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LeetCode 3044: Most Frequent Prime Solution

Master LeetCode problem 3044 (Most Frequent Prime), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we need to traverse the 2D matrix and generate all possible numbers by moving in 8 directions from each cell. Then, we count the frequency of each prime number greater than 10 and return the most frequent prime number or -1 if no such prime number exists.

  1. Create a helper function to check if a number is prime.
  2. Initialize a map to store the frequency of prime numbers greater than 10.
  3. Traverse the matrix and generate all possible numbers from each cell by moving in 8 directions.
  4. Count the frequency of prime numbers greater than 10.
  5. Find the most frequent prime number greater than 10 or return -1 if none exists.

Time Complexity: O(m * n * 8 * log(max_element)) Space Complexity: O(m * n)

:

Solution Code

class Solution {
    public int mostFrequentPrime(int[][] mat) {
        // Helper function to check if a number is prime
        boolean isPrime(int num) {
            if (num <= 1) return false;
            for (int i = 2; i <= Math.sqrt(num); i++) {
                if (num % i == 0) return false;
            }
            return true;
        }
        
        Map<Integer, Integer> freqMap = new HashMap<>();
        
        int m = mat.length;
        int n = mat[0].length;
        
        int[] dx = {1, 1, 0, -1, -1, -1, 0, 1};
        int[] dy = {0, 1, 1, 1, 0, -1, -1, -1};
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < 8; k++) {
                    int num = 0;
                    int x = i, y = j;
                    while (x >= 0 && x < m && y >= 0 && y < n) {
                        num = num * 10 + mat[x][y];
                        if (num > 10 && isPrime(num)) {
                            freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
                        }
                        x += dx[k];
                        y += dy[k];
                    }
                }
            }
        }
        
        int maxFreq = 0;
        int mostFreqPrime = -1;
        for (int prime : freqMap.keySet()) {
            if (freqMap.get(prime) > maxFreq || (freqMap.get(prime) == maxFreq && prime > mostFreqPrime)) {
                maxFreq = freqMap.get(prime);
                mostFreqPrime = prime;
            }
        }
        
        return mostFreqPrime;
    }
}

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