Solution #2523 - Mridul/Edited - 07/06/2025 #22
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JRS296
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Mar 21, 2025
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| private static boolean[] isPrime = new boolean[1000001]; This boolean array is used to mark prime numbers up to 1000000. The size of the array is 1000001 to include the number 1000000. | ||
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| Static Block: | ||
| The static block is used to initialize the isPrime array and precompute the prime numbers. | ||
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| Arrays.fill(isPrime, true); This method sets all elements of the isPrime array to true, assuming all numbers are prime initially. | ||
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| precompute(); This method is called to mark non-prime numbers in the isPrime array. | ||
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| Method: precompute() | ||
| This method uses the Sieve of Eratosthenes algorithm to mark non-prime numbers in the isPrime array. | ||
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| isPrime[0] = isPrime[1] = false; Numbers 0 and 1 are not prime. | ||
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| A loop is used to iterate from 2 to sqrt(1000000). | ||
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| If isPrime[i] is true, all multiples of i are marked as false (not prime) starting from i*i. | ||
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| Method: closestPrimes(int left, int right) | ||
| This method finds the closest prime numbers within the range [left, right]. | ||
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| int prev = -1, minDiff = Integer.MAX_VALUE; Variables to track the previous prime number and the minimum difference between consecutive primes. | ||
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| int num1 = -1, num2 = -1; Variables to store the closest prime numbers. | ||
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| A loop iterates through the range [left, right]. | ||
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| If isPrime[i] is true, the number i is a prime. | ||
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| If prev is not -1 and the difference (i - prev) is less than minDiff, update minDiff and set num1 to prev and num2 to i. | ||
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| Update prev to i. | ||
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| If no prime numbers are found, return new int[]{-1, -1}. Otherwise, return the closest prime numbers num1 and num2. |
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This can be improved quite a bit! Use Markdown, and explain the time and space complexity as well!
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added solution.java and explanation.md file