HackerRank Cavity Map problem solution

In this 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.

HackerRank Cavity Map problem solution in Python.

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))

Cavity Map 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("%sn",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);
});