How do you solve Most Frequent Prime on LeetCode?

Most Frequent Prime (LeetCode #3044) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n²) space complexity. Key insight: Understanding how to traverse a 2D matrix in multiple directions is crucial for this problem.

What is the brute force solution for Most Frequent Prime?

The brute force approach for Most Frequent Prime is Brute Force, with O(n²) time complexity and O(1) space complexity. We will explore all possible paths from each cell in the matrix to generate numbers. This approach is straightforward but may be inefficient due to the large number of potential paths. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Most Frequent Prime?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases and how your solution will handle them. Discuss your thought process and approach with the interviewer to demonstrate your understanding.

What are common mistakes when solving Most Frequent Prime?

Failing to check for prime numbers correctly, especially for edge cases like numbers less than 2. Not handling the memoization correctly, leading to unnecessary recalculations.

#3044

Most Frequent Prime

Medium

ArrayHash TableMathMatrixCountingEnumerationNumber TheoryHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n²)

💡

Intuition

Time O(n²)Space O(n²)

We can optimize the brute force approach by using memoization to avoid recalculating paths from the same cell, thus reducing redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Use a memoization table to store results of previously computed paths from each cell.
2Step 2: For each cell, if the result is already computed, return it instead of recalculating.
3Step 3: After generating all numbers, filter for primes and find the most frequent one.

solution.py37 lines

1from collections import defaultdict
2
3class Solution:
4    def mostFrequentPrime(self, mat):
5        directions = [(0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1), (-1, 0), (-1, 1)]
6        prime_count = defaultdict(int)
7        primes = set()
8        memo = {}  # Memoization table
9
10        def is_prime(num):
11            if num < 2:
12                return False
13            for i in range(2, int(num**0.5) + 1):
14                if num % i == 0:
15                    return False
16            return True
17
18        def dfs(x, y, num):
19            if (x, y, num) in memo:
20                return memo[(x, y, num)]
21            if num > 10 and is_prime(num):
22                prime_count[num] += 1
23                primes.add(num)
24            for dx, dy in directions:
25                nx, ny = x + dx, y + dy
26                if 0 <= nx < len(mat) and 0 <= ny < len(mat[0]):
27                    dfs(nx, ny, num * 10 + mat[nx][ny])
28            memo[(x, y, num)] = prime_count
29            return prime_count
30
31        for i in range(len(mat)):
32            for j in range(len(mat[0])):
33                dfs(i, j, mat[i][j])
34
35        if not primes:
36            return -1
37        return max(prime_count, key=lambda x: (prime_count[x], x))

ℹ

Complexity note: The time complexity remains O(n²) as we still explore all cells, but memoization helps avoid redundant calculations. The space complexity increases to O(n²) due to the memoization table storing results for each cell.

1Understanding how to traverse a 2D matrix in multiple directions is crucial for this problem.
2Recognizing prime numbers and efficiently counting their occurrences can significantly impact performance.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.