can be calculated by multiplying column number with value in hist[i][j]. This step also takes O(R * C) time.
Below is the implementation based of above idea.
C++
#include <bits/stdc++.h>
#define R 3
#define C 5
using namespace std;
int maxArea( bool mat[R][C])
{
int hist[R + 1][C + 1];
for ( int i = 0; i < C; i++) {
hist[0][i] = mat[0][i];
for ( int j = 1; j < R; j++)
hist[j][i] = (mat[j][i] == 0) ? 0 : hist[j - 1][i] + 1;
}
for ( int i = 0; i < R; i++) {
int count[R + 1] = { 0 };
for ( int j = 0; j < C; j++)
count[hist[i][j]]++;
int col_no = 0;
for ( int j = R; j >= 0; j--) {
if (count[j] > 0) {
for ( int k = 0; k < count[j]; k++) {
hist[i][col_no] = j;
col_no++;
}
}
}
}
int curr_area, max_area = 0;
for ( int i = 0; i < R; i++) {
for ( int j = 0; j < C; j++) {
curr_area = (j + 1) * hist[i][j];
if (curr_area > max_area)
max_area = curr_area;
}
}
return max_area;
}
int main()
{
bool mat[R][C] = { { 0, 1, 0, 1, 0 },
{ 0, 1, 0, 1, 1 },
{ 1, 1, 0, 1, 0 } };
cout << "Area of the largest rectangle is " << maxArea(mat);
return 0;
}
|
Java
import java.util.*;
class GFG {
static final int R = 3 ;
static final int C = 5 ;
static int maxArea( int mat[][])
{
int hist[][] = new int [R + 1 ][C + 1 ];
for ( int i = 0 ; i < C; i++)
{
hist[ 0 ][i] = mat[ 0 ][i];
for ( int j = 1 ; j < R; j++)
{
hist[j][i] = (mat[j][i] == 0 ) ? 0 : hist[j - 1 ][i] + 1 ;
}
}
for ( int i = 0 ; i < R; i++)
{
int count[] = new int [R + 1 ];
for ( int j = 0 ; j < C; j++)
{
count[hist[i][j]]++;
}
int col_no = 0 ;
for ( int j = R; j >= 0 ; j--)
{
if (count[j] > 0 )
{
for ( int k = 0 ; k < count[j]; k++)
{
hist[i][col_no] = j;
col_no++;
}
}
}
}
int curr_area, max_area = 0 ;
for ( int i = 0 ; i < R; i++)
{
for ( int j = 0 ; j < C; j++)
{
curr_area = (j + 1 ) * hist[i][j];
if (curr_area > max_area)
{
max_area = curr_area;
}
}
}
return max_area;
}
public static void main(String[] args)
{
int mat[][] = {{ 0 , 1 , 0 , 1 , 0 },
{ 0 , 1 , 0 , 1 , 1 },
{ 1 , 1 , 0 , 1 , 0 }};
System.out.println( "Area of the largest rectangle is " + maxArea(mat));
}
}
|
Python3
R = 3
C = 5
def maxArea(mat):
hist = [[ 0 for i in range (C + 1 )]
for i in range (R + 1 )]
for i in range ( 0 , C, 1 ):
hist[ 0 ][i] = mat[ 0 ][i]
for j in range ( 1 , R, 1 ):
if ((mat[j][i] = = 0 )):
hist[j][i] = 0
else :
hist[j][i] = hist[j - 1 ][i] + 1
for i in range ( 0 , R, 1 ):
count = [ 0 for i in range (R + 1 )]
for j in range ( 0 , C, 1 ):
count[hist[i][j]] + = 1
col_no = 0
j = R
while (j > = 0 ):
if (count[j] > 0 ):
for k in range ( 0 , count[j], 1 ):
hist[i][col_no] = j
col_no + = 1
j - = 1
max_area = 0
for i in range ( 0 , R, 1 ):
for j in range ( 0 , C, 1 ):
curr_area = (j + 1 ) * hist[i][j]
if (curr_area > max_area):
max_area = curr_area
return max_area
if __name__ = = '__main__' :
mat = [[ 0 , 1 , 0 , 1 , 0 ],
[ 0 , 1 , 0 , 1 , 1 ],
[ 1 , 1 , 0 , 1 , 0 ]]
print ( "Area of the largest rectangle is" ,
maxArea(mat))
|
C#
using System;
class GFG
{
static readonly int R = 3;
static readonly int C = 5;
static int maxArea( int [,]mat)
{
int [,]hist = new int [R + 1, C + 1];
for ( int i = 0; i < C; i++)
{
hist[0, i] = mat[0, i];
for ( int j = 1; j < R; j++)
{
hist[j, i] = (mat[j, i] == 0) ? 0 :
hist[j - 1, i] + 1;
}
}
for ( int i = 0; i < R; i++)
{
int []count = new int [R + 1];
for ( int j = 0; j < C; j++)
{
count[hist[i, j]]++;
}
int col_no = 0;
for ( int j = R; j >= 0; j--)
{
if (count[j] > 0)
{
for ( int k = 0; k < count[j]; k++)
{
hist[i, col_no] = j;
col_no++;
}
}
}
}
int curr_area, max_area = 0;
for ( int i = 0; i < R; i++)
{
for ( int j = 0; j < C; j++)
{
curr_area = (j + 1) * hist[i, j];
if (curr_area > max_area)
{
max_area = curr_area;
}
}
}
return max_area;
}
public static void Main()
{
int [,]mat = {{0, 1, 0, 1, 0},
{0, 1, 0, 1, 1},
{1, 1, 0, 1, 0}};
Console.WriteLine( "Area of the largest rectangle is " +
maxArea(mat));
}
}
|
Javascript
<script>
var R = 3;
var C = 5;
function maxArea(mat)
{
var hist = Array.from(Array(R+1), ()=>Array(C+1));
for ( var i = 0; i < C; i++)
{
hist[0][i] = mat[0][i];
for ( var j = 1; j < R; j++)
{
hist[j][i] = (mat[j][i] == 0) ? 0 :
hist[j - 1][i] + 1;
}
}
for ( var i = 0; i < R; i++)
{
var count = Array(R+1).fill(0);
for ( var j = 0; j < C; j++)
{
count[hist[i][j]]++;
}
var col_no = 0;
for ( var j = R; j >= 0; j--)
{
if (count[j] > 0)
{
for ( var k = 0; k < count[j]; k++)
{
hist[i][col_no] = j;
col_no++;
}
}
}
}
var curr_area, max_area = 0;
for ( var i = 0; i < R; i++)
{
for ( var j = 0; j < C; j++)
{
curr_area = (j + 1) * hist[i][j];
if (curr_area > max_area)
{
max_area = curr_area;
}
}
}
return max_area;
}
var mat = [[0, 1, 0, 1, 0],
[0, 1, 0, 1, 1],
[1, 1, 0, 1, 0]];
document.write( "Area of the largest rectangle is " +
maxArea(mat));
</script>
|
Output
Area of the largest rectangle is 6
Time complexity of above solution is O(R * (R + C)) where R is number of rows and C is number of columns in input matrix. Extra space: O(R * C)
Last Updated :
12 Dec, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...