Get started

Prime In Diagonal — Easy Problem & Solution

Return the largest prime that lies on either diagonal of the square matrix nums, or 0 if neither diagonal holds a prime.

Problem statement

Return the largest prime that lies on either diagonal of the square matrix nums, or 0 if neither diagonal holds a prime.

The diagonals are the cells with i == j and the cells with i + j == n - 1.

Example 1

Input: nums = [[1,2,3],[5,6,7],[9,10,11]]
Output: 11
Explanation: The diagonals hold 1, 6, 11, 3, 6, 9; the largest prime is 11.

Example 2

Input: nums = [[1,2,3],[5,17,7],[9,11,10]]
Output: 17

Example 3

Input: nums = [[1,4],[6,8]]
Output: 0
Explanation: No diagonal entry is prime.

Constraints

  • 1 <= nums.length <= 300
  • nums.length == nums[i].length
  • 1 <= nums[i][j] <= 4000000

How to solve Prime In Diagonal

Visit the 2n diagonal cells and keep the largest prime. Checking value > best before testing primality keeps the expensive part rare.

Approach

  1. For each row i, look at nums[i][i] and nums[i][n-1-i].
  2. If a value exceeds the current best, test it for primality by trial division up to its square root.
  3. Return the best found, or 0.

Why it works

Trial division to sqrt(x) is enough because a composite x must have a factor at or below its square root. Writing the loop bound as d * d <= x avoids a floating-point square root entirely, which matters where the harness has no math library. The > best guard means most cells cost one comparison.

Complexity

  • Time — O(n · sqrt(V))
  • Space — O(1)

Pitfalls

  • The two diagonals overlap at the centre of an odd-sized matrix — harmless here, since taking a maximum is idempotent.
  • 1 is not prime, and the answer for 'no prime found' is 0.
  • A sieve up to 4,000,000 also works but is far more memory than the 2n cells need.

Reference solution

Python

from typing import List

def diagonalPrime(nums: List[List[int]]) -> int:
    def is_prime(x: int) -> bool:
        if x < 2:
            return False
        d = 2
        while d * d <= x:
            if x % d == 0:
                return False
            d += 1
        return True

    n = len(nums)
    best = 0
    for i in range(n):
        for v in (nums[i][i], nums[i][n - 1 - i]):
            if v > best and is_prime(v):
                best = v
    return best

JavaScript

var diagonalPrime = function(nums) {
    var isPrime = function(x) {
        if (x < 2) return false;
        for (var d = 2; d * d <= x; d++) if (x % d === 0) return false;
        return true;
    };
    var n = nums.length;
    var best = 0;
    for (var i = 0; i < n; i++) {
        var a = nums[i][i], b = nums[i][n - 1 - i];
        if (a > best && isPrime(a)) best = a;
        if (b > best && isPrime(b)) best = b;
    }
    return best;
};

Also on the editorial tab: C, C#, C++, Go, Java, Kotlin, PHP, Ruby, Rust, Swift, TypeScript.

All 667 arrays problems · the whole catalogue