In this HackerRank Cavity Map problem, 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 a 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.

HackerRank Cavity Map problem solution


Problem solution in Python programming.

n = int(input())
a = [list(input()) for _ in range(n)]

def getval(p, off):
    px = p[0] + off[0]
    py = p[1] + off[1]
    pi = a[px][py]
    return 10 if pi == 'X' else int(pi)

for i in range(1, n-1):
    for j in range(1, n-1):
        p = (i, j)
        if getval(p, (0, 0)) > max(getval(p, off) for off in (
            (0, 1), (0, -1), (-1, 0), (1, 0))):
            a[i][j] = 'X'
            
for row in a:
    print(''.join(row))


Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner in = new Scanner(System.in);
        int n = Integer.parseInt(in.nextLine());
        int[][] array = new int[n][n];
        int[][] target = new int[n][n];
        for(int i=0; i<n; i++){
        	String line = in.nextLine();
        	char[] data = line.toCharArray();
            for(int j=0; j<n; j++){
                array[i][j] = Character.getNumericValue(data[j]);
                //System.out.println(array[i][j]);
            }
        }
        cavityMap(array, target);
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(target[i][j] == -1)
                    System.out.print("X");
                else
                    System.out.print(array[i][j]);
            }
            System.out.println("");
        }
    }
    
    public static void cavityMap(int[][] array, int[][] target) {
        int row = array.length;
        int col = array[0].length;
        for(int i=1; i<row-1; i++){
            for(int j=1; j<col-1; j++){
                if(isMaximum(array, i, j)){
                    target[i][j] = -1; // will be translate to X later
                }
            }
        }
    }
    
    public static boolean isMaximum(int[][] array, int i, int j) {
        int max = Math.max(Math.max(array[i-1][j], array[i+1][j]), Math.max(array[i][j-1], array[i][j+1]));
        if(max >= array[i][j])
            return false;
        else
            return true;
    }
}


Problem solution in C++ programming.

#include <iostream>
#include <ctime>
#include <fstream>
#include <cmath>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <complex>
#include <utility>
#include <cctype>
#include <list>
#include <deque>

using namespace std;

#define FORALL(i,a,b) for(int i=(a);i<=(b);++i)
#define FOR(i,n) for(int i=0;i<(n);++i)
#define FORB(i,a,b) for(int i=(a);i>=(b);--i)

typedef long long ll;
typedef long double ld;
typedef complex<ld> vec;

typedef pair<int,int> pii;
typedef map<int,int> mii;

#define pb push_back
#define mp make_pair

#define MV 4
#define MAXN 102
char A[MAXN][MAXN];
char B[MAXN][MAXN];
int dr[] = {-1,0,1,0};
int dc[] = {0,-1,0,1};
int N;
bool is_cavity(int i, int j)  {
	if (i==0 || i==N-1) return false;
	if (j==0 || j==N-1) return false;
	FOR(mv,MV) {
		int i2 = i + dr[mv];
		int j2 = j + dc[mv];
		if (A[i2][j2] >= A[i][j]) return false;
	}
	return true;
}

int main() {
	cin >> N;
	FOR(i,N) scanf("%s",&A[i][0]);
	FOR(i,N) FOR(j,N)
		if (is_cavity(i,j)) B[i][j] = 'X';
		else B[i][j] = A[i][j];
	FOR(i,N) cout << B[i] << endl;
}


Problem solution in C programming.

#include<stdio.h>

int main()
{
    int t;
        int i,j,n;
        scanf("%d",&n);
        char a[n+2][n+2];
        char b[n+2][n+2];
        for(i=0; i<n; i++)
        {
            scanf("%s",a[i]);
        }

        for(i=0; i<n ;i++)
        {
            for(j=0; j<n; j++)
            {
                if( i > 0 && i < n-1 && j > 0 && j < n-1)
                {
                    char ch = a[i][j];
                    if(ch > a[i+1][j] && ch > a[i][j+1] && ch > a[i-1][j] && ch > a[i][j-1])
                        b[i][j] = 'X';
                    else
                        b[i][j] = a[i][j];
                }
                else
                    b[i][j] = a[i][j];
            }
            b[i][j] = 0;
        }
        for(i=0;i<n;i++)
        {
            printf("%s\n",b[i]);
        }

    return 0;
}


Problem solution in JavaScript programming.

function processData(input) {
    input = input.split('\n')
    input.shift()
    for (var i = 0; i < input.length ; i++ ){
        input[i] = input[i].split("");
    }
    
    var output = cavityMap(input);
    
    output.forEach(function(line){
        console.log( line.join("") );
    })
}

function cavityMap (map) {

    var size = map.length-1;
    var cavs = [];

    for (var i = 1; i < size; i++) {
        for (var j = 1; j < size; j++) {
            var curr = map[i][j];
            if ( curr > map[i-1][j] && curr > map[i+1][j] ){
                if (curr > map[i][j-1] && curr > map[i][j+1]) {
                    cavs.push([i,j]);
                }
            }

        };
    };

    cavs.forEach(function (cavity){
        map[cavity[0]][cavity[1]] = 'X';
    })

    return map;
}




process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});