Given a grid of size N * M consisting of O’s and X’s. The task is to find the number of ‘X’ total shapes.
Note: ‘X’ shape consists of one or more adjacents X’s (diagonals not included).
Examples:
Input: grid = {{X, O, X}, {O, X, O}, {X, X, X}}
Output: 3
Explanation: The grid is:
X O X | X O X | X O X
O X O | O X O | O X O
X X X | X X X | X X X
So, X with bold color are adjacent to each other vertically for horizontally (diagonals not included). So, there are 3 different groups in the given grid.Input: grid = {{X, X}, {X, X}}
Output: 1
Explanation: The grid is:
X X
X X
So, X with bold color are adjacent to each other vertically for horizontally (diagonals not included). So, there is only 1 group in the given grid.
Approach: To solve the problem follow the below idea:
The idea is to apply dfs from every cell having ‘X’ and not visited, mark all its adjacent cells visited if their value is X and increase the answer by one.
Follow the steps to solve this problem:
- Instead of using an extra vector vis of size n*m to keep track of what cell is visited and not, we can simply convert a visited cell having ‘X’ to ‘O’
- Initialize variable ans with 0, it will keep count of connected components having only ‘X‘.
- Make a function dfs(x, y) to mark the whole connected component having only ‘X’ as visited by converting all ‘X’ to ‘O’.
- Run 2 nested loops for iterating over the whole grid
- Check if the current cell is having ‘X‘, then run a dfs from it and convert all cells in its connected component as ‘O’ and
- Increase the variable ans by 1.
- Return variable ans as the answer.
Below is the implementation of the above approach:
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to run DFSvoid dfs(vector<vector<char> >& grid, int x, int y){ // Checking whether the cell is within // the matrix bounds if (x < 0 || y < 0 || x >= grid.size() || grid[x].size() <= y) { return; } // If grid value is X, we update the // grid value as 0 and call the // function recursively for // adjacent nodes. if (grid[x][y] == 'X') { grid[x][y] = 'O'; dfs(grid, x + 1, y); dfs(grid, x, y + 1); dfs(grid, x - 1, y); dfs(grid, x, y - 1); }}// Function to find the number of// 'X' total shapes.int xShape(vector<vector<char> >& grid){ int n = grid.size(); int m = grid[0].size(); int i, j, ans = 0; // Traversing all the cells of the matrix. for (i = 0; i < grid.size(); i++) { for (j = 0; j < grid[i].size(); j++) { // If grid value is X, we increment // the counter and call the // dfs function. if (grid[i][j] == 'X') { ans++; dfs(grid, i, j); } } } // Returning the count. return ans;}// Driver Codeint main(){ vector<vector<char> > grid = { { 'X', 'O', 'X' }, { 'O', 'X', 'O' }, { 'X', 'X', 'X' } }; // Function Call cout << xShape(grid) << endl; return 0;} |
Java
// Java code to implement the approachimport java.io.*;class GFG { // Function to run DFS static void dfs(char[][] grid, int x, int y) { // Checking whether the cell is within // the matrix bounds if (x < 0 || y < 0 || x >= grid.length || grid[x].length <= y) { return; } // If grid value is X, we update the // grid value as 0 and call the // function recursively for // adjacent nodes. if (grid[x][y] == 'X') { grid[x][y] = 'O'; dfs(grid, x + 1, y); dfs(grid, x, y + 1); dfs(grid, x - 1, y); dfs(grid, x, y - 1); } } // Function to find the number of // 'X' total shapes. static int xShape(char[][] grid) { int n = grid.length; int m = grid[0].length; int i, j, ans = 0; // Traversing all the cells of the matrix. for (i = 0; i < n; i++) { for (j = 0; j < m; j++) { // If grid value is X, we increment // the counter and call the // dfs function. if (grid[i][j] == 'X') { ans++; dfs(grid, i, j); } } } // Returning the count. return ans; } public static void main(String[] args) { char[][] grid = { { 'X', 'O', 'X' }, { 'O', 'X', 'O' }, { 'X', 'X', 'X' } }; // Function call System.out.println(xShape(grid)); }}// This code is contributed by lokeshmvs21. |
Python3
# python code# Function to run DFSdef dfs(grid, x, y): # Checking whether the cell is within # the matrix bounds if x < 0 or y < 0 or x >= len(grid) or len(grid[x]) <= y: return # If grid value is X, we update the # grid value as 0 and call the # function recursively for # adjacent nodes. if grid[x][y] == 'X': grid[x][y] = 'O' dfs(grid, x + 1, y) dfs(grid, x, y + 1) dfs(grid, x - 1, y) dfs(grid, x, y - 1)# Function to find the number of# 'X' total shapes.def xShape(grid): n = len(grid) m = len(grid[0]) ans = 0 # Traversing all the cells of the matrix. for i in range(len(grid)): for j in range(len(grid[i])): # If grid value is X, we increment # the counter and call the # dfs function. if grid[i][j] == 'X': ans += 1 dfs(grid, i, j) # Returning the count. return ans# Driver Codegrid = [ ['X', 'O', 'X'], ['O', 'X', 'O'], ['X', 'X', 'X']]# Function Callprint(xShape(grid))# This code is contributed by ksam24000 |
C#
// C# code to implement the approachusing System;public class GFG { // Function to run DFS static void dfs(char[, ] grid, int x, int y) { // Checking whether the cell is within // the matrix bounds if (x < 0 || y < 0 || x >= grid.GetLength(0) || grid.GetLength(1) <= y) { return; } // If grid value is X, we update the // grid value as 0 and call the // function recursively for // adjacent nodes. if (grid[x, y] == 'X') { grid[x, y] = 'O'; dfs(grid, x + 1, y); dfs(grid, x, y + 1); dfs(grid, x - 1, y); dfs(grid, x, y - 1); } } // Function to find the number of // 'X' total shapes. static int xShape(char[, ] grid) { int n = grid.GetLength(0); int m = grid.GetLength(1); int i, j, ans = 0; // Traversing all the cells of the matrix. for (i = 0; i < n; i++) { for (j = 0; j < m; j++) { // If grid value is X, we increment // the counter and call the // dfs function. if (grid[i, j] == 'X') { ans++; dfs(grid, i, j); } } } // Returning the count. return ans; } static public void Main() { // Code char[, ] grid = { { 'X', 'O', 'X' }, { 'O', 'X', 'O' }, { 'X', 'X', 'X' } }; // Function call Console.WriteLine(xShape(grid)); }}// This code is contributed by lokesh. |
Javascript
// JavaScript code for the above approach // Function to run DFS function dfs(grid, x, y) { // Checking whether the cell is within // the matrix bounds if (x < 0 || y < 0 || x >= grid.length || grid[x].length <= y) { return; } // If grid value is X, we update the // grid value as 0 and call the // function recursively for // adjacent nodes. if (grid[x][y] == 'X') { grid[x][y] = 'O'; dfs(grid, x + 1, y); dfs(grid, x, y + 1); dfs(grid, x - 1, y); dfs(grid, x, y - 1); } } // Function to find the number of // 'X' total shapes. function xShape(grid) { let n = grid.length; let m = grid[0].length; let i, j, ans = 0; // Traversing all the cells of the matrix. for (i = 0; i < grid.length; i++) { for (j = 0; j < grid[i].length; j++) { // If grid value is X, we increment // the counter and call the // dfs function. if (grid[i][j] == 'X') { ans++; dfs(grid, i, j); } } } // Returning the count. return ans; } // Driver Code let grid = [['X', 'O', 'X'], ['O', 'X', 'O'], ['X', 'X', 'X']]; // Function Call console.log(xShape(grid));// This code is contributed by Potta Lokesh. |
Output
3
Time Complexity: O(N * M), as the vector grid is having N * M cells, and each cell is visited at a max of one time.
Auxiliary Space: O(1) If the recursion stack is considered the time complexity is O(N * M)
Related Articles:
- Introduction to Matrix or Grid – Data Structures and Algorithms Tutorials
- Depth First Search or DFS for a Graph
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