Added

Author: Karnika06

Data Structures/Stacks/Maximum_size_rectangle_binary_sub-matrix.cpp

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+
+/*Implementation of Maximum size rectangle binary sub-matrix with all 1s GFG
+
+Given a binary matrix, find the maximum size rectangle binary-sub-matrix with all 1’s. 
+
+Link of the problem: https://www.geeksforgeeks.org/maximum-size-rectangle-binary-sub-matrix-1s/
+
+*/
+
+#include <iostream>
+#include <vector>
+#include <stack>
+using namespace std;
+
+class MaximalRectangle 
+{
+	public:
+    //In this function, we find next smallest element of each element
+	vector<int> NSE(vector<int> v)
+	{
+		vector<int> nse(v.size(), v.size());
+		stack<int> stk;
+		for(int i=0; i<v.size(); i++)
+		{
+			while(!stk.empty() && v[stk.top()]>v[i])
+			{
+				nse[stk.top()]=i;
+				stk.pop();
+			}
+			stk.push(i);
+		}
+		return nse;
+	}
+	
+	//In this function, we will find previous smallest element of each element
+	vector<int> PSE(vector<int> v)
+	{
+		vector<int> pse(v.size(), -1);
+		stack<int> stk;
+		for(int i=v.size()-1; i>=0; i--)
+		{
+			while(!stk.empty() && v[stk.top()]>v[i])
+			{
+				pse[stk.top()]=i;		
+				stk.pop();
+			}
+			stk.push(i);
+		}
+		return pse;
+	}
+    
+    int maximalRectangle(vector<vector<char>>& matrix) 
+	{
+        if(matrix.empty()) return {};
+        
+        int r= matrix.size();
+        int c= matrix[0].size();
+        
+        if(r==0)
+            return 0;
+            
+        if(r == 1 && c == 1) 
+			return matrix[0][0] - '0';
+        
+        vector<vector<int>> v(r, vector<int>(c));
+        
+        for (int i = 0; i <r; i++) 
+		{
+        	for (int j = 0; j < c; j++)
+                v[i][j]= matrix[i][j]=='1'?1: 0;
+        }
+        
+        for (int i = 1; i < r; i++) 
+		{
+ 
+        	for (int j = 0; j < c; j++)
+            	if (v[i][j])
+                	v[i][j] += v[i - 1][j];
+        }
+		
+        int area;
+		for(int i=0; i<r ;i++)
+		{
+			vector<int> nse=NSE(v[i]);	
+			vector<int> pse=PSE(v[i]);
+		
+			for(int j=0;j<c;j++)
+				area = max(area, v[i][j]*(nse[j]-pse[j]-1));
+		}
+		return area;
+    }
+    
+    
+};
+
+int main()
+{	
+	MaximalRectangle ob;
+
+	int r,c;
+	char value;
+	cin>>r;
+	cin>>c;
+	vector<vector<char>> v(r);
+	
+	for(int i=0;i<r;i++)
+	{
+		for(int j=0;j<c;j++)
+		{
+			cin>>value;
+			v[i].push_back(value);
+		}
+	}
+	int area = ob.maximalRectangle(v);
+	cout<<area;
+}
+
+/*
+
+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)
+
+Example 1: 
+
+Input:
+	0 1 1 0
+	1 1 1 1
+	1 1 1 1
+	1 1 0 0
+	
+Output : 8
+
+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 
+
+Example 2:
+
+Input:
+	0 1 1
+	1 1 1
+	0 1 1 
+	     
+Output: 6
+
+Explanation : 
+The largest rectangle with only 1's is from 
+(0, 1) to (2, 2) which is
+1 1
+1 1
+1 1
+
+*/