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Find number of cavities in a matrix
• Difficulty Level : Medium
• Last Updated : 13 May, 2021

Count the number of the cavity in the 2d matrix, a cavity is defined as all the surrounding number 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 ``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));``}``}`

## 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

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## Javascript

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Output:

`1`

Optimizations We can avoid use of extra space and extra conditions by following below steps.
1) Explicitly check for four corner elements, remaining elements of first row, last row, first column and last column.
2) Check for remaining elements using above logic.

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