<p><strong>Attention</strong>: In this version, the number of operations that can be performed, has been increased to <strong>twice</strong>.<!-- notionvc: 278e7cb2-3b05-42fa-8ae9-65f5fd6f7585 --></p>

<p>You are given an array <code>nums</code> consisting of positive integers.</p>

<p>We call two integers <code>x</code> and <code>y</code> <strong>almost equal</strong> if both integers can become equal after performing the following operation <strong>at most <u>twice</u></strong>:</p>

<ul>
	<li>Choose <strong>either</strong> <code>x</code> or <code>y</code> and swap any two digits within the chosen number.</li>
</ul>

<p>Return the number of indices <code>i</code> and <code>j</code> in <code>nums</code> where <code>i &lt; j</code> such that <code>nums[i]</code> and <code>nums[j]</code> are <strong>almost equal</strong>.</p>

<p><strong>Note</strong> that it is allowed for an integer to have leading zeros after performing an operation.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1023,2310,2130,213]</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p>The almost equal pairs of elements are:</p>

<ul>
	<li>1023 and 2310. By swapping the digits 1 and 2, and then the digits 0 and 3 in 1023, you get 2310.</li>
	<li>1023 and 213. By swapping the digits 1 and 0, and then the digits 1 and 2 in 1023, you get 0213, which is 213.</li>
	<li>2310 and 213. By swapping the digits 2 and 0, and then the digits 3 and 2 in 2310, you get 0213, which is 213.</li>
	<li>2310 and 2130. By swapping the digits 3 and 1 in 2310, you get 2130.</li>
</ul>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,10,100]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>The almost equal pairs of elements are:</p>

<ul>
	<li>1 and 10. By swapping the digits 1 and 0 in 10, you get 01 which is 1.</li>
	<li>1 and 100. By swapping the second 0 with the digit 1 in 100, you get 001, which is 1.</li>
	<li>10 and 100. By swapping the first 0 with the digit 1 in 100, you get 010, which is 10.</li>
</ul>
</div>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 5000</code></li>
	<li><code>1 &lt;= nums[i] &lt; 10<sup>7</sup></code></li>
</ul>

from collections import defaultdict

class Solution(object):
    def countPairs(self, nums):
        freq = defaultdict(int)
        comb = defaultdict(set)
        max_length = len(str(max(nums)))
        for i in range(len(nums)):
            freq[nums[i]] += 1
        
        count = 0
        freq_key = list(freq.keys())
        for i in range(len(freq_key)):
            s = str(freq_key[i])
            s = '0' * (max_length - len(s)) + s
            for j in range(i):
                if s in comb[freq_key[j]]:
                    count += freq[freq_key[i]] * freq[freq_key[j]] 
            if i != len(freq_key) - 1:
                comb[freq_key[i]] = self.get_comb(s, max_length)
            
            count += freq[freq_key[i]] * (freq[freq_key[i]]-1) // 2

        return count

    def get_comb(self, s, length):
        if length == 1: return set([s]) 
        visited = set()
        visited.add(s)

        for i in range(length):
            for j in range(i, length):
                if i == j: continue
                ss = s[:i] + s[j] + s[i+1:j] + s[i] + s[j+1:]
                visited.add(ss)

        temp_visited = visited.copy()
        for x in temp_visited:
            for i in range(length):
                for j in range(i, length):
                    if i == j: continue
                    xx = x[:i] + x[j] + x[i+1:j] + x[i] + x[j+1:]
                    visited.add(xx)
        return visited
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