Find number of cavities in a matrix

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Count the number of the cavity in the 2d matrix, a cavity is defined as all the surrounding numbers are greater than the mid number.

Examples:

Input : a = {{4, 5, 6}, {7, 1, 5}, {4, 5, 6}}
Output : 1

Input : a = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Output : 1

Source: Ola Interview Experience Set 13

Below is the implementation of above approach.

C++

// C++ program find number of cavities in a matrix
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 100;
 
int countCavities(int a[][MAX], int n)
{
    int A[n + 2][n + 2];
    int coun = 0;
 
    // form another matrix with one extra layer of 
    // boundary elements.
    // Boundary elements will contain max value.
    for (int i = 0; i < n + 2; i++) {
        for (int j = 0; j < n + 2; j++) {
            if ((i == 0) || (j == 0) || (i == n + 1) ||
                                         (j == n + 1))
                A[i][j] = INT_MAX;
            else
                A[i][j] = a[i - 1][j - 1];
        }
    }
 
    // Check for cavities in the modified matrix
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
 
            // check for all  directions
            if ((A[i][j] < A[i - 1][j]) && (A[i][j] < A[i + 1][j]) && 
                (A[i][j] < A[i][j - 1]) && (A[i][j] < A[i][j + 1]) && 
                (A[i][j] < A[i - 1][j - 1]) && (A[i][j] < A[i + 1][j + 1]) &&
                (A[i][j] < A[i - 1][j + 1]) && (A[i][j] < A[i + 1][j - 1]))
                coun++;
        }
    }
 
    return coun;
}
 
int main()
{
    int a[][MAX] = { { 4, 5, 6 }, { 7, 1, 5 }, { 4, 5, 6 } };
    int n = 3;
    cout << countCavities(a, n);
    return 0;
}

Java

// Java program find number of cavities in a matrix
class GfG { 
 
static int MAX = 100; 
 
static int countCavities(int a[][], int n) 
{ 
    int A[][] = new int[n + 2][n + 2]; 
    int coun = 0; 
 
    // form another matrix with one extra layer of 
    // boundary elements. 
    // Boundary elements will contain max value. 
    for (int i = 0; i < n + 2; i++) { 
        for (int j = 0; j < n + 2; j++) { 
            if ((i == 0) || (j == 0) || (i == n + 1) || (j == n + 1)) 
                A[i][j] = Integer.MAX_VALUE; 
            else
                A[i][j] = a[i - 1][j - 1]; 
        } 
    } 
 
    // Check for cavities in the modified matrix 
    for (int i = 1; i <= n; i++) { 
        for (int j = 1; j <= n; j++) { 
 
            // check for all directions 
            if ((A[i][j] < A[i - 1][j]) && (A[i][j] < A[i + 1][j]) && 
                (A[i][j] < A[i][j - 1]) && (A[i][j] < A[i][j + 1]) && 
                (A[i][j] < A[i - 1][j - 1]) && (A[i][j] < A[i + 1][j + 1]) && 
                (A[i][j] < A[i - 1][j + 1]) && (A[i][j] < A[i + 1][j - 1])) 
                coun++; 
        } 
    } 
 
    return coun; 
} 
 
public static void main(String[] args) 
{ 
    int a[][] = new int[][]{{ 4, 5, 6 }, { 7, 1, 5 }, { 4, 5, 6 }}; 
    int n = 3; 
System.out.println(countCavities(a, n)); 
}
} 

Python3

# Python program find number of cavities in a matrix
import sys
MAX = 100
 
def countCavities(a, n):
 
    A = [[0 for i in range(n + 2)] for j in range(n + 2)]
    count = 0
 
    # form another matrix with one extra layer of
    # boundary elements.
    # Boundary elements will contain max value.
    for i in range(n+2):
        for j in range(n+2):
            if ((i == 0) or (j == 0) or (i == n + 1) or (j == n + 1)):
                A[i][j] = sys.maxsize
            else:
                A[i][j] = a[i - 1][j - 1]
 
    # Check for cavities in the modified matrix
    for i in range(1,n):
        for j in range(1,n):
 
            # check for all directions
            if ((A[i][j] < A[i - 1][j]) and (A[i][j] < A[i + 1][j]) and
                (A[i][j] < A[i][j - 1]) and (A[i][j] < A[i][j + 1]) and
                (A[i][j] < A[i - 1][j - 1]) and (A[i][j] < A[i + 1][j + 1]) and
                (A[i][j] < A[i - 1][j + 1]) and (A[i][j] < A[i + 1][j - 1])):
                count += 1
 
    return count
 
# driver program
a = [ [ 4, 5, 6 ], [ 7, 1, 5 ], [ 4, 5, 6 ] ]
n = 3
print(countCavities(a, n))
 
# This code is contributed by shinjanpatra

C#

// C# program find number of cavities in a matrix
using System;
 
class GfG 
{ 
 
    static int MAX = 100; 
 
    static int countCavities(int [,]a, int n) 
    { 
        int [,]A = new int[n + 2, n + 2]; 
        int coun = 0; 
 
        // form another matrix with one extra layer of 
        // boundary elements. 
        // Boundary elements will contain max value. 
        for (int i = 0; i < n + 2; i++) 
        { 
            for (int j = 0; j < n + 2; j++) 
            { 
                if ((i == 0) || (j == 0) || (i == n + 1) || (j == n + 1)) 
                    A[i, j] = int.MaxValue; 
                else
                    A[i, j] = a[i - 1, j - 1]; 
            } 
        } 
 
        // Check for cavities in the modified matrix 
        for (int i = 1; i <= n; i++) 
        { 
            for (int j = 1; j <= n; j++)
            { 
 
                // check for all directions 
                if ((A[i, j] < A[i - 1, j]) && (A[i, j] < A[i + 1, j]) && 
                    (A[i, j] < A[i, j - 1]) && (A[i, j] < A[i, j + 1]) && 
                    (A[i, j] < A[i - 1, j - 1]) && (A[i, j] < A[i + 1, j + 1]) && 
                    (A[i, j] < A[i - 1, j + 1]) && (A[i, j] < A[i + 1, j - 1])) 
                    coun++; 
            } 
        } 
        return coun; 
    } 
 
    public static void Main(String[] args) 
    { 
        int [,]a = new int[,]{{ 4, 5, 6 }, { 7, 1, 5 }, { 4, 5, 6 }}; 
        int n = 3; 
        Console.WriteLine(countCavities(a, n)); 
    }
}
 
// This code contributed by Rajput-Ji

PHP

<?php
// PHP program find number of cavities
// in a matrix
function countCavities($a, $n) 
{ 
    $A = array(); 
    $coun = 0; 
 
    // form another matrix with one extra 
    // layer of boundary elements. 
    // Boundary elements will contain 
    // max value. 
    for ($i = 0; $i < $n + 2; $i++) 
    { 
        for ($j = 0; $j < $n + 2; $j++) 
        { 
            if (($i == 0) || ($j == 0) || 
                ($i == $n + 1) || ($j == $n + 1)) 
                $A[$i][$j] = 100; 
            else
                $A[$i][$j] = $a[$i - 1][$j - 1]; 
        } 
    } 
 
    // Check for cavities in the modified matrix 
    for ($i = 1; $i <= $n; $i++)
    { 
        for ($j = 1; $j <= $n; $j++) 
        { 
 
            // check for all directions 
            if (($A[$i][$j] < $A[$i - 1][$j]) && 
                ($A[$i][$j] < $A[$i + 1][$j]) && 
                ($A[$i][$j] < $A[$i][$j - 1]) && 
                ($A[$i][$j] < $A[$i][$j + 1]) && 
                ($A[$i][$j] < $A[$i - 1][$j - 1]) && 
                ($A[$i][$j] < $A[$i + 1][$j + 1]) && 
                ($A[$i][$j] < $A[$i - 1][$j + 1]) && 
                ($A[$i][$j] < $A[$i + 1][$j - 1])) 
                $coun++; 
        } 
    } 
 
    return $coun; 
} 
 
// Driver Code
$a = array(array(4, 5, 6), 
           array(7, 1, 5), 
           array(4, 5, 6)); 
$n = 3; 
echo(countCavities($a, $n)); 
 
// This code is contributed by Code_Mech

Javascript

<script>
// Javascript program find number of cavities in a matrix
 
MAX = 100;
 
function countCavities( a, n)
{
    var A = new Array(n+2).fill(0).map(() => new Array(n+2).fill(0));
    var coun = 0;
 
    // form another matrix with one extra layer of 
    // boundary elements.
    // Boundary elements will contain max value.
    for (var i = 0; i < n + 2; i++) {
        for (var j = 0; j < n + 2; j++) {
            if ((i == 0) || (j == 0) || (i == n + 1) ||
                                         (j == n + 1))
                A[i][j] = Number.MAX_VALUE;
            else
                A[i][j] = a[i - 1][j - 1];
        }
    }
 
    // Check for cavities in the modified matrix
    for (var i = 1; i <= n; i++) {
        for (var j = 1; j <= n; j++) {
 
            // check for all  directions
            if ((A[i][j] < A[i - 1][j]) && (A[i][j] < A[i + 1][j]) && 
                (A[i][j] < A[i][j - 1]) && (A[i][j] < A[i][j + 1]) && 
                (A[i][j] < A[i - 1][j - 1]) && (A[i][j] < A[i + 1][j + 1]) &&
                (A[i][j] < A[i - 1][j + 1]) && (A[i][j] < A[i + 1][j - 1]))
                coun++;
        }
    }
 
    return coun;
}
 
 
 var a = [ [ 4, 5, 6 ],[ 7, 1, 5 ], [ 4, 5, 6 ] ];
 var n = 3;
 document.write( countCavities(a, n));
 
// This code is contributed by SoumikMondal
</script>

Output

Optimizations We can avoid use of extra space and extra conditions by following below steps.

Explicitly check for four corner elements, remaining elements of first row, last row, first column and last column.
Check for remaining elements using above logic.

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