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Count of cells in a matrix whose adjacent cells’s sum is prime Number
• Last Updated : 17 May, 2021

Given a M x N matrix mat[][], the task is to count the number of cells which have the sum of its adjacent cells equal to a prime number. For a cell x[i][j], only x[i+1][j], x[i-1][j], x[i][j+1] and x[i][j-1] are the adjacent cells.
Examples:

Input : mat[][] = {{1, 3}, {2, 5}}
Output :
Explanation: Only the cells mat[0][0] and mat[1][1] satisfying the condition.
i.e for mat[0][0]:(3+2) = 5, for mat[1][1]: (3+2) = 5
Input : mat[][] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}
Output :
Explanation: Cells mat[0][0], mat[0][2], mat[0][3], mat[1][3], mat[2][2] and mat[2][3] are satisfying the condition.

Prerequisites: Sieve of Eratosthenes
Approach:

• Precompute and store the prime numbers using Sieve.
• Iterate the entire matrix and for each cell find the sum of all adjacent cells.
• If the sum of adjacent cells equal to a prime number then increments the count.
• Return the value of the count.

Below is the implementation of the above approach.

## C++

 // CPP program to find the cells whose// adjacent cells's sum is prime Number#include using namespace std;#define MAX 100005 bool prime[MAX]; void SieveOfEratosthenes(){    // Create a boolean array "prime[0..MAX-1]"    // and initialize all entries it as true.    // A value in prime[i] will finally    // be false if i is Not a prime, else true.    memset(prime, true, sizeof(prime));     prime[0] = prime[1] = false;     for (int p = 2; p * p < MAX; p++) {        // If prime[p] is not changed,        // then it is a prime        if (prime[p] == true) {            // Update all multiples of p            // greater than or            // equal to the square of it            // numbers which are multiple of p and are            // less than p^2 are already been marked.            for (int i = p * p; i < MAX; i += p)                prime[i] = false;        }    }} // Function to count the cells having// adjacent cell's sum// is equal to primeint PrimeSumCells(vector >& mat){    int count = 0;     int N = mat.size();    int M = mat[0].size();     // Traverse for all the cells    for (int i = 0; i < N; i++) {        for (int j = 0; j < M; j++) {             int sum = 0;             // i-1, j            if (i - 1 >= 0)                sum += mat[i - 1][j];             // i+1, j            if (i + 1 < N)                sum += mat[i + 1][j];             // i, j-1            if (j - 1 >= 0)                sum += mat[i][j - 1];             // i, j+1            if (j + 1 < M)                sum += mat[i][j + 1];             // If the sum is a prime number            if (prime[sum])                count++;        }    }     // Return the count    return count;} // Driver Programint main(){    SieveOfEratosthenes();     vector > mat = { { 1, 2, 3, 4 },                                 { 5, 6, 7, 8 },                                 { 9, 10, 11, 12 } };     // Function call    cout << PrimeSumCells(mat) << endl;}

## Java

 // Java program to find the cells whose// adjacent cells's sum is prime Numberclass GFG{static final int MAX = 100005; static boolean []prime = new boolean[MAX]; static void SieveOfEratosthenes(){    // Create a boolean array "prime[0..MAX-1]"    // and initialize all entries it as true.    // A value in prime[i] will finally    // be false if i is Not a prime, else true.    for (int i = 0; i < prime.length; i++)    prime[i] = true;     prime[0] = prime[1] = false;     for (int p = 2; p * p < MAX; p++)    {        // If prime[p] is not changed,        // then it is a prime        if (prime[p] == true)        {            // Update all multiples of p            // greater than or            // equal to the square of it            // numbers which are multiple of p and are            // less than p^2 are already been marked.            for (int i = p * p; i < MAX; i += p)                prime[i] = false;        }    }} // Function to count the cells having// adjacent cell's sum// is equal to primestatic int PrimeSumCells(int [][]mat){    int count = 0;     int N = mat.length;    int M = mat[0].length;     // Traverse for all the cells    for (int i = 0; i < N; i++)    {        for (int j = 0; j < M; j++)        {            int sum = 0;             // i-1, j            if (i - 1 >= 0)                sum += mat[i - 1][j];             // i+1, j            if (i + 1 < N)                sum += mat[i + 1][j];             // i, j-1            if (j - 1 >= 0)                sum += mat[i][j - 1];             // i, j+1            if (j + 1 < M)                sum += mat[i][j + 1];             // If the sum is a prime number            if (prime[sum])                count++;        }    }     // Return the count    return count;} // Driver Codepublic static void main(String[] args){    SieveOfEratosthenes();     int [][]mat = { { 1, 2, 3, 4 },                    { 5, 6, 7, 8 },                    { 9, 10, 11, 12 } };     // Function call    System.out.print(PrimeSumCells(mat) + "\n");}} // This code is contributed by sapnasingh4991

## Python3

 # Python 3 program to# find the cells whose# adjacent cells's# sum is prime NumberMAX = 100005prime = [True] * MAX def SieveOfEratosthenes():     # Create a boolean array "prime[0..MAX-1]"    # and initialize all entries it as true.    # A value in prime[i] will finally    # be false if i is Not a prime, else true.    global prime         prime[0] = prime[1] = False     p = 2    while p * p < MAX:               # If prime[p] is not changed,        # then it is a prime        if (prime[p] == True):                       # Update all multiples of p            # greater than or            # equal to the square of it            # numbers which are multiple of            # p and are less than p^2 are            # already been marked.            for i in range (p * p, MAX, p):                prime[i] = False                       p += 1       # Function to count the# cells having adjacent# cell's sum is equal to primedef PrimeSumCells(mat):     count = 0    N = len(mat)    M = len(mat[0])     # Traverse for all the cells    for i in range (N):        for j in range (M):             sum = 0             # i - 1, j            if (i - 1 >= 0):                sum += mat[i - 1][j]             # i + 1, j            if (i + 1 < N):                sum += mat[i + 1][j]             # i, j - 1            if (j - 1 >= 0):                sum += mat[i][j - 1]             # i, j + 1            if (j + 1 < M):                sum += mat[i][j + 1]             # If the sum is a prime number            if (prime[sum]):                count += 1        # Return the count    return count # Driver codeif __name__ =="__main__":           SieveOfEratosthenes()    mat = [[1, 2, 3, 4],           [5, 6, 7, 8],           [9, 10, 11, 12]]     # Function call    print (PrimeSumCells(mat))     # This code is contributed by Chitranayal

## C#

 // C# program to find the cells whose// adjacent cells's sum is prime Numberusing System;class GFG{     static readonly int MAX = 100005;static bool []prime = new bool[MAX]; static void SieveOfEratosthenes(){    // Create a bool array "prime[0..MAX-1]"    // and initialize all entries it as true.    // A value in prime[i] will finally    // be false if i is Not a prime, else true.    for (int i = 0; i < prime.Length; i++)    prime[i] = true;     prime[0] = prime[1] = false;     for (int p = 2; p * p < MAX; p++)    {        // If prime[p] is not changed,        // then it is a prime        if (prime[p] == true)        {            // Update all multiples of p            // greater than or            // equal to the square of it            // numbers which are multiple of p and are            // less than p^2 are already been marked.            for (int i = p * p; i < MAX; i += p)                prime[i] = false;        }    }} // Function to count the cells having// adjacent cell's sum// is equal to primestatic int PrimeSumCells(int [,]mat){    int count = 0;     int N = mat.GetLength(0);    int M = mat.GetLength(1);     // Traverse for all the cells    for (int i = 0; i < N; i++)    {        for (int j = 0; j < M; j++)        {            int sum = 0;             // i-1, j            if (i - 1 >= 0)                sum += mat[i - 1, j];             // i+1, j            if (i + 1 < N)                sum += mat[i + 1, j];             // i, j-1            if (j - 1 >= 0)                sum += mat[i, j - 1];             // i, j+1            if (j + 1 < M)                sum += mat[i, j + 1];             // If the sum is a prime number            if (prime[sum])                count++;        }    }     // Return the count    return count;} // Driver Codepublic static void Main(String[] args){    SieveOfEratosthenes();     int [,]mat = { { 1, 2, 3, 4 },                   { 5, 6, 7, 8 },                   { 9, 10, 11, 12 } };     // Function call    Console.Write(PrimeSumCells(mat) + "\n");}} // This code is contributed by sapnasingh4991

## Javascript


Output:
6

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