Medium
You are given a m x n
0-indexed 2D matrix mat
. From every cell, you can create numbers in the following way:
8
paths from the cells namely: east, south-east, south, south-west, west, north-west, north, and north-east.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
import kotlin.math.max
import kotlin.math.sqrt
@Suppress("NAME_SHADOWING")
class Solution {
private var max = 0
private var freqNum = -1
fun mostFrequentPrime(mat: Array<IntArray>): Int {
val nexts =
arrayOf(
intArrayOf(1, 1),
intArrayOf(-1, -1),
intArrayOf(1, -1),
intArrayOf(-1, 1),
intArrayOf(0, 1),
intArrayOf(0, -1),
intArrayOf(1, 0),
intArrayOf(-1, 0),
)
val m = mat.size
val n = mat[0].size
val primeFreq: MutableMap<Int, Int> = HashMap()
for (i in 0 until m) {
for (j in 0 until n) {
for (next in nexts) {
getPrime(i, j, mat, 0, next, primeFreq)
}
}
}
return freqNum
}
private fun getPrime(
i: Int,
j: Int,
mat: Array<IntArray>,
num: Int,
next: IntArray,
primeFreq: MutableMap<Int, Int>,
) {
var num = num
val m = mat.size
val n = mat[0].size
if (i < 0 || j < 0 || i == m || j == n) {
return
}
num = num * 10 + mat[i][j]
if (num > 10 && isPrime(num)) {
val count = primeFreq.getOrDefault(num, 0) + 1
if ((count == max && freqNum < num) || count > max) {
freqNum = num
}
max = max(max, count)
primeFreq[num] = count
}
getPrime(i + next[0], j + next[1], mat, num, next, primeFreq)
}
private fun isPrime(num: Int): Boolean {
if (num == 2) {
return true
}
if (num == 1 || (num and 1) == 0) {
return false
}
val n = sqrt(num.toDouble()).toInt()
var i = 3
while (i <= n) {
if ((num % i) == 0) {
return false
}
i += 2
}
return true
}
}