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 :2
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 : 6
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 <bits/stdc++.h>using namespace std;#define MAX 100005bool 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<vector<int> >& 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<vector<int> > 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] * MAXdef 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
<script>// Javascript program to find the cells whose// adjacent cells's sum is prime Numberlet MAX = 100005let prime = new Array(MAX);function 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. prime.fill(true) prime[0] = prime[1] = false; for (let 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 (let i = p * p; i < MAX; i += p) prime[i] = false; } }}// Function to count the cells having// adjacent cell's sum// is equal to primefunction PrimeSumCells(mat){ let count = 0; let N = mat.length; let M = mat[0].length; // Traverse for all the cells for (let i = 0; i < N; i++) { for (let j = 0; j < M; j++) { let 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 Program SieveOfEratosthenes(); let mat = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ] ]; // Function call document.write(PrimeSumCells(mat) + "<br>");// This code is contributed by gfgking</script> |
Output:
6
Time Complexity: O(N*M)
Auxiliary Space: O(MAX)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Find More Information here to that Topic: geeksforgeeks.org/count-of-cells-in-a-matrix-whose-adjacent-cellss-sum-is-prime-number/ […]