LeetCode #2523 — MEDIUM

Closest Prime Numbers in Range

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

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The Problem

Problem Statement

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

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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].

Baseline thinking

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

Pattern signal: Math

Example 1

10
19

Example 2

4
6

Related Problems

  • Count Ways to Make Array With Product (count-ways-to-make-array-with-product)
Step 02

Core Insight

What unlocks the optimal approach

  • Use Sieve of Eratosthenes to mark numbers that are primes.
  • Iterate from right to left and find pair with the minimum distance between marked numbers.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #2523: Closest Prime Numbers in Range
class Solution {
    public int[] closestPrimes(int left, int right) {
        int cnt = 0;
        boolean[] st = new boolean[right + 1];
        int[] prime = new int[right + 1];
        for (int i = 2; i <= right; ++i) {
            if (!st[i]) {
                prime[cnt++] = i;
            }
            for (int j = 0; prime[j] <= right / i; ++j) {
                st[prime[j] * i] = true;
                if (i % prime[j] == 0) {
                    break;
                }
            }
        }
        int i = -1, j = -1;
        for (int k = 0; k < cnt; ++k) {
            if (prime[k] >= left && prime[k] <= right) {
                if (i == -1) {
                    i = k;
                }
                j = k;
            }
        }
        int[] ans = new int[] {-1, -1};
        if (i == j || i == -1) {
            return ans;
        }
        int mi = 1 << 30;
        for (int k = i; k < j; ++k) {
            int d = prime[k + 1] - prime[k];
            if (d < mi) {
                mi = d;
                ans[0] = prime[k];
                ans[1] = prime[k + 1];
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(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.

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