HackerRank Cavity Map Problem Solution Yashwant Parihar, April 19, 2023April 19, 2023 In this Post, we will solve HackerRank Cavity Map Problem Solution. You are given a square map as a matrix of integer strings. Each cell of the map has a value denoting its depth. We will call a cell of the map a cavity if and only if this cell is not on the border of the map and each cell adjacent to it has strictly smaller depth. Two cells are adjacent if they have a common side, or edge.Find all the cavities on the map and replace their depths with the uppercase character X.Examplegrid [‘989′,’ 191′,’ 111′]The grid is rearranged for clarity: 989 191 111 Return: 989 1X1 111 The center cell was deeper than those on its edges: [8,1,1,1]. The deep cells in the top two corners do not share an edge with the center cell, and none of the border cells is eligible. Function Description Complete the cavityMap function in the editor below. cavityMap has the following parameter(s): string grid[n]: each string represents a row of the grid Returns string{n}: the modified grid Input FormatThe first line contains an integer n, the number of rows and columns in the grid.Each of the following n lines (rows) contains n positive digits without spaces (columns) thatrepresent the depth at grid[row, column]. Sample Input STDIN Function ----- -------- 4 grid[] size n = 4 1112 grid = ['1112', '1912', '1892', '1234'] 1912 1892 1234 Sample Output 1112 1X12 18X2 1234 Explanation The two cells with the depth of 9 are not on the border and are surrounded on all sides by shallower cells. Their values are replaced by X. HackerRank Cavity Map Problem Solution Cavity Map C Solution #include<stdio.h> int main(void) { int N; scanf("%d",&N); int mat[N][N]; int i=0,j; for(;i<N;i++) { for(j=0;j<N;j++) { scanf("%1d",&mat[i][j]); } } for(i=0;i<N;i++) { for(j=0;j<N;j++) { if((i!=0 && i!=N-1 && j!=0 && j!=N-1) && mat[i][j]>mat[i+1][j] && mat[i][j]>mat[i][j+1] && mat[i][j]>mat[i-1][j] &&mat[i][j]>mat[i][j-1]) printf("X"); else printf("%d",mat[i][j]); } printf("\n"); } return 0; } Cavity Map C++ Solution #include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { int n; cin >> n; vector<string> grid(n); for (string& s : grid) cin >> s; vector<string> grid2 = grid; for (int i = 1; i < n - 1; i++) { for (int j = 1; j < n - 1; j++) { if (grid[i][j] > grid[i - 1][j] && grid[i][j] > grid[i + 1][j] && grid[i][j] > grid[i][j - 1] && grid[i][j] > grid[i][j + 1]) grid2[i][j] = 'X'; } } for (string& s : grid2) cout << s << endl; return 0; } Cavity Map C Sharp Solution using System; using System.Collections.Generic; using System.IO; using System.Linq; class Solution { static void Main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */ var sizeOfMap = Convert.ToInt32(Console.ReadLine()); var intArray = new int[sizeOfMap][]; var outCharArray = new char[sizeOfMap][]; for (int i = 0; i < sizeOfMap; i++) { var input = Console.ReadLine().ToCharArray(); intArray[i] = input.Select(c => Convert.ToInt32(c.ToString())).ToArray(); outCharArray[i] = input; } for (int i = 1; i < sizeOfMap - 1; i++) { for (int j = 1; j < sizeOfMap - 1; j++) { var item = intArray[i][j]; var lFlag = intArray[i][j - 1] < item; var rFlag = intArray[i][j + 1] < item; var bFlag = intArray[i + 1][j] < item; var tFlag = intArray[i - 1][j] < item; var xFlag = lFlag && rFlag && bFlag && tFlag; outCharArray[i][j] = xFlag ? 'X' : char.Parse(intArray[i][j].ToString()); } } for (int i = 0; i < sizeOfMap; i++) { Console.WriteLine(outCharArray[i]); } } } Cavity Map Java Solution import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.function.*; import java.util.regex.*; import java.util.stream.*; import static java.util.stream.Collectors.joining; import static java.util.stream.Collectors.toList; class Result { /* * Complete the 'cavityMap' function below. * * The function is expected to return a STRING_ARRAY. * The function accepts STRING_ARRAY grid as parameter. */ public static List<String> cavityMap(List<String> grid) { // Write your code here List<String> output = new ArrayList<>(); StringBuilder sb = new StringBuilder(); output.add(grid.get(0)); for(int i=1;i<grid.size()-1;i++){ sb.setLength(0); sb.append(grid.get(i)); for(int j=1;j<grid.get(i).length()-1;j++){ int element = grid.get(i).charAt(j) - '0'; int side1 = grid.get(i).charAt(j-1) - '0'; int side2 = grid.get(i).charAt(j+1) - '0'; int edge1 = grid.get(i+1).charAt(j) - '0'; int edge2 = grid.get(i-1).charAt(j) - '0'; if(side1 < element && side2 < element && edge1 < element && edge2 < element){ sb.setCharAt(j, 'X'); } } output.add(sb.toString()); } if(grid.size()>1){ output.add(grid.get(grid.size() - 1)); } return output; } } public class Solution { public static void main(String[] args) throws IOException { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in)); BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); int n = Integer.parseInt(bufferedReader.readLine().trim()); List<String> grid = IntStream.range(0, n).mapToObj(i -> { try { return bufferedReader.readLine(); } catch (IOException ex) { throw new RuntimeException(ex); } }) .collect(toList()); List<String> result = Result.cavityMap(grid); bufferedWriter.write( result.stream() .collect(joining("\n")) + "\n" ); bufferedReader.close(); bufferedWriter.close(); } } Cavity Map JavaScript Solution function processData(input) { //Enter your code here var lines=input.split('\n'); var n = parseInt( lines.shift() ); // print the resulting map for( var y=0; y<n; y++ ) { if( y==0 || y==n-1 ) { // shortcut the first and last lines process.stdout.write(lines[y] + "\n"); } else { for( var x=0; x<n; x++ ) { if( x==0 || x==n-1 ) { // shortcut the first and last character in each line process.stdout.write( lines[y].charAt(x) ); } else { // check for a "cavity" var me = lines[y].charAt(x); if( me>lines[y-1].charAt(x) && me>lines[y+1].charAt(x) && me>lines[y].charAt(x-1) && me>lines[y].charAt(x+1) ) { process.stdout.write('X'); } else { process.stdout.write( lines[y].charAt(x) ); } } } process.stdout.write('\n'); } } } process.stdin.resume(); process.stdin.setEncoding("ascii"); _input = ""; process.stdin.on("data", function (input) { _input += input; }); process.stdin.on("end", function () { processData(_input); }); Cavity Map Python Solution n = int(input()) a = [] for i in range(n): a.append(input()) s = a[0] s += "\n" for i in range(1, n - 1): s += a[i][0] for j in range(1, n - 1): c = a[i][j] if a[i-1][j] < c and a[i+1][j] < c and a[i][j+1] < c and a[i][j-1] < c: s += "X" else: s += c s += a[i][-1] s += "\n" if n > 1: s += a[-1] + "\n" print(s) Other Solutions HackerRank Manasa and Stones Problem Solution HackerRank The Grid Search Problem Solution c C# C++ HackerRank Solutions java javascript python CcppCSharpHackerrank Solutionsjavajavascriptpython