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 : 1Input : 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 matrixclass 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 matriximport sysMAX = 100def 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 programa = [ [ 4, 5, 6 ], [ 7, 1, 5 ], [ 4, 5, 6 ] ]n = 3print(countCavities(a, n))# This code is contributed by shinjanpatra |
C#
// C# program find number of cavities in a matrixusing 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 matrixfunction 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 matrixMAX = 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
1
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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