HackerRank DFS: Connected Cell in a Grid solution YASH PAL, 31 July 2024 In this HackerRank DFS: Connected Cell in a Grid Interview preparation kit problem there is Given an n x m matrix, find and print the number of cells in the largest region in the matrix. Problem solution in Python programming. def getAdjacentNodes(matrix, i, j): s = set() if i - 1 >= 0: if matrix[i-1][j] == 1: s = s.union([(i-1, j)]) if j - 1 >= 0: if matrix[i-1][j-1] == 1: s = s.union([(i-1, j-1)]) if j + 1 < len(matrix[i]): if matrix[i-1][j+1] == 1: s = s.union([(i-1, j+1)]) if i + 1 < len(matrix): if matrix[i+1][j] == 1: s = s.union([(i+1, j)]) if j - 1 >= 0: if matrix[i+1][j-1] == 1: s = s.union([(i+1, j-1)]) if j + 1 < len(matrix[i]): if matrix[i+1][j+1] == 1: s = s.union([(i+1, j+1)]) if j - 1 >= 0: if matrix[i][j-1] == 1: s = s.union([(i, j-1)]) if j + 1 < len(matrix[i]): if matrix[i][j+1] == 1: s = s.union([(i, j+1)]) return s def matrixToAdjacencyList(matrix): g = {} for i in range(len(matrix)): for j in range(len(matrix[i])): if matrix[i][j] == 1: g[(i, j)] = getAdjacentNodes(matrix, i, j) return g def dfs(graph): def helper(start): visited, stack = set(), [start] while stack: vertex = stack.pop() if vertex not in visited: visited.add(vertex) stack.extend(graph[vertex] - visited) return len(visited) return helper def getBiggestRegion(grid): graph = matrixToAdjacencyList(grid) sizes = [] gdfs = dfs(graph) for i in range(len(grid)): for j in range(len(grid[i])): if grid[i][j] == 1: sizes.append(gdfs((i, j))) return max(sizes) n = int(input().strip()) m = int(input().strip()) grid = [] for grid_i in range(n): grid_t = list(map(int, input().strip().split(' '))) grid.append(grid_t) print(getBiggestRegion(grid)) Problem solution in Java Programming. import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { static boolean[] marked; static int[] id; static int[] size; static int count; private static int largestConnectedComponentCount(Graph g) { List<Integer> nodes = new ArrayList<>(); int i = 0; for (boolean e : g.enabled) { if (e) { nodes.add(i); } i++; } marked = new boolean[g.enabled.length]; id = new int[g.enabled.length]; size = new int[g.enabled.length]; count = 0; for (int v : nodes) { if (!marked[v]) { dfs(g, v); count++; } } return Arrays.stream(size).max().orElse(0); } // depth-first search private static void dfs(Graph g, int v) { marked[v] = true; id[v] = count; size[count]++; g.adj.get(v) .stream() .filter(w -> !marked[w]) .forEach(w -> dfs(g, w)); } private static class Graph { boolean[] enabled; List<Set<Integer>> adj; Graph(int size) { enabled = new boolean[size]; adj = new ArrayList<>(); for (int i = 0; i < size; i++) { adj.add(new HashSet<>()); } } } public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int m = in.nextInt(); int i = 0; Graph g = new Graph(n * m); for(int row = 0; row < n; row++){ for(int col = 0; col < m; col++) { int val = in.nextInt(); if (val == 1) { g.enabled[i] = true; int prev = i - 1; int upLeft = i - m - 1; int up = i - m; int upRight = i - m + 1; if (col > 0 && g.enabled[prev]) { g.adj.get(prev).add(i); g.adj.get(i).add(prev); } if (col > 0 && row > 0 && g.enabled[upLeft]) { g.adj.get(upLeft).add(i); g.adj.get(i).add(upLeft); } if (row > 0 && g.enabled[up]) { g.adj.get(up).add(i); g.adj.get(i).add(up); } if (col < m - 1 && row > 0 && g.enabled[upRight]) { g.adj.get(upRight).add(i); g.adj.get(i).add(upRight); } } i++; } } System.out.println(largestConnectedComponentCount(g)); } } Problem solution in C++ programming. #include<bits/stdc++.h> using namespace std; int a[10][10]; int count(int i,int j,int n, int m) { int answer = 0; if(a[i][j] == 0) return 0; if(i<0 or j<0 or i>= n or j >=m) return 0; if(a[i][j] == 1) { answer++; a[i][j]--; } answer += count(i-1,j-1,n,m); answer += count(i-1,j,n,m); answer += count(i-1,j+1,n,m); answer += count(i,j-1,n,m); answer += count(i,j+1,n,m); answer += count(i+1,j-1,n,m); answer += count(i+1,j,n,m); answer += count(i+1,j+1,n,m); return answer; } int main() { int n,m; cin>>n>>m; for(int i=0;i<n;i++) for(int j=0;j<m;j++) cin>>a[i][j]; int max = -1,ans; for(int i=0;i<n;i++) for(int j=0;j<m;j++) { ans = count(i,j,n,m); if(ans > max) max = ans; } cout<<max; return 0; } Problem solution in C programming. #include <stdio.h> #include <stdint.h> #include <string.h> #define RUN_TEST_CASES(VAR) int VAR##_total; scanf("%d", & VAR##_total); for(int VAR=1; VAR<=VAR##_total; ++VAR) typedef unsigned long long ull; int get_area_size(int matrix[10][10], int rows, int cols, int row, int col) { if (row < 0 || row >= rows) return 0; if (col < 0 || col >= cols) return 0; if (matrix[row][col] != 1) return 0; matrix[row][col] = 0; // visited return 1 + get_area_size(matrix, rows, cols, row - 1, col - 1) + get_area_size(matrix, rows, cols, row - 1, col) + get_area_size(matrix, rows, cols, row - 1, col + 1) + get_area_size(matrix, rows, cols, row, col - 1) + get_area_size(matrix, rows, cols, row, col + 1) + get_area_size(matrix, rows, cols, row + 1, col - 1) + get_area_size(matrix, rows, cols, row + 1, col) + get_area_size(matrix, rows, cols, row + 1, col + 1) ; } int main() { #ifdef _DEBUG char FNAME[250]; strcpy(FNAME, __FILE__); strcpy(strchr(FNAME, '.'), ".txt"); freopen(FNAME, "rt", stdin); #endif int matrix[10][10]; int rows, cols; scanf("%d %d", &rows, &cols); for (int row = 0; row < rows; ++row) for (int col = 0; col < cols; ++col) scanf("%d", &matrix[row][col]); int largest_area = 0; for (int row = 0; row < rows; ++row) { for (int col = 0; col < cols; ++col) { if (matrix[row][col] != 1) continue; int area_size = get_area_size(matrix, rows, cols, row, col); if (area_size > largest_area) largest_area = area_size; } } printf("%d", largest_area); } Problem solution in JavaScript programming. process.stdin.resume(); process.stdin.setEncoding('ascii'); var input_stdin = ""; var input_stdin_array = ""; var input_currentline = 0; process.stdin.on('data', function (data) { input_stdin += data; }); process.stdin.on('end', function () { input_stdin_array = input_stdin.split("n"); main(); }); function readLine() { return input_stdin_array[input_currentline++]; } /////////////// ignore above this line //////////////////// var isValid = function(x, y, n, m) { return (x >= 0 && y >= 0 && x < n && y < m); } var visit = function(x, y, grid, visited, n, m) { if (grid[x][y] === 1 && !visited[x][y]) { var count = 1; visited[x][y] = true; for (var i = x - 1; i <= x + 1; i++) { for (var j = y - 1; j <= y + 1; j++) { if (isValid(i, j, n, m)) { count += visit(i, j, grid, visited, n, m); } } } return count; } else { return 0; } } function main() { var n = parseInt(readLine()); var m = parseInt(readLine()); var grid = []; for(grid_i = 0; grid_i < n; grid_i++){ grid[grid_i] = readLine().split(' '); grid[grid_i] = grid[grid_i].map(Number); } var visited = []; for (var i = 0; i < grid.length; i++) { visited.push([]); } var max = 0; for (var i = 0; i < n; i++) { for (var j = 0; j < m; j++) { var count = visit(i, j, grid, visited, n, m); max = Math.max(count, max); } } console.log(max); } coding problems interview prepration kit