Consider a matrix with rows and columns, where each cell contains either a ‘0’ or a ‘1’ and any cell containing a 1 is called a filled cell. Two cells are said to be connected if they are adjacent to each other horizontally, vertically, or diagonally .If one or more filled cells are also connected, they form a region. find the length of the largest region.
Examples:
Input : M[][5] = { 0 0 1 1 0
1 0 1 1 0
0 1 0 0 0
0 0 0 0 1 }
Output : 6
Ex: in the following example, there are 2 regions one with length 1 and the other as 6.
so largest region : 6
Asked in : Amazon interview
Idea is based on the problem or finding number of islands in Boolean 2D-matrix
A cell in 2D matrix can be connected to at most 8 neighbors. So in DFS, we make recursive calls for 8 neighbors. We keep track of the visited 1’s in every DFS and update maximum length region.
Below is implementation of above idea.
C++
// Program to find the length of the largest // region in boolean 2D-matrix #include<bits/stdc++.h> using namespace std; #define ROW 4 #define COL 5 // A function to check if a given cell (row, col) // can be included in DFS int isSafe(int M[][COL], int row, int col, bool visited[][COL]) { // row number is in range, column number is in // range and value is 1 and not yet visited return (row >= 0) && (row < ROW) && (col >= 0) && (col < COL) && (M[row][col] && !visited[row][col]); } // A utility function to do DFS for a 2D boolean // matrix. It only considers the 8 neighbours as // adjacent vertices void DFS(int M[][COL], int row, int col, bool visited[][COL], int &count) { // These arrays are used to get row and column // numbers of 8 neighbours of a given cell static int rowNbr[] = {-1, -1, -1, 0, 0, 1, 1, 1}; static int colNbr[] = {-1, 0, 1, -1, 1, -1, 0, 1}; // Mark this cell as visited visited[row][col] = true; // Recur for all connected neighbours for (int k = 0; k < 8; ++k) { if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited)) { // increment region length by one count++; DFS(M, row + rowNbr[k], col + colNbr[k], visited, count); } } } // The main function that returns largest length region // of a given boolean 2D matrix int largestRegion(int M[][COL]) { // Make a bool array to mark visited cells. // Initially all cells are unvisited bool visited[ROW][COL]; memset(visited, 0, sizeof(visited)); // Initialize result as 0 and travesle through the // all cells of given matrix int result = INT_MIN; for (int i = 0; i < ROW; ++i) { for (int j = 0; j < COL; ++j) { // If a cell with value 1 is not if (M[i][j] && !visited[i][j]) { // visited yet, then new region found int count = 1 ; DFS(M, i, j, visited , count); // maximum region result = max(result , count); } } } return result ; } // Driver program to test above function int main() { int M[][COL] = { {0, 0, 1, 1, 0}, {1, 0, 1, 1, 0}, {0, 1, 0, 0, 0}, {0, 0, 0, 0, 1}}; cout << largestRegion(M); return 0; } |
Java
// Java program to find the length of the largest // region in boolean 2D-matrix import java.io.*; import java.util.*; class GFG { static int ROW, COL, count; // A function to check if a given cell (row, col) // can be included in DFS static boolean isSafe(int[][] M, int row, int col, boolean[][] visited) { // row number is in range, column number is in // range and value is 1 and not yet visited return ((row >= 0) && (row < ROW) && (col >= 0) && (col < COL) && (M[row][col] == 1 && !visited[row][col])); } // A utility function to do DFS for a 2D boolean // matrix. It only considers the 8 neighbours as // adjacent vertices static void DFS(int[][] M, int row, int col, boolean[][] visited) { // These arrays are used to get row and column // numbers of 8 neighbours of a given cell int[] rowNbr = {-1, -1, -1, 0, 0, 1, 1, 1}; int[] colNbr = {-1, 0, 1, -1, 1, -1, 0, 1}; // Mark this cell as visited visited[row][col] = true; // Recur for all connected neighbours for (int k = 0; k < 8; k++) { if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited)) { // increment region length by one count++; DFS(M, row + rowNbr[k], col + colNbr[k], visited); } } } // The main function that returns largest length region // of a given boolean 2D matrix static int largestRegion(int[][] M) { // Make a boolean array to mark visited cells. // Initially all cells are unvisited boolean[][] visited = new boolean[ROW][COL]; // Initialize result as 0 and traverse through the // all cells of given matrix int result = 0; for (int i = 0; i < ROW; i++) { for (int j = 0; j < COL; j++) { // If a cell with value 1 is not if (M[i][j] == 1 && !visited[i][j]) { // visited yet, then new region found count = 1; DFS(M, i, j, visited); // maximum region result = Math.max(result, count); } } } return result; } // Driver code public static void main(String args[]) { int M[][] = { {0, 0, 1, 1, 0}, {1, 0, 1, 1, 0}, {0, 1, 0, 0, 0}, {0, 0, 0, 0, 1}}; ROW = 4; COL = 5; System.out.println(largestRegion(M)); } } // This code is contributed by rachana soma |
Python3
# Python3 program to find the length of the # largest region in boolean 2D-matrix # A function to check if a given cell # (row, col) can be included in DFS def isSafe(M, row, col, visited): global ROW, COL # row number is in range, column number is in # range and value is 1 and not yet visited return ((row >= 0) and (row < ROW) and (col >= 0) and (col < COL) and (M[row][col] and not visited[row][col])) # A utility function to do DFS for a 2D # boolean matrix. It only considers # the 8 neighbours as adjacent vertices def DFS(M, row, col, visited, count): # These arrays are used to get row and column # numbers of 8 neighbours of a given cell rowNbr = [-1, -1, -1, 0, 0, 1, 1, 1] colNbr = [-1, 0, 1, -1, 1, -1, 0, 1] # Mark this cell as visited visited[row][col] = True # Recur for all connected neighbours for k in range(8): if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited)): # increment region length by one count[0] += 1 DFS(M, row + rowNbr[k], col + colNbr[k], visited, count) # The main function that returns largest # length region of a given boolean 2D matrix def largestRegion(M): global ROW, COL # Make a bool array to mark visited cells. # Initially all cells are unvisited visited = [[0] * COL for i in range(ROW)] # Initialize result as 0 and travesle # through the all cells of given matrix result = -999999999999 for i in range(ROW): for j in range(COL): # If a cell with value 1 is not if (M[i][j] and not visited[i][j]): # visited yet, then new region found count = [1] DFS(M, i, j, visited , count) # maximum region result = max(result , count[0]) return result # Driver Code ROW = 4COL = 5 M = [[0, 0, 1, 1, 0], [1, 0, 1, 1, 0], [0, 1, 0, 0, 0], [0, 0, 0, 0, 1]] print(largestRegion(M)) # This code is contributed by PranchalK |
Output:
6
Time complexity: O(ROW x COL)
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