LeetCode #3044 — MEDIUM

Most Frequent Prime

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math

Example 1

[[1,1],[9,9],[1,1]]

Example 2

[[7]]

Example 3

[[9,7,8],[4,6,5],[2,8,6]]

Step 02

Core Insight

What unlocks the optimal approach

Use recursion to find all possible numbers for each cell and then check for prime.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3044: Most Frequent Prime
class Solution {
    public int mostFrequentPrime(int[][] mat) {
        int m = mat.length, n = mat[0].length;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int a = -1; a <= 1; ++a) {
                    for (int b = -1; b <= 1; ++b) {
                        if (a == 0 && b == 0) {
                            continue;
                        }
                        int x = i + a, y = j + b, v = mat[i][j];
                        while (x >= 0 && x < m && y >= 0 && y < n) {
                            v = v * 10 + mat[x][y];
                            if (isPrime(v)) {
                                cnt.merge(v, 1, Integer::sum);
                            }
                            x += a;
                            y += b;
                        }
                    }
                }
            }
        }
        int ans = -1, mx = 0;
        for (var e : cnt.entrySet()) {
            int v = e.getKey(), x = e.getValue();
            if (mx < x || (mx == x && ans < v)) {
                mx = x;
                ans = v;
            }
        }
        return ans;
    }

    private boolean isPrime(int n) {
        for (int i = 2; i <= n / i; ++i) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #3044: Most Frequent Prime
func mostFrequentPrime(mat [][]int) int {
	m, n := len(mat), len(mat[0])
	cnt := make(map[int]int)
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			for a := -1; a <= 1; a++ {
				for b := -1; b <= 1; b++ {
					if a == 0 && b == 0 {
						continue
					}
					x, y, v := i+a, j+b, mat[i][j]
					for x >= 0 && x < m && y >= 0 && y < n {
						v = v*10 + mat[x][y]
						if isPrime(v) {
							cnt[v]++
						}
						x += a
						y += b
					}
				}
			}
		}
	}
	ans, mx := -1, 0
	for v, x := range cnt {
		if mx < x || (mx == x && ans < v) {
			mx = x
			ans = v
		}
	}
	return ans
}

func isPrime(n int) bool {
	for i := 2; i <= n/i; i++ {
		if n%i == 0 {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #3044: Most Frequent Prime
class Solution:
    def mostFrequentPrime(self, mat: List[List[int]]) -> int:
        def is_prime(x: int) -> int:
            return all(x % i != 0 for i in range(2, isqrt(x) + 1))

        m, n = len(mat), len(mat[0])
        cnt = Counter()
        for i in range(m):
            for j in range(n):
                for a in range(-1, 2):
                    for b in range(-1, 2):
                        if a == 0 and b == 0:
                            continue
                        x, y, v = i + a, j + b, mat[i][j]
                        while 0 <= x < m and 0 <= y < n:
                            v = v * 10 + mat[x][y]
                            if is_prime(v):
                                cnt[v] += 1
                            x, y = x + a, y + b
        ans, mx = -1, 0
        for v, x in cnt.items():
            if mx < x:
                mx = x
                ans = v
            elif mx == x:
                ans = max(ans, v)
        return ans

// Accepted solution for LeetCode #3044: Most Frequent Prime
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3044: Most Frequent Prime
// class Solution {
//     public int mostFrequentPrime(int[][] mat) {
//         int m = mat.length, n = mat[0].length;
//         Map<Integer, Integer> cnt = new HashMap<>();
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 for (int a = -1; a <= 1; ++a) {
//                     for (int b = -1; b <= 1; ++b) {
//                         if (a == 0 && b == 0) {
//                             continue;
//                         }
//                         int x = i + a, y = j + b, v = mat[i][j];
//                         while (x >= 0 && x < m && y >= 0 && y < n) {
//                             v = v * 10 + mat[x][y];
//                             if (isPrime(v)) {
//                                 cnt.merge(v, 1, Integer::sum);
//                             }
//                             x += a;
//                             y += b;
//                         }
//                     }
//                 }
//             }
//         }
//         int ans = -1, mx = 0;
//         for (var e : cnt.entrySet()) {
//             int v = e.getKey(), x = e.getValue();
//             if (mx < x || (mx == x && ans < v)) {
//                 mx = x;
//                 ans = v;
//             }
//         }
//         return ans;
//     }
// 
//     private boolean isPrime(int n) {
//         for (int i = 2; i <= n / i; ++i) {
//             if (n % i == 0) {
//                 return false;
//             }
//         }
//         return true;
//     }
// }

// Accepted solution for LeetCode #3044: Most Frequent Prime
function mostFrequentPrime(mat: number[][]): number {
    const m: number = mat.length;
    const n: number = mat[0].length;
    const cnt: Map<number, number> = new Map();
    const isPrime = (x: number): boolean => {
        for (let i = 2; i <= x / i; ++i) {
            if (x % i === 0) {
                return false;
            }
        }
        return true;
    };

    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            for (let a = -1; a <= 1; ++a) {
                for (let b = -1; b <= 1; ++b) {
                    if (a === 0 && b === 0) {
                        continue;
                    }
                    let [x, y, v] = [i + a, j + b, mat[i][j]];
                    while (x >= 0 && x < m && y >= 0 && y < n) {
                        v = v * 10 + mat[x][y];
                        if (isPrime(v)) {
                            cnt.set(v, (cnt.get(v) || 0) + 1);
                        }
                        x += a;
                        y += b;
                    }
                }
            }
        }
    }

    let [ans, mx] = [-1, 0];
    cnt.forEach((x, v) => {
        if (mx < x || (mx === x && ans < v)) {
            mx = x;
            ans = v;
        }
    });
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n × \max(m, n)

Space

O(m × n × \max(m, n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.