Easy
Prime In Diagonal — C++
Full explanation · Time precompute: O(MAXN) runtime: O(n) · Space O(MAXN)
// Time: precompute: O(MAX_N)
// runtime: O(n)
// Space: O(MAX_N)
// number theory
vector<int> linear_sieve_of_eratosthenes(int n) { // Time: O(n), Space: O(n)
vector<int> spf(n + 1, -1);
vector<int> primes;
for (int i = 2; i <= n; ++i) {
if (spf[i] == -1) {
spf[i] = i;
primes.emplace_back(i);
}
for (const auto& p : primes) {
if (i * p > n || p > spf[i]) {
break;
}
spf[i * p] = p;
}
}
return primes; // len(primes) = O(n/(logn-1)), reference: https://math.stackexchange.com/questions/264544/how-to-find-number-of-prime-numbers-up-to-to-n
}
const int MAX_N = 4 * 1e6;
const auto& PRIMES = linear_sieve_of_eratosthenes(MAX_N);
const unordered_set<int> PRIMES_SET(cbegin(PRIMES), cend(PRIMES));
class Solution {
public:
int diagonalPrime(vector<vector<int>>& nums) {
int result = 0;
for (int i = 0; i < size(nums); ++i) {
if (PRIMES_SET.count(nums[i][i])) {
result = max(result, nums[i][i]);
}
if (PRIMES_SET.count(nums[i][size(nums) - 1 - i])) {
result = max(result, nums[i][size(nums) - 1 - i ]);
}
}
return result;
}
};