LeetCode #2614 — EASY

Prime In Diagonal

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed two-dimensional integer array nums.

Return the largest prime number that lies on at least one of the diagonals of nums. In case, no prime is present on any of the diagonals, return 0.

Note that:

An integer is prime if it is greater than 1 and has no positive integer divisors other than 1 and itself.
An integer val is on one of the diagonals of nums if there exists an integer i for which nums[i][i] = val or an i for which nums[i][nums.length - i - 1] = val.

In the above diagram, one diagonal is [1,5,9] and another diagonal is [3,5,7].

Example 1:

Input: nums = [[1,2,3],[5,6,7],[9,10,11]]
Output: 11
Explanation: The numbers 1, 3, 6, 9, and 11 are the only numbers present on at least one of the diagonals. Since 11 is the largest prime, we return 11.

Example 2:

Input: nums = [[1,2,3],[5,17,7],[9,11,10]]
Output: 17
Explanation: The numbers 1, 3, 9, 10, and 17 are all present on at least one of the diagonals. 17 is the largest prime, so we return 17.

Constraints:

1 <= nums.length <= 300
nums.length == nums_i.length
1 <= nums[i][j] <= 4*10⁶

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed two-dimensional integer array nums. Return the largest prime number that lies on at least one of the diagonals of nums. In case, no prime is present on any of the diagonals, return 0. Note that: An integer is prime if it is greater than 1 and has no positive integer divisors other than 1 and itself. An integer val is on one of the diagonals of nums if there exists an integer i for which nums[i][i] = val or an i for which nums[i][nums.length - i - 1] = val. In the above diagram, one diagonal is [1,5,9] and another diagonal is [3,5,7].

Baseline thinking

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

Pattern signal: Array · Math

Example 1

[[1,2,3],[5,6,7],[9,10,11]]

Example 2

[[1,2,3],[5,17,7],[9,11,10]]

Step 02

Core Insight

What unlocks the optimal approach

Iterate over the diagonals of the matrix and check for each element.
Check if the element is prime or not in O(sqrt(n)) time.

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 #2614: Prime In Diagonal
class Solution {
    public int diagonalPrime(int[][] nums) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (isPrime(nums[i][i])) {
                ans = Math.max(ans, nums[i][i]);
            }
            if (isPrime(nums[i][n - i - 1])) {
                ans = Math.max(ans, nums[i][n - i - 1]);
            }
        }
        return ans;
    }

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

// Accepted solution for LeetCode #2614: Prime In Diagonal
func diagonalPrime(nums [][]int) (ans int) {
	n := len(nums)
	for i, row := range nums {
		if isPrime(row[i]) {
			ans = max(ans, row[i])
		}
		if isPrime(row[n-i-1]) {
			ans = max(ans, row[n-i-1])
		}
	}
	return
}

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

# Accepted solution for LeetCode #2614: Prime In Diagonal
class Solution:
    def diagonalPrime(self, nums: List[List[int]]) -> int:
        def is_prime(x: int) -> bool:
            if x < 2:
                return False
            return all(x % i for i in range(2, int(sqrt(x)) + 1))

        n = len(nums)
        ans = 0
        for i, row in enumerate(nums):
            if is_prime(row[i]):
                ans = max(ans, row[i])
            if is_prime(row[n - i - 1]):
                ans = max(ans, row[n - i - 1])
        return ans

// Accepted solution for LeetCode #2614: Prime In Diagonal
impl Solution {
    pub fn diagonal_prime(nums: Vec<Vec<i32>>) -> i32 {
        let mut ans = 0;
        let n = nums.len();

        for (i, row) in nums.iter().enumerate() {
            if Self::is_prime(row[i]) && row[i] > ans {
                ans = row[i];
            }
            if Self::is_prime(row[n - i - 1]) && row[n - i - 1] > ans {
                ans = row[n - i - 1];
            }
        }

        ans
    }

    fn is_prime(n: i32) -> bool {
        if n < 2 {
            return false;
        }

        let upper = (n as f64).sqrt() as i32;
        for i in 2..=upper {
            if n % i == 0 {
                return false;
            }
        }

        true
    }
}

// Accepted solution for LeetCode #2614: Prime In Diagonal
function diagonalPrime(nums: number[][]): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        if (isPrime(nums[i][i])) {
            ans = Math.max(ans, nums[i][i]);
        }
        if (isPrime(nums[i][n - i - 1])) {
            ans = Math.max(ans, nums[i][n - i - 1]);
        }
    }
    return ans;
}

function isPrime(x: number): boolean {
    if (x < 2) {
        return false;
    }
    for (let i = 2; i <= Math.floor(x / i); ++i) {
        if (x % i === 0) {
            return false;
        }
    }
    return true;
}

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(n)

Space

O(1)

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.

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.