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Check if Number is Prime

Certification: Basic Level Accuracy: 0% Submissions: 0 Points: 5

Write a JavaScript program to determine if a given positive integer is a prime number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The first few prime numbers are 2, 3, 5, 7, 11, 13, etc.

Example 1
    
      
  • Input: num = 17
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  • Output: true
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  • Explanation: 
    • Check if 17 is greater than 1 (it is). 
    • Check if 17 is divisible by any number from 2 to √17 (approximately 4.12). 
    • Test divisibility: 17 % 2 = 1, 17 % 3 = 2, 17 % 4 = 1. 
    • None of these divisions result in 0, so 17 has no divisors other than 1 and itself.
    • Therefore, 17 is a prime number.
    •
Example 2
    
      
  • Input: num = 15
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  • Output: false
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  • Explanation: 
    • Check if 15 is greater than 1 (it is). 
    • Check if 15 is divisible by any number from 2 to √15 (approximately 3.87). 
    • Test divisibility: 15 % 2 = 1, 15 % 3 = 0. 
    • Since 15 % 3 = 0, we found that 15 is divisible by 3. 
    • Therefore, 15 is not a prime number.
    •
Constraints
    
      
  • 1 ≤ num ≤ 10^6
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  • The input will always be a positive integer
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  • Numbers less than or equal to 1 should return false
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  • Time Complexity: O(√n) where n is the input number
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  • Space Complexity: O(1)
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Hint
Solution

Solution Hints

    
      
  • Handle edge cases first: numbers less than or equal to 1 are not prime
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  • Check if the number is 2, which is the only even prime number
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  • If the number is even and greater than 2, it's not prime
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  • For odd numbers, check divisibility from 3 up to the square root of the number
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  • Only check odd divisors to optimize the algorithm
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  • If any divisor is found, the number is not prime
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  • If no divisors are found, the number is prime
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Steps to solve by this approach:

 Step 1: Handle edge cases by checking if the number is less than or equal to 1, return false
 Step 2: Check if the number is exactly 2, which is the only even prime number, return true
 Step 3: Check if the number is even and greater than 2, return false since even numbers > 2 are not prime
 Step 4: Calculate the square root of the number to determine the upper limit for checking divisors
 Step 5: Use a loop starting from 3 and increment by 2 to check only odd potential divisors
 Step 6: For each odd number in the range, check if it divides the input number evenly
 Step 7: If any divisor is found, return false; if no divisors are found after the loop, return true

Submitted Code :