All divisors of a Number

Problem Statement:

Given an integer n, print all the divisors of N in the ascending order.

• Example:

  Examples:

  Input :n =20
  Output:1 2 4 5 10 20
  Explanation:20 is completely divisible by 1, 2, 4, 5, 10 and 20.
  

  Input:n =21191
  Output:1 21191
  Explanation: As 21191 is a prime number, it has only 2 factors(1 and the number itself).

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✅ Solution: Divisor Enumeration up to √N

class Solution {
  public:
    void print_divisors(int n) {
        vector<int> res;

        // Iterate from 1 to √n
        for(int i = 1; i <= sqrt(n); i++){
            if(n % i == 0){
                if(n / i == i){
                    // i is a perfect square divisor (e.g., √n)
                    res.push_back(i);
                }
                else{
                    // Add both divisors: i and n/i
                    res.push_back(i);
                    res.push_back(n / i);
                }
            }
        }

        // Sort all divisors before printing
        sort(res.begin(), res.end());

        // Print each divisor
        for(auto it : res) cout << it << " ";
    }
};

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📝 How It Works

• For each number i from 1 to √n, check if i divides n.
• If yes, both i and n/i are divisors.
• Take care not to insert duplicates (when i == n/i, like for perfect squares).
• Finally, sort and print all collected divisors.

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⏱️ Time & Space Complexity

Metric	Value
⏱️ Time	O(√N + K log K) where K is number of divisors
💾 Space	O(K) to store divisors

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⚠️ Edge Cases

• n = 1 → output is just 1
• n = prime → output is 1 and n
• Handles perfect squares correctly (e.g., 36 includes 6 only once)

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🔁 Related Problems

• Count number of divisors
• Sum of all divisors
• GCD / LCM based problems
• Efficient Sieve + divisor functions