2523. Closest Prime Numbers in Range LeetCode Solution

Last updated: March 5, 2025

In this guide, you will get 2523. Closest Prime Numbers in Range LeetCode Solution with the best time and space complexity. The solution to Closest Prime Numbers in Range problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Closest Prime Numbers in Range solution in C++
Closest Prime Numbers in Range solution in Java
Closest Prime Numbers in Range solution in Python
Additional Resources

2523. Closest Prime Numbers in Range LeetCode Solution image

Problem Statement of Closest Prime Numbers in Range

Given two positive integers left and right, find the two integers num1 and num2 such that:

left <= num1 < num2 <= right .
Both num1 and num2 are prime numbers.
num2 – num1 is the minimum amongst all other pairs satisfying the above conditions.

Return the positive integer array ans = [num1, num2]. If there are multiple pairs satisfying these conditions, return the one with the smallest num1 value. If no such numbers exist, return [-1, -1].

Example 1:

Input: left = 10, right = 19
Output: [11,13]
Explanation: The prime numbers between 10 and 19 are 11, 13, 17, and 19.
The closest gap between any pair is 2, which can be achieved by [11,13] or [17,19].
Since 11 is smaller than 17, we return the first pair.

Example 2:

Input: left = 4, right = 6
Output: [-1,-1]
Explanation: There exists only one prime number in the given range, so the conditions cannot be satisfied.

Constraints:

1 <= left <= right <= 106

Complexity Analysis

Time Complexity: O(n\log(\log n))
Space Complexity: O(n)

2523. Closest Prime Numbers in Range LeetCode Solution in C++

class Solution {
 public:
  vector<int> closestPrimes(int left, int right) {
    const vector<bool> isPrime = sieveEratosthenes(right + 1);
    vector<int> primes;

    for (int i = left; i <= right; ++i)
      if (isPrime[i])
        primes.push_back(i);

    if (primes.size() < 2)
      return {-1, -1};

    int minDiff = INT_MAX;
    int num1 = -1;
    int num2 = -1;

    for (int i = 1; i < primes.size(); ++i) {
      const int diff = primes[i] - primes[i - 1];
      if (diff < minDiff) {
        minDiff = diff;
        num1 = primes[i - 1];
        num2 = primes[i];
      }
    }

    return {num1, num2};
  }

 private:
  vector<bool> sieveEratosthenes(int n) {
    vector<bool> isPrime(n, true);
    isPrime[0] = false;
    isPrime[1] = false;
    for (int i = 2; i * i < n; ++i)
      if (isPrime[i])
        for (int j = i * i; j < n; j += i)
          isPrime[j] = false;
    return isPrime;
  }
};
/* code provided by PROGIEZ */

2523. Closest Prime Numbers in Range LeetCode Solution in Java

class Solution {
  public int[] closestPrimes(int left, int right) {
    final boolean[] isPrime = sieveEratosthenes(right + 1);
    List<Integer> primes = new ArrayList<>();

    for (int i = left; i <= right; ++i)
      if (isPrime[i])
        primes.add(i);

    if (primes.size() < 2)
      return new int[] {-1, -1};

    int minDiff = Integer.MAX_VALUE;
    int num1 = -1;
    int num2 = -1;

    for (int i = 1; i < primes.size(); ++i) {
      final int diff = prime.get(i) - prime.get(i - 1);
      if (diff < minDiff) {
        minDiff = diff;
        num1 = prime.get(i - 1);
        num2 = prime.get(i);
      }
    }

    return new int[] {num1, num2};
  }

  private boolean[] sieveEratosthenes(int n) {
    boolean[] isPrime = new boolean[n];
    Arrays.fill(isPrime, true);
    isPrime[0] = false;
    isPrime[1] = false;
    for (int i = 2; i * i < n; ++i)
      if (isPrime[i])
        for (int j = i * i; j < n; j += i)
          isPrime[j] = false;
    return isPrime;
  }
}
// code provided by PROGIEZ

2523. Closest Prime Numbers in Range LeetCode Solution in Python

class Solution:
  def closestPrimes(self, left: int, right: int) -> list[int]:
    isPrime = self._sieveEratosthenes(right + 1)
    primes = [i for i in range(left, right + 1) if isPrime[i]]

    if len(primes) < 2:
      return [-1, -1]

    minDiff = math.inf
    num1 = -1
    num2 = -1

    for a, b in zip(primes, primes[1:]):
      diff = b - a
      if diff < minDiff:
        minDiff = diff
        num1 = a
        num2 = b

    return [num1, num2]

  def _sieveEratosthenes(self, n: int) -> list[bool]:
    isPrime = [True] * n
    isPrime[0] = False
    isPrime[1] = False
    for i in range(2, int(n**0.5) + 1):
      if isPrime[i]:
        for j in range(i * i, n, i):
          isPrime[j] = False
    return isPrime
 # code by PROGIEZ

Additional Resources

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