LeetCode-in-Java

3044. Most Frequent Prime

Medium

You are given a m x n 0-indexed 2D matrix mat. From every cell, you can create numbers in the following way:

• There could be at most 8 paths from the cells namely: east, south-east, south, south-west, west, north-west, north, and north-east.
• Select a path from them and append digits in this path to the number being formed by traveling in this direction.
• Note that numbers are generated at every step, for example, if the digits along the path are 1, 9, 1, then there will be three numbers generated along the way: 1, 19, 191.

Return the most frequent prime number greater than 10 out of all the numbers created by traversing the matrix or -1 if no such prime number exists. If there are multiple prime numbers with the highest frequency, then return the largest among them.

Note: It is invalid to change the direction during the move.

Example 1:

Input: mat = [[1,1],[9,9],[1,1]]

Output: 19

Explanation:

From cell (0,0) there are 3 possible directions and the numbers greater than 10 which can be created in those directions are:

East: [11], South-East: [19], South: [19,191].

Numbers greater than 10 created from the cell (0,1) in all possible directions are: [19,191,19,11].

Numbers greater than 10 created from the cell (1,0) in all possible directions are: [99,91,91,91,91].

Numbers greater than 10 created from the cell (1,1) in all possible directions are: [91,91,99,91,91].

Numbers greater than 10 created from the cell (2,0) in all possible directions are: [11,19,191,19].

Numbers greater than 10 created from the cell (2,1) in all possible directions are: [11,19,19,191].

The most frequent prime number among all the created numbers is 19.

Example 2:

Input: mat = [[7]]

Output: -1

Explanation: The only number which can be formed is 7. It is a prime number however it is not greater than 10, so return -1.

Example 3:

Input: mat = [[9,7,8],[4,6,5],[2,8,6]]

Output: 97

Explanation:

Numbers greater than 10 created from the cell (0,0) in all possible directions are: [97,978,96,966,94,942].

Numbers greater than 10 created from the cell (0,1) in all possible directions are: [78,75,76,768,74,79].

Numbers greater than 10 created from the cell (0,2) in all possible directions are: [85,856,86,862,87,879].

Numbers greater than 10 created from the cell (1,0) in all possible directions are: [46,465,48,42,49,47].

Numbers greater than 10 created from the cell (1,1) in all possible directions are: [65,66,68,62,64,69,67,68].

Numbers greater than 10 created from the cell (1,2) in all possible directions are: [56,58,56,564,57,58].

Numbers greater than 10 created from the cell (2,0) in all possible directions are: [28,286,24,249,26,268].

Numbers greater than 10 created from the cell (2,1) in all possible directions are: [86,82,84,86,867,85].

Numbers greater than 10 created from the cell (2,2) in all possible directions are: [68,682,66,669,65,658].

The most frequent prime number among all the created numbers is 97.

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n <= 6
• 1 <= mat[i][j] <= 9

Solution

import java.util.HashMap;
import java.util.Map;

public class Solution {
    private int max = 0;
    private int freqNum = -1;

    public int mostFrequentPrime(int[][] mat) {
        int[][] nexts =
                new int[][] { {1, 1}, {-1, -1}, {1, -1}, {-1, 1}, {0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        int m = mat.length;
        int n = mat[0].length;
        Map<Integer, Integer> primeFreq = new HashMap<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                for (int[] next : nexts) {
                    getPrime(i, j, mat, 0, next, primeFreq);
                }
            }
        }
        return freqNum;
    }

    private void getPrime(
            int i, int j, int[][] mat, int num, int[] next, Map<Integer, Integer> primeFreq) {
        int m = mat.length;
        int n = mat[0].length;
        if (i < 0 || j < 0 || i == m || j == n) {
            return;
        }
        num = num * 10 + mat[i][j];
        if (num > 10 && isPrime(num)) {
            int count = primeFreq.getOrDefault(num, 0) + 1;
            if ((count == max && freqNum < num) || count > max) {
                freqNum = num;
            }
            max = Math.max(max, count);
            primeFreq.put(num, count);
        }
        getPrime(i + next[0], j + next[1], mat, num, next, primeFreq);
    }

    private boolean isPrime(int num) {
        if (num == 2) {
            return true;
        }
        if (num == 1 || (num & 1) == 0) {
            return false;
        }
        int n = (int) Math.sqrt(num);
        for (int i = 3; i <= n; i += 2) {
            if ((num % i) == 0) {
                return false;
            }
        }
        return true;
    }
}

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