Algorithm

Instead of checking all numbers from 1 to N, observe that divisors come in pairs (i, N / i). This means we only need to check up to √N because for every divisor less than √N, there is a corresponding divisor greater than √N. For each number i from 1 to √N, if N % i == 0, then i is a divisor and N / i is also a divisor (except when both are the same, i.e., perfect squares). This reduces the number of checks significantly.

Code

#include <bits/stdc++.h>
using namespace std;

// Class containing the method to find divisors
class Solution {
public:
    // Function to find and print all divisors
    void findDivisors(int n) {
        // Create a vector to store divisors
        vector<int> divisors;

        // Loop from 1 to sqrt(n)
        for (int i = 1; i * i <= n; i++) {
            // If i divides n completely
            if (n % i == 0) {
                // Add i as a divisor
                divisors.push_back(i);
                // Add the paired divisor if different
                if (i != n / i) {
                    divisors.push_back(n / i);
                }
     