<p>Given an array of integers <code>nums</code>, sort the array in <strong>increasing</strong> order based on the frequency of the values. If multiple values have the same frequency, sort them in <strong>decreasing</strong> order.</p>

<p>Return the <em>sorted array</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,2,2,2,3]
<strong>Output:</strong> [3,1,1,2,2,2]
<strong>Explanation:</strong> &#39;3&#39; has a frequency of 1, &#39;1&#39; has a frequency of 2, and &#39;2&#39; has a frequency of 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,1,3,2]
<strong>Output:</strong> [1,3,3,2,2]
<strong>Explanation:</strong> &#39;2&#39; and &#39;3&#39; both have a frequency of 2, so they are sorted in decreasing order.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,1,-6,4,5,-6,1,4,1]
<strong>Output:</strong> [5,-1,4,4,-6,-6,1,1,1]</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution:
    def frequencySort(self, nums: List[int]) -> List[int]:
        for i in range(0,len(nums)):
            for j in range(0,len(nums)-1) :
                if nums.count(nums[j+1])<nums.count(nums[j]) :
                    a=nums[j]
                    nums[j]=nums[j+1]
                    nums[j+1]=a
        for i in range(0,len(nums)):
            for j in range(0,len(nums)-1) :
                if nums.count(nums[j])==nums.count(nums[j+1]) :
                    if nums[j]<nums[j+1]:
                        a=nums[j+1]
                        nums[j+1]=nums[j]
                        nums[j]=a
        return nums 
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