Largest Submatrix With Rearrangements of Columns Allowed
Last Updated : 17 Sep, 2024
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: 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: mat[][] =
{ {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> using namespace std; // Returns area of the largest rectangle of 1's int maxArea(vector<vector<bool>>& mat) { int R = mat.size(); int C = mat[0].size(); // Auxiliary matrix to store count of consecutive 1's in every column. vector<vector<int>> hist(R, vector<int>(C, 0)); // Step 1: Fill the auxiliary array hist[][] for (int i = 0; i < C; i++) { // First row in hist is a copy of the first row in mat hist[0][i] = mat[0][i]; // Fill remaining rows of hist for (int j = 1; j < R; j++) hist[j][i] = (mat[j][i] == 0) ? 0 : hist[j - 1][i] + 1; } // Step 2: Sort columns of hist[][] in non-increasing order for (int i = 0; i < R; i++) { vector<int> count(R + 1, 0); // Counting occurrence for (int j = 0; j < C; j++) count[hist[i][j]]++; // Traverse the count array from right side 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++; } } } } // Step 3: Traverse the sorted hist[][] to find maximum area int curr_area, max_area = 0; for (int i = 0; i < R; i++) { for (int j = 0; j < C; j++) { // Since values are in decreasing order, // The area ending with cell (i, j) can be obtained by // multiplying column number with value of hist[i][j] curr_area = (j + 1) * hist[i][j]; max_area = max(max_area, curr_area); } } return max_area; } // Driver program int main() { vector<vector<bool>> mat = { { 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; } // C program to find the largest rectangle of 1's with swapping // of columns allowed. #include <stdio.h> #include <stdbool.h> #define R 3 #define C 5 // Returns area of the largest rectangle of 1's int maxArea(bool mat[R][C]) { // An auxiliary array to store count of consecutive 1's // in every column. int hist[R + 1][C + 1]; // Step 1: Fill the auxiliary array hist[][] for (int i = 0; i < C; i++) { // First row in hist[][] is copy of first row in mat[][] hist[0][i] = mat[0][i]; // Fill remaining rows of hist[][] for (int j = 1; j < R; j++) hist[j][i] = (mat[j][i] == 0) ? 0 : hist[j - 1][i] + 1; } // Step 2: Sort columns of hist[][] in non-increasing order for (int i = 0; i < R; i++) { int count[R + 1] = { 0 }; // counting occurrence for (int j = 0; j < C; j++) count[hist[i][j]]++; // Traverse the count array from right side 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++; } } } } // Step 3: Traverse the sorted hist[][] to find maximum area int curr_area, max_area = 0; for (int i = 0; i < R; i++) { for (int j = 0; j < C; j++) { // Since values are in decreasing order, // The area ending with cell (i, j) can // be obtained by multiplying column number // with value of hist[i][j] curr_area = (j + 1) * hist[i][j]; if (curr_area > max_area) max_area = curr_area; } } return max_area; } // Driver program int main() { bool mat[R][C] = { { 0, 1, 0, 1, 0 }, { 0, 1, 0, 1, 1 }, { 1, 1, 0, 1, 0 } }; printf("Area of the largest rectangle is %d", maxArea(mat)); return 0; } public class GfG { // Returns the area of the largest rectangle of 1's public static int maxArea(int[][] mat) { int R = mat.length; int C = mat[0].length; // Auxiliary matrix to store the count of // consecutive 1's in every column. int[][] hist = new int[R][C]; // Step 1: Fill the auxiliary array hist[][] for (int i = 0; i < C; i++) { // First row in hist is a copy of the first row in mat hist[0][i] = mat[0][i]; // Fill remaining rows of hist for (int j = 1; j < R; j++) { hist[j][i] = (mat[j][i] == 0) ? 0 : hist[j - 1][i] + 1; } } // Step 2: Sort columns of hist[][] in // non-increasing order for (int i = 0; i < R; i++) { int[] count = new int[R + 1]; // Counting occurrence for (int j = 0; j < C; j++) { count[hist[i][j]]++; } // Traverse the count array from the right side 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++; } } } } // Step 3: Traverse the sorted hist[][] to find the maximum area int curr_area, max_area = 0; for (int i = 0; i < R; i++) { for (int j = 0; j < C; j++) { // The area ending with cell (i, j) curr_area = (j + 1) * hist[i][j]; max_area = Math.max(max_area, curr_area); } } return max_area; } // Driver program 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)); } } def maxArea(mat): R = len(mat) C = len(mat[0]) # Auxiliary matrix to store count of consecutive 1's in every column. hist = [[0] * C for _ in range(R)] # Step 1: Fill the auxiliary array hist[][] for i in range(C): # First row in hist is a copy of the first row in mat hist[0][i] = mat[0][i] # Fill remaining rows of hist for j in range(1, R): hist[j][i] = hist[j - 1][i] + 1 if mat[j][i] == 1 else 0 # Step 2: Sort columns of hist[][] in non-increasing order for i in range(R): count = [0] * (R + 1) # Counting occurrence for j in range(C): count[hist[i][j]] += 1 # Traverse the count array from the right side col_no = 0 for j in range(R, -1, -1): if count[j] > 0: for k in range(count[j]): hist[i][col_no] = j col_no += 1 # Step 3: Traverse the sorted hist[][] to find the maximum area max_area = 0 for i in range(R): for j in range(C): # The area ending with cell (i, j) curr_area = (j + 1) * hist[i][j] max_area = max(max_area, curr_area) return max_area # Driver code 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)) using System; class GfG { // Returns the area of the largest rectangle of 1's public static int MaxArea(int[,] mat) { int R = mat.GetLength(0); int C = mat.GetLength(1); // Auxiliary matrix to store count of consecutive 1's in every column. int[,] hist = new int[R, C]; // Step 1: Fill the auxiliary array hist[][] for (int i = 0; i < C; i++) { // First row in hist is a copy of the first row in mat hist[0, i] = mat[0, i]; // Fill remaining rows of hist for (int j = 1; j < R; j++) { hist[j, i] = (mat[j, i] == 0) ? 0 : hist[j - 1, i] + 1; } } // Step 2: Sort columns of hist[][] in non-increasing order for (int i = 0; i < R; i++) { int[] count = new int[R + 1]; // Counting occurrence for (int j = 0; j < C; j++) { count[hist[i, j]]++; } // Traverse the count array from the right side 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++; } } } } // Step 3: Traverse the sorted hist[][] to find the maximum area int max_area = 0; for (int i = 0; i < R; i++) { for (int j = 0; j < C; j++) { // The area ending with cell (i, j) int curr_area = (j + 1) * hist[i, j]; max_area = Math.Max(max_area, curr_area); } } return max_area; } // Driver code 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)); } } function maxArea(mat) { const R = mat.length; const C = mat[0].length; // Auxiliary matrix to store count of consecutive 1's in every column. let hist = Array.from({ length: R }, () => Array(C).fill(0)); // Step 1: Fill the auxiliary array hist[][] for (let i = 0; i < C; i++) { // First row in hist is a copy of the first row in mat hist[0][i] = mat[0][i]; // Fill remaining rows of hist for (let j = 1; j < R; j++) { hist[j][i] = (mat[j][i] === 1) ? hist[j - 1][i] + 1 : 0; } } // Step 2: Sort columns of hist[][] in non-increasing order for (let i = 0; i < R; i++) { let count = Array(R + 1).fill(0); // Counting occurrence for (let j = 0; j < C; j++) { count[hist[i][j]]++; } // Traverse the count array from the right side let col_no = 0; for (let j = R; j >= 0; j--) { if (count[j] > 0) { for (let k = 0; k < count[j]; k++) { hist[i][col_no] = j; col_no++; } } } } // Step 3: Traverse the sorted hist[][] to find the maximum area let max_area = 0; for (let i = 0; i < R; i++) { for (let j = 0; j < C; j++) { // The area ending with cell (i, j) let curr_area = (j + 1) * hist[i][j]; max_area = Math.max(max_area, curr_area); } } return max_area; } // Driver code let mat = [ [0, 1, 0, 1, 0], [0, 1, 0, 1, 1], [1, 1, 0, 1, 0] ]; console.log("Area of the largest rectangle is " + maxArea(mat)); OutputArea 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)
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