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Divisors of a Given Number

Certification: Basic Level Accuracy: 66.67% Submissions: 3 Points: 15

Write a C++ program that prints all the divisors of a given positive integer in ascending order.

Example 1
    
  
  • Input: n = 12
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  • Output: 1, 2, 3, 4, 6, 12
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  • Explanation: 
      
    
    • Step 1: Find all integers from 1 to sqrt(n) that divide n without a remainder.
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    • Step 2: For each divisor i found in Step 1, n/i is also a divisor.
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    • Step 3: The divisors of 12 are 1, 2, 3, 4, 6, and 12 since each divides 12 without a remainder.
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    • Step 4: Return the list of divisors in ascending order.
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    •
Example 2
    
  
  • Input: n = 17
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  • Output: 1, 17
    • 
  
  • Explanation: 
      
    
    • Step 1: Find all integers from 1 to sqrt(n) that divide n without a remainder.
      • 
    
    • Step 2: For each divisor i found in Step 1, n/i is also a divisor.
      • 
    
    • Step 3: Since 17 is a prime number, its only divisors are 1 and itself (17).
      • 
    
    • Step 4: Return the list of divisors in ascending order.
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    •
Constraints
    
  
  • 1 ≤ n ≤ 10^9
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  • Input is a positive integer
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  • Time Complexity: O(sqrt(n))
    • 
  
  • Space Complexity: O(sqrt(n)) for storing the divisors
    •
Control StructuresFunctions / MethodsIBMDropbox
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Hint
Solution

Solution Hints

    
  
  • A naive approach would be to check all numbers from 1 to n, but that's inefficient for large n
    • 
  
  • For every divisor d below or equal to sqrt(n), n/d is also a divisor
    • 
  
  • Collect divisors from both sides and sort them before printing
    • 
  
  • Handle perfect squares carefully to avoid duplicate divisors
    •

Steps to solve by this approach:

 Step 1: Create a vector to store all divisors of the number.

 Step 2: Iterate from 1 to the square root of the number.

 Step 3: Check if the current number is a divisor (n % i == 0).

 Step 4: If it is a divisor, add both i and n/i to the vector (unless they're equal).

 Step 5: Sort the vector of divisors to get them in ascending order.

 Step 6: Return the sorted vector containing all divisors.

Submitted Code :