# Maximum size rectangle binary sub-matrix with all 1s

Last Updated : 30 Apr, 2024

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

Example:

`Input:0 1 1 01 1 1 11 1 1 11 1 0 0Output :8Explanation : The largest rectangle with only 1's is from (1, 0) to (2, 3) which is1 1 1 11 1 1 1 Input:0 1 11 1 10 1 1      Output:6Explanation : The largest rectangle with only 1's is from (0, 1) to (2, 2) which is1 11 11 1`
Recommended Practice

There is already an algorithm discussed a dynamic programming based solution for finding the 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 01 1 1 11 1 1 11 1 0 0Step 1: 0 1 1 0  maximum area  = 2Step 2:row 1  1 2 2 1  area = 4, maximum area becomes 4row 2  2 3 3 2  area = 8, maximum area becomes 8row 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 to this post first as most of the code is taken from there.

Implementation

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. 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 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 ``` Python3 ```# Python3 program to find largest rectangle # with all 1s in a binary matrix # Finds the maximum area under the # histogram represented # by histogram. See below article for details. class Solution(): def maxHist(self, 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 of stack top_val = 0 # Initialize max area in current max_area = 0 # row (or histogram) area = 0 # Initialize area with current top # Run through all bars of given # histogram (or row) i = 0 while (i < len(row)): # If this bar is higher than the # bar on top stack, push it to stack if (len(result) == 0) or (row[result[-1]] <= 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.pop()] area = top_val * i if (len(result)): area = top_val * (i - result[-1] - 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.pop()] area = top_val * i if (len(result)): area = top_val * (i - result[-1] - 1) max_area = max(area, max_area) return max_area # Returns area of the largest rectangle # with all 1s in A def maxRectangle(self, A): # Calculate area for first row and # initialize it as result result = self.maxHist(A[0]) # iterate over row to find maximum rectangular # area considering each row as histogram for i in range(1, len(A)): for j in range(len(A[i])): # 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, self.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]] ans = Solution() print("Area of maximum rectangle is", ans.maxRectangle(A)) # This code is contributed # by Aaryaman Sharma ``` C# ```// 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. 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 ``` Javascript ```<script> // Javascript program to find largest rectangle // with all 1s in a binary matrix // Finds the maximum area under the // histogram represented by histogram. function maxHist(R, C, 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. let result = []; let top_val; // Top of stack let max_area = 0; // Initialize max area in // current row (or histogram) let area = 0; // Initialize area with // current top // Run through all bars of // given histogram (or row) let i = 0; while (i < C) { // If this bar is higher than the // bar on top stack, push it to stack if (result.length == 0 || row[result[result.length - 1]] <= 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[result.length - 1]]; result.pop(); area = top_val * i; if (result.length > 0) { area = top_val * (i - result[result.length - 1] - 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.length > 0) { top_val = row[result[result.length - 1]]; result.pop(); area = top_val * i; if (result.length > 0) { area = top_val * (i - result[result.length - 1] - 1); } max_area = Math.max(area, max_area); } return max_area; } // Returns area of the largest // rectangle with all 1s in A[][] function maxRectangle(R, C, A) { // Calculate area for first row // and initialize it as result let result = maxHist(R, C, A[0]); // iterate over row to find // maximum rectangular area // considering each row as histogram for (let i = 1; i < R; i++) { for (let 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; } let R = 4; let C = 4; let A = [ [ 0, 1, 1, 0 ], [ 1, 1, 1, 1 ], [ 1, 1, 1, 1 ], [ 1, 1, 0, 0 ] ]; document.write("Area of maximum rectangle is " + maxRectangle(R, C, A)); // This code is contributed by decode2207. </script> ```

Output
`Area of maximum rectangle is 8`

Complexity Analysis:

• Time Complexity: O(R x C x C).
One traversal of the matrix is required, another is traversal of every column, so the time complexity is O(R X C X C)
• Auxiliary Space: O(C).
Stack is required to store the columns, so space complexity is O(C)

### Approach : Dynamic Programming

`For each cell, use dynamic programming to get the maximum height of the rectangle that ends at the cell. To get the maximum width of the rectangle with the cell as its bottom-right corner, expand leftwards. To calculate the maximum area for each cell, expand leftward. Keep track of the maximum`

Follow the given steps to solve the problem :

• Create two auxiliary matrices.
• One to store the maximum height of consecutive 1s above a cell
• One to store the maximum height of consecutive 1s below a cell
• For each cell in the matrix calculate the maximum area by considering the heights stored in the auxiliary matrices
• Keep track of the maximum area found

Below is the implementation of the above approach:

C++ ```#include <iostream> #include <vector> #include <algorithm> using namespace std; int maximalRectangle(vector<vector<char>>& matrix) { if (matrix.empty() || matrix[0].empty()) { return 0; } int m = matrix.size(); int n = matrix[0].size(); vector<int> left(n, 0); vector<int> right(n, n); vector<int> height(n, 0); int maxArea = 0; for (const auto& row : matrix) { int curLeft = 0, curRight = n; // Update height array for (int j = 0; j < n; j++) { if (row[j] == '1') { height[j]++; } else { height[j] = 0; } } // Update left boundary array for (int j = 0; j < n; j++) { if (row[j] == '1') { left[j] = max(left[j], curLeft); } else { left[j] = 0; curLeft = j + 1; } } // Update right boundary array for (int j = n - 1; j >= 0; j--) { if (row[j] == '1') { right[j] = min(right[j], curRight); } else { right[j] = n; curRight = j; } } // Calculate maximum area for each cell for (int j = 0; j < n; j++) { maxArea = max(maxArea, (right[j] - left[j]) * height[j]); } } return maxArea; } int main() { vector<vector<char>> matrix1 = { {'0', '1', '1', '0'}, {'1', '1', '1', '1'}, {'1', '1', '1', '1'}, {'1', '1', '0', '0'} }; cout << maximalRectangle(matrix1) << endl; return 0; } //This code is contributed by ayush panwar ``` Java ```// Java Code import java.io.*; import java.util.Arrays; public class GFG{ public static int maximalRectangle(char[][] matrix) { if (matrix == null || matrix.length == 0) { return 0; } int m = matrix.length; int n = matrix[0].length; int[] left = new int[n]; // Array to store the left boundary of consecutive 1's int[] right = new int[n]; // Array to store the right boundary of consecutive 1's int[] height = new int[n]; // Array to store the height of consecutive 1's Arrays.fill(right, n); int maxArea = 0; for (char[] row : matrix) { int curLeft = 0, curRight = n; // Update height array for (int j = 0; j < n; j++) { if (row[j] == '1') { height[j]++; } else { height[j] = 0; } } // Update left boundary array for (int j = 0; j < n; j++) { if (row[j] == '1') { left[j] = Math.max(left[j], curLeft); } else { left[j] = 0; curLeft = j + 1; } } // Update right boundary array for (int j = n - 1; j >= 0; j--) { if (row[j] == '1') { right[j] = Math.min(right[j], curRight); } else { right[j] = n; curRight = j; } } // Calculate maximum area for each cell for (int j = 0; j < n; j++) { maxArea = Math.max(maxArea, (right[j] - left[j]) * height[j]); } } return maxArea; } public static void main(String[] args) { char[][] matrix1 = { {'0', '1', '1', '0'}, {'1', '1', '1', '1'}, {'1', '1', '1', '1'}, {'1', '1', '0', '0'} }; System.out.println(maximalRectangle(matrix1)); } } // This code is contributed by guptapratik ``` Python3 ```def maximalRectangle(matrix): if not matrix: return 0 m, n = len(matrix), len(matrix[0]) left = [0] * n # Array to store the left boundary of consecutive 1's right = [n] * n # Array to store the right boundary of consecutive 1's height = [0] * n # Array to store the height of consecutive 1's max_area = 0 for i in range(m): cur_left, cur_right = 0, n # Update height array for j in range(n): if matrix[i][j] == '1': height[j] += 1 else: height[j] = 0 # Update left boundary array for j in range(n): if matrix[i][j] == '1': left[j] = max(left[j], cur_left) else: left[j] = 0 cur_left = j + 1 # Update right boundary array for j in range(n-1, -1, -1): if matrix[i][j] == '1': right[j] = min(right[j], cur_right) else: right[j] = n cur_right = j # Calculate maximum area for each cell for j in range(n): max_area = max(max_area, (right[j] - left[j]) * height[j]) return max_area matrix1 = [ ["0", "1", "1", "0"], ["1", "1", "1", "1"], ["1", "1", "1", "1"], ["1", "1", "0", "0"] ] print(maximalRectangle(matrix1)) ``` JavaScript ```function maximalRectangle(matrix) { if (!matrix || matrix.length === 0) { return 0; } const m = matrix.length; const n = matrix[0].length; const left = new Array(n).fill(0); // Array to store the left boundary of consecutive 1's const right = new Array(n).fill(n); // Array to store the right boundary of consecutive 1's const height = new Array(n).fill(0); // Array to store the height of consecutive 1's let maxArea = 0; for (const row of matrix) { let curLeft = 0, curRight = n; // Update height array for (let j = 0; j < n; j++) { if (row[j] === '1') { height[j]++; } else { height[j] = 0; } } // Update left boundary array for (let j = 0; j < n; j++) { if (row[j] === '1') { left[j] = Math.max(left[j], curLeft); } else { left[j] = 0; curLeft = j + 1; } } // Update right boundary array for (let j = n - 1; j >= 0; j--) { if (row[j] === '1') { right[j] = Math.min(right[j], curRight); } else { right[j] = n; curRight = j; } } // Calculate maximum area for each cell for (let j = 0; j < n; j++) { maxArea = Math.max(maxArea, (right[j] - left[j]) * height[j]); } } return maxArea; } const matrix1 = [ ['0', '1', '1', '0'], ['1', '1', '1', '1'], ['1', '1', '1', '1'], ['1', '1', '0', '0'] ]; console.log(maximalRectangle(matrix1)); ```

Output
```8
```

Time Complexity: O(m * n), where m is the number of rows & n is the number of columns in the input matrix.

Auxiliary Complexity: O(n)

Previous Article
Next Article
Article Tags :
Practice Tags :