LeetCode Java - Frequency of the most frequent element

Question

So I was trying to solve the LeetCode question n.1838 (Frequency of the most frequent element) and it works except for a test case with an input array of 500000 elements (1 <= nums.length <= 10^5): it gives me TimeLimitExceeded and I don't understand why. I tried to explain my approach with comments in the code, but it's very easy: I sort the array first so I can start from the last element (the bigger one), I loop the array and for each element in the array i calculate the frequency for different cases:

• if nums[i] == nums[i - 1] increment the frequency
• if k < nums[i] - nums[i - 1] there's no need to do anything, so i just update j to get outside of the loop
• else I add to nums[i - 1] the difference so it can be nums[i], I update q and num.

I then update the max frequency and once I'm outside the while loop I do the same for each element of the array.

Here's my method:

public int maxFrequency(int[] nums, int k) {
    //corner case
    if (nums.length == 1)
        return 1;
    
    //order array
    Arrays.sort(nums);
    
    int max = 0; 
    int freq = 0;
    int num = 0; //will always be the int at the index - j (j will be incremented each time)
    int j = 0;
    
    //start from the last element of the array
    for (int i = nums.length - 1; i >= 0; i--) {
        num = nums[i - j];
        int q = k;
        while (q >= 0 && i - j >= 0) {
            //if element equals itself, increment frequency
            if (nums[i] - num == 0) {
                freq++;
                j++;
            } // if k < nums[i] - num (aka even if we add k to num we won't have a new nums[i])
            else if (nums[i] - num > q) {
                j = i + 1; //to get out of the while loop
            } else { //case where i can add k because k >= nums[i] - num
                q -= nums[i] - num;
                freq++;
                j++;
            } // update num only if i - j > 0 so we dont have ArrayOutOfBoundException
            if (i - j >= 0) {
                num = nums[i - j];
            } //update max frequency
            max = Math.max(freq, max);
        }
        freq = 0;
        j = 0;
    }
    
    return max;
}

Any advice or hint would be helpful, thanks in advance.