Binary Matrix after flipping submatrices in given range for Q queries
Given a binary matrix arr[][] of dimensions M x N and Q queries of the form (x1, y1, x2, y2), where (x1, y1) and (x2, y2) denotes the top-left and bottom-right indices of the submatrix required to be flipped(convert 0s to 1s and vice versa) respectively. The task is to print the final matrix obtained after performing the given Q queries.
Examples:
Input: arr[][] = {{0, 1, 0}, {1, 1, 0}}, queries[][] = {{1, 1, 2, 3}}
Output: [[1, 0, 1], [0, 0, 1]]
Explanation:
The submatrix to be flipped is equal to {{0, 1, 0}, {1, 1, 0}}
The flipped matrix is {{1, 0, 1}, {0, 0, 1}}.Input: arr[][] = {{0, 1, 0}, {1, 1, 0}}, queries[][] = {{1, 1, 2, 3}, {1, 1, 1, 1], {1, 2, 2, 3}}
Output: [[0, 1, 0], [0, 1, 0]]
Explanation:
Query 1:
Submatrix to be flipped = [[0, 1, 0], [1, 1, 0]]
Flipped submatrix is [[1, 0, 1], [0, 0, 1]].
Therefore, the modified matrix is [[1, 0, 1], [0, 0, 1]].
Query 2:
Submatrix to be flipped = [[1]]
Flipped submatrix is [[0]]
Therefore, the matrix is [[0, 0, 1], [0, 0, 1]].
Query 3:
Submatrix to be flipped = [[0, 1], [0, 1]]
Flipped submatrix is [[1, 0], [1, 0]]
Therefore, the modified matrix is [[0, 1, 0], [0, 1, 0]].
Naive Approach: The simplest approach to solve the problem for each query is to iterate over the given submatrices and for every element, check if it is 0 or 1, and flip accordingly. After completing these operations for all the queries, print the final matrix obtained.
Below is the implementation of the above approach :
C++
// C++ program to implement // the above approach #include<bits/stdc++.h> using namespace std; // Function to flip a submatrices void manipulation(vector<vector< int >> &matrix, vector< int > &q) { // Boundaries of the submatrix int x1 = q[0], y1 = q[1], x2 = q[2], y2 = q[3]; // Iterate over the submatrix for ( int i = x1 - 1; i < x2; i++) { for ( int j = y1 - 1; j < y2; j++) { // Check for 1 or 0 // and flip accordingly if (matrix[i][j] == 1) matrix[i][j] = 0; else matrix[i][j] = 1; } } } // Function to perform the queries void queries_fxn(vector<vector< int >> &matrix, vector<vector< int >> &queries) { for ( auto q : queries) manipulation(matrix, q); } // Driver code int main() { vector<vector< int >> matrix = { { 0, 1, 0 }, { 1, 1, 0 } }; vector<vector< int >> queries = { { 1, 1, 2, 3 }, { 1, 1, 1, 1 }, { 1, 2, 2, 3 } }; // Function call queries_fxn(matrix, queries); cout << "[" ; for ( int i = 0; i < matrix.size(); i++) { cout << "[" ; for ( int j = 0; j < matrix[i].size(); j++) cout << matrix[i][j] << " " ; if (i == matrix.size() - 1) cout << "]" ; else cout << "], " ; } cout << "]" ; } // This code is contributed by bgangwar59 |
Java
// Java program to implement // the above approach import java.util.*; import java.lang.*; class GFG{ // Function to flip a submatrices static void manipulation( int [][] matrix, int [] q) { // Boundaries of the submatrix int x1 = q[ 0 ], y1 = q[ 1 ], x2 = q[ 2 ], y2 = q[ 3 ]; // Iterate over the submatrix for ( int i = x1 - 1 ; i < x2; i++) { for ( int j = y1 - 1 ; j < y2; j++) { // Check for 1 or 0 // and flip accordingly if (matrix[i][j] == 1 ) matrix[i][j] = 0 ; else matrix[i][j] = 1 ; } } } // Function to perform the queries static void queries_fxn( int [][] matrix, int [][] queries) { for ( int [] q : queries) manipulation(matrix, q); } // Driver code public static void main (String[] args) { int [][] matrix = {{ 0 , 1 , 0 }, { 1 , 1 , 0 }}; int [][] queries = {{ 1 , 1 , 2 , 3 }, { 1 , 1 , 1 , 1 }, { 1 , 2 , 2 , 3 }}; // Function call queries_fxn(matrix, queries); System.out.print( "[" ); for ( int i = 0 ; i < matrix.length; i++) { System.out.print( "[" ); for ( int j = 0 ; j < matrix[i].length; j++) System.out.print(matrix[i][j] + " " ); if (i == matrix.length - 1 ) System.out.print( "]" ); else System.out.print( "], " ); } System.out.print( "]" ); } } // This code is contributed by offbeat |
Python3
# Python3 Program to implement # the above approach # Function to flip a submatrices def manipulation(matrix, q): # Boundaries of the submatrix x1, y1, x2, y2 = q # Iterate over the submatrix for i in range (x1 - 1 , x2): for j in range (y1 - 1 , y2): # Check for 1 or 0 # and flip accordingly if matrix[i][j]: matrix[i][j] = 0 else : matrix[i][j] = 1 # Function to perform the queries def queries_fxn(matrix, queries): for q in queries: manipulation(matrix, q) # Driver Code matrix = [[ 0 , 1 , 0 ], [ 1 , 1 , 0 ]] queries = [[ 1 , 1 , 2 , 3 ], \ [ 1 , 1 , 1 , 1 ], \ [ 1 , 2 , 2 , 3 ]] # Function call queries_fxn(matrix, queries) print (matrix) |
C#
// C# program to implement // the above approach using System; class GFG{ // Function to flip a submatrices static void manipulation( int [,] matrix, int [] q) { // Boundaries of the submatrix int x1 = q[0], y1 = q[1], x2 = q[2], y2 = q[3]; // Iterate over the submatrix for ( int i = x1 - 1; i < x2; i++) { for ( int j = y1 - 1; j < y2; j++) { // Check for 1 or 0 // and flip accordingly if (matrix[i, j] == 1) matrix[i, j] = 0; else matrix[i, j] = 1; } } } public static int [] GetRow( int [,] matrix, int row) { var rowLength = matrix.GetLength(1); var rowVector = new int [rowLength]; for ( var i = 0; i < rowLength; i++) rowVector[i] = matrix[row, i]; return rowVector; } // Function to perform the queries static void queries_fxn( int [,] matrix, int [,] queries) { for ( int i = 0; i < queries.GetLength(0); i++) manipulation(matrix, GetRow(queries, i)); } // Driver code public static void Main(String[] args) { int [,] matrix = { { 0, 1, 0 }, { 1, 1, 0 } }; int [,] queries = { { 1, 1, 2, 3 }, { 1, 1, 1, 1 }, { 1, 2, 2, 3 } }; // Function call queries_fxn(matrix, queries); Console.Write( "[" ); for ( int i = 0; i < matrix.GetLength(0); i++) { Console.Write( "[" ); for ( int j = 0; j < matrix.GetLength(1); j++) Console.Write(matrix[i, j] + ", " ); if (i == matrix.Length - 1) Console.Write( "]" ); else Console.Write( "], " ); } Console.Write( "]" ); } } // This code is contributed by Princi Singh |
Javascript
<script> // Javascript program to implement // the above approach // Function to flip a submatrices function manipulation(matrix, q) { // Boundaries of the submatrix let x1 = q[0], y1 = q[1], x2 = q[2], y2 = q[3]; // Iterate over the submatrix for (let i = x1 - 1; i < x2; i++) { for (let j = y1 - 1; j < y2; j++) { // Check for 1 or 0 // and flip accordingly if (matrix[i][j] == 1) matrix[i][j] = 0; else matrix[i][j] = 1; } } } // Function to perform the queries function queries_fxn(matrix, queries) { for (let q of queries) manipulation(matrix, q); } // Driver code let matrix = [[0, 1, 0], [1, 1, 0]]; let queries = [[1, 1, 2, 3], [1, 1, 1, 1], [1, 2, 2, 3]]; // Function call queries_fxn(matrix, queries); document.write( "[" ); for (let i = 0; i < matrix.length; i++) { document.write( "[" ); for (let j = 0; j < matrix[i].length; j++) { if (j < matrix[i].length - 1) { document.write(matrix[i][j] + ", " ); } else { document.write(matrix[i][j] + " " ); } } if (i == matrix.length - 1) document.write( "]" ); else document.write( "], " ); } document.write( "]" ); // This code is contributed by _saurabh_jaiswal </script> |
[[0, 1, 0], [0, 1, 0]]
Time complexity: O( N * M * Q), The time complexity of the given program is O(N*M*Q), where N is the number of queries, M is the number of rows in the matrix, and Q is the number of columns in the matrix. This is because we need to iterate over each element in the submatrix for each query, which takes O(N*M) time, and we have Q queries to perform.
Auxiliary Space: O(1),
Efficient Approach: The above approach can be optimized using Dynamic programming and Prefix Sum technique. Mark the boundaries of the submatrices involved in each query and then calculate the prefix sum of the operations involved in the matrix and update the matrix accordingly. Follow the steps below to solve the problem:
- Initialize a 2D state space table dp[][] to store the count of flips at respective indices of the matrix
- For each query {x1, y1, x2, y2, K}, update the dp[][] matrix by the following operations:
- dp[x1][y1] += 1
- dp[x2 + 1][y1] -= 1
- dp[x2 + 1][y2 + 1] += 1
- dp[x1][y2 + 1] -= 1
- Now, traverse over the dp[][] matrix and update dp[i][j] by calculating the prefix sum of the rows and columns and diagonals by the following relation:
dp[i][j] = dp[i][j] + dp[i-1][j] + dp[i][j – 1] – dp[i – 1][j – 1]
- If dp[i][j] is found to be odd, reduce mat[i – 1][j – 1] by 1.
- Finally, print the updated matrix mat[][] as the result.
Below is the implementation of the above approach :
C++
// C++ code for the above approach #include <iostream> #include <vector> using namespace std; // Function to modify dp[][] array by // generating prefix sum void modifyDP(vector<vector< int > >& matrix, vector<vector< int > >& dp) { int row = matrix.size(); int col = matrix[0].size(); for ( int j = 1; j <= row; j++) { for ( int k = 1; k <= col; k++) { // Update the tabular data dp[j][k] = dp[j][k] + dp[j - 1][k] + dp[j][k - 1] - dp[j - 1][k - 1]; // If the count of flips is even if (dp[j][k] % 2 != 0) { matrix[j - 1][k - 1] = matrix[j - 1][k - 1] ^ 1; } } } } // Function to update dp[][] matrix // for each query void queries_fxn(vector<vector< int > >& matrix, vector<vector< int > >& queries, vector<vector< int > >& dp) { for ( int i = 0; i < queries.size(); i++) { int x1 = queries[i][0]; int y1 = queries[i][1]; int x2 = queries[i][2]; int y2 = queries[i][3]; // Update the table dp[x1][y1] += 1; dp[x2 + 1][y2 + 1] += 1; dp[x1][y2 + 1] -= 1; dp[x2 + 1][y1] -= 1; } modifyDP(matrix, dp); } // Driver Code int main() { vector<vector< int > > matrix = { { 0, 1, 0 }, { 1, 1, 0 } }; vector<vector< int > > queries = { { 1, 1, 2, 3 }, { 1, 1, 1, 1 }, { 1, 2, 2, 3 } }; // Initialize dp table vector<vector< int > > dp( matrix.size() + 2, vector< int >(matrix[0].size() + 2, 0)); queries_fxn(matrix, queries, dp); cout << "[" ; for ( int i = 0; i < matrix.size(); i++) { cout << "[" ; for ( int j = 0; j < matrix[i].size(); j++) { cout << matrix[i][j] << " " ; } if (i == matrix.size() - 1) { cout << "]" ; } else { cout << "], " ; } } } // This code is contributed by lokeshpotta20. |
Java
import java.util.*; import java.io.*; // Java program for the above approach class GFG{ // Function to modify dp[][] array by // generating prefix sum static void modifyDP( int matrix[][], int dp[][]){ for ( int j = 1 ; j <= matrix.length ; j++){ for ( int k = 1 ; k <= matrix[ 0 ].length ; k++){ // Update the tabular data dp[j][k] = dp[j][k] + dp[j- 1 ][k] + dp[j][k- 1 ]- dp[j- 1 ][k- 1 ]; // If the count of flips is even if (dp[j][k] % 2 != 0 ){ matrix[j- 1 ][k- 1 ] = matrix[j- 1 ][k- 1 ] ^ 1 ; } } } } // Function to update dp[][] matrix // for each query static void queries_fxn( int matrix[][], int queries[][], int dp[][]){ for ( int i = 0 ; i < queries.length ; i++){ int x1 = queries[i][ 0 ]; int y1 = queries[i][ 1 ]; int x2 = queries[i][ 2 ]; int y2 = queries[i][ 3 ]; // Update the table dp[x1][y1] += 1 ; dp[x2 + 1 ][y2 + 1 ] += 1 ; dp[x1][y2 + 1 ] -= 1 ; dp[x2 + 1 ][y1] -= 1 ; } modifyDP(matrix, dp); } // Driver Code public static void main(String args[]) { int matrix[][] = new int [][]{ new int []{ 0 , 1 , 0 }, new int []{ 1 , 1 , 0 } }; int queries[][] = new int [][]{ new int []{ 1 , 1 , 2 , 3 }, new int []{ 1 , 1 , 1 , 1 }, new int []{ 1 , 2 , 2 , 3 } }; // Initialize dp table int dp[][] = new int [(matrix.length + 2 )][(matrix[ 0 ].length + 2 )]; queries_fxn(matrix, queries, dp); System.out.print( "[" ); for ( int i = 0 ; i < matrix.length ; i++) { System.out.print( "[" ); for ( int j = 0 ; j < matrix[i].length ; j++){ System.out.print(matrix[i][j] + " " ); } if (i == matrix.length - 1 ){ System.out.print( "]" ); } else { System.out.print( "], " ); } } System.out.print( "]" ); } } // This code is contributed by entertain2022. |
Python3
# Python3 program to implement # the above approach # Function to modify dp[][] array by # generating prefix sum def modifyDP(matrix, dp): for j in range ( 1 , len (matrix) + 1 ): for k in range ( 1 , len (matrix[ 0 ]) + 1 ): # Update the tabular data dp[j][k] = dp[j][k] + dp[j - 1 ][k] \ + dp[j][k - 1 ] - dp[j - 1 ][k - 1 ] # If the count of flips is even if dp[j][k] % 2 ! = 0 : matrix[j - 1 ][k - 1 ] = int (matrix[j - 1 ][k - 1 ]) ^ 1 # Function to update dp[][] matrix # for each query def queries_fxn(matrix, queries, dp): for q in queries: x1, y1, x2, y2 = q # Update the table dp[x1][y1] + = 1 dp[x2 + 1 ][y2 + 1 ] + = 1 dp[x1][y2 + 1 ] - = 1 dp[x2 + 1 ][y1] - = 1 modifyDP(matrix, dp) # Driver Code matrix = [[ 0 , 1 , 0 ], [ 1 , 1 , 0 ]] queries = [[ 1 , 1 , 2 , 3 ], \ [ 1 , 1 , 1 , 1 ], \ [ 1 , 2 , 2 , 3 ]] # Initialize dp table dp = [[ 0 for i in range ( len (matrix[ 0 ]) + 2 )] \ for j in range ( len (matrix) + 2 )] queries_fxn(matrix, queries, dp) print (matrix) |
C#
// C# code for the above approach using System; class GFG { // Function to modify dp[][] array by generating prefix sum static void ModifyDP( int [,] matrix, int [,] dp) { for ( int j = 1; j <= matrix.GetLength(0); j++) { for ( int k = 1; k <= matrix.GetLength(1); k++) { // Update the tabular data dp[j, k] = dp[j, k] + dp[j - 1, k] + dp[j, k - 1] - dp[j - 1, k - 1]; // If the count of flips is even if (dp[j, k] % 2 != 0) { matrix[j - 1, k - 1] = matrix[j - 1, k - 1] ^ 1; } } } } // Function to update dp[][] matrix for each query static void QueriesFxn( int [,] matrix, int [,] queries, int [,] dp) { for ( int i = 0; i < queries.GetLength(0); i++) { int x1 = queries[i, 0]; int y1 = queries[i, 1]; int x2 = queries[i, 2]; int y2 = queries[i, 3]; // Update the table dp[x1, y1] += 1; dp[x2 + 1, y2 + 1] += 1; dp[x1, y2 + 1] -= 1; dp[x2 + 1, y1] -= 1; } ModifyDP(matrix, dp); } // Driver Code public static void Main( string [] args) { int [,] matrix = new int [,] { { 0, 1, 0 }, { 1, 1, 0 } }; int [,] queries = new int [,] { { 1, 1, 2, 3 }, { 1, 1, 1, 1 }, { 1, 2, 2, 3 } }; // Initialize dp table int [,] dp = new int [(matrix.GetLength(0) + 2, matrix.GetLength(1) + 2)]; QueriesFxn(matrix, queries, dp); Console.Write( "[" ); for ( int i = 0; i < matrix.GetLength(0); i++) { Console.Write( "[" ); for ( int j = 0; j < matrix.GetLength(1); j++) { Console.Write(matrix[i, j] + " " ); } if (i == matrix.GetLength(0) - 1) { Console.Write( "]" ); } else { Console.Write( "], " ); } } Console.Write( "]" ); } } // This code is contributed by pradeepkumarppk2003. |
Javascript
// Javascript code for the above approach // Function to modify dp[][] array by // generating prefix sum function modifyDP(matrix, dp) { let row = matrix.length; let col = matrix[0].length; for (let j = 1; j <= row; j++) { for (let k = 1; k <= col; k++) { // Update the tabular data dp[j][k] = dp[j][k] + dp[j - 1][k] + dp[j][k - 1] - dp[j - 1][k - 1]; // If the count of flips is even if (dp[j][k] % 2 != 0) { matrix[j - 1][k - 1] = matrix[j - 1][k - 1] ^ 1; } } } } // Function to update dp[][] matrix // for each query function queries_fxn(matrix, queries, dp) { for (let i = 0; i < queries.length; i++) { let x1 = queries[i][0]; let y1 = queries[i][1]; let x2 = queries[i][2]; let y2 = queries[i][3]; // Update the table dp[x1][y1] += 1; dp[x2 + 1][y2 + 1] += 1; dp[x1][y2 + 1] -= 1; dp[x2 + 1][y1] -= 1; } modifyDP(matrix, dp); } // Driver Code let matrix = [[0, 1, 0 ], [ 1, 1, 0]]; let queries = [ [ 1, 1, 2, 3 ],[1, 1, 1, 1 ],[1, 2, 2, 3 ]]; // Initialize dp table let dp= new Array(matrix.length + 2); for (let i=0; i<matrix.length+2; i++) dp[i]= new Array(matrix[0].length+2).fill(0); queries_fxn(matrix, queries, dp); document.write( "[" ); for (let i = 0; i < matrix.length; i++) { document.write( "[" ); for (let j = 0; j < matrix[i].length; j++) { document.write(matrix[i][j] + " " ); } if (i == matrix.length - 1) { document.write( "]" ); } else { document.write( "], " ); } } |
[[0, 1, 0], [0, 1, 0]]
Time Complexity: O(N * M )
Auxiliary Space: O(N * M)
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