Find number of times X appears in N by N multiplication Table

Suppose you have a multiplication table that is N by N. That is, a 2D array where the value at the i-th row and j-th column is (i + 1) * (j + 1) (if 0-indexed) or i * j (if 1-indexed).

Given integers N and X, write a function that returns the number of times X appears as a value in an N by N multiplication table.

For example, given N = 6 and X = 12, you should return 4, since the multiplication table looks like this:

| 1 | 2 | 3 | 4 | 5 | 6 |

| 2 | 4 | 6 | 8 | 10 | 12 |

| 3 | 6 | 9 | 12 | 15 | 18 |

| 4 | 8 | 12 | 16 | 20 | 24 |

| 5 | 10 | 15 | 20 | 25 | 30 |

| 6 | 12 | 18 | 24 | 30 | 36 |

And there are 4 12's in the table.

Example:

Input:  n = 6, x= 12
Output: 4

Approach 1: 

C++

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

int findXAppearsinTable(int n, int x)
{
    if (x <= 0 || x > n * n)
    {
        return 0;
    }

    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        if (x % i == 0 && x / i <= n)
        {
            ans++;
        }
    }
    return ans;
}
int main()
{

    int n = 6;
    int x = 12;

    cout << findXAppearsinTable(n, x);

    return 0;
}

Approach 2:

C++

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

int findXAppearsinTable(int n, int x)
{
    vector<int> factors;
    if (x <= 0 || x > n * n)
    {
        return 0;
    }

    //find all the factors of x
    for (int i = 1; i * i <= x; i++)
    {
        if (x % i == 0)
        {
            if (i == x / i)
            {
                factors.push_back(i);
            }
            else
            {
                factors.push_back(i);
                factors.push_back(x / i);
            }
        }
    }

    //sort the factors
    sort(factors.begin(), factors.end());
    int ans = 0;

    //if x is perfect square
    if (sqrt(x) == int(sqrt(x)))
    {
        ans++;
    }
    for (int i = 0; i < factors.size() / 2; i++)
    {
        int a = factors[i];
        int b = factors[factors.size() - 1 - i];

        //if in the range then update answer
        //anwer will add 2 because the value occurs in
        //the pair
        if (a >= 1 && b <= n)
        {
            ans += 2;
        }
    }
    return ans;
}
int main()
{

    int n = 6;
    int x = 12;

    cout << findXAppearsinTable(n, x);
    return 0;
}