LeetCode #3765 — MEDIUM

Complete Prime Number

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer num.

A number num is called a Complete Prime Number if every prefix and every suffix of num is prime.

Return true if num is a Complete Prime Number, otherwise return false.

Note:

A prefix of a number is formed by the first k digits of the number.
A suffix of a number is formed by the last k digits of the number.
Single-digit numbers are considered Complete Prime Numbers only if they are prime.

Example 1:

Input: num = 23

Output: true

Explanation:

Prefixes of num = 23 are 2 and 23, both are prime.
Suffixes of num = 23 are 3 and 23, both are prime.
All prefixes and suffixes are prime, so 23 is a Complete Prime Number and the answer is true.

Example 2:

Input: num = 39

Output: false

Explanation:

Prefixes of num = 39 are 3 and 39. 3 is prime, but 39 is not prime.
Suffixes of num = 39 are 9 and 39. Both 9 and 39 are not prime.
At least one prefix or suffix is not prime, so 39 is not a Complete Prime Number and the answer is false.

Example 3:

Input: num = 7

Output: true

Explanation:

7 is prime, so all its prefixes and suffixes are prime and the answer is true.

Constraints:

1 <= num <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer num. A number num is called a Complete Prime Number if every prefix and every suffix of num is prime. Return true if num is a Complete Prime Number, otherwise return false. Note: A prefix of a number is formed by the first k digits of the number. A suffix of a number is formed by the last k digits of the number. Single-digit numbers are considered Complete Prime Numbers only if they are prime.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Example 3

Step 02

Core Insight

What unlocks the optimal approach

Check primality for all prefixes and all suffixes of <code>num</code> and return true only if every one is 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 #3765: Complete Prime Number
class Solution {
    public boolean completePrime(int num) {
        char[] s = String.valueOf(num).toCharArray();
        int x = 0;
        for (int i = 0; i < s.length; i++) {
            x = x * 10 + (s[i] - '0');
            if (!isPrime(x)) {
                return false;
            }
        }
        x = 0;
        int p = 1;
        for (int i = s.length - 1; i >= 0; i--) {
            x = p * (s[i] - '0') + x;
            p *= 10;
            if (!isPrime(x)) {
                return false;
            }
        }
        return true;
    }

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

// Accepted solution for LeetCode #3765: Complete Prime Number
func completePrime(num int) bool {
    isPrime := func(x int) bool {
        if x < 2 {
            return false
        }
        for i := 2; i*i <= x; i++ {
            if x%i == 0 {
                return false
            }
        }
        return true
    }

    s := strconv.Itoa(num)

    x := 0
    for i := 0; i < len(s); i++ {
        x = x*10 + int(s[i]-'0')
        if !isPrime(x) {
            return false
        }
    }

    x = 0
    p := 1
    for i := len(s) - 1; i >= 0; i-- {
        x = p*int(s[i]-'0') + x
        p *= 10
        if !isPrime(x) {
            return false
        }
    }

    return true
}

# Accepted solution for LeetCode #3765: Complete Prime Number
class Solution:
    def completePrime(self, num: int) -> bool:
        def is_prime(x: int) -> bool:
            if x < 2:
                return False
            return all(x % i for i in range(2, int(sqrt(x)) + 1))

        s = str(num)
        x = 0
        for c in s:
            x = x * 10 + int(c)
            if not is_prime(x):
                return False
        x, p = 0, 1
        for c in s[::-1]:
            x = p * int(c) + x
            p *= 10
            if not is_prime(x):
                return False
        return True

// Accepted solution for LeetCode #3765: Complete Prime Number
// 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 #3765: Complete Prime Number
// class Solution {
//     public boolean completePrime(int num) {
//         char[] s = String.valueOf(num).toCharArray();
//         int x = 0;
//         for (int i = 0; i < s.length; i++) {
//             x = x * 10 + (s[i] - '0');
//             if (!isPrime(x)) {
//                 return false;
//             }
//         }
//         x = 0;
//         int p = 1;
//         for (int i = s.length - 1; i >= 0; i--) {
//             x = p * (s[i] - '0') + x;
//             p *= 10;
//             if (!isPrime(x)) {
//                 return false;
//             }
//         }
//         return true;
//     }
// 
//     private boolean isPrime(int x) {
//         if (x < 2) {
//             return false;
//         }
//         for (int i = 2; i * i <= x; i++) {
//             if (x % i == 0) {
//                 return false;
//             }
//         }
//         return true;
//     }
// }

// Accepted solution for LeetCode #3765: Complete Prime Number
function completePrime(num: number): boolean {
    const isPrime = (x: number): boolean => {
        if (x < 2) return false;
        for (let i = 2; i * i <= x; i++) {
            if (x % i === 0) {
                return false;
            }
        }
        return true;
    };

    const s = String(num);

    let x = 0;
    for (let i = 0; i < s.length; i++) {
        x = x * 10 + (s.charCodeAt(i) - 48);
        if (!isPrime(x)) {
            return false;
        }
    }

    x = 0;
    let p = 1;
    for (let i = s.length - 1; i >= 0; i--) {
        x = p * (s.charCodeAt(i) - 48) + x;
        p *= 10;
        if (!isPrime(x)) {
            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(sqrtn × log n)

Space

O(log n)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.