<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>

import numpy as np
class Solution:
    def countPairs(self, nums: List[int]) -> int:
        n = len(nums)
        result = 0
        mp = defaultdict(lambda:0)
        max_digit = int(math.log10(max(nums)) + 1)
        a = []
        for number in nums:
            a.append(str(number).zfill(max_digit))
        a = np.array(a)
        for i in range(n):
            curr_st = set()
            curr_l = list(a[i])
            for l1 in range(0 , max_digit):
                for r1 in range(l1 , max_digit):
                    curr_l[l1] , curr_l[r1] = curr_l[r1] , curr_l[l1]
                    for l2 in range(0 , max_digit):
                        for r2 in range(l2 , max_digit):
                            curr_l[l2] , curr_l[r2] = curr_l[r2] , curr_l[l2]
                            number = "".join(curr_l)
                            if number not in curr_st:
                                curr_st.add(number)
                                result += mp[number]
                            curr_l[l2] , curr_l[r2] = curr_l[r2] , curr_l[l2]
                    curr_l[l1] , curr_l[r1] = curr_l[r1] , curr_l[l1]
            mp[a[i]]+=1
        return result

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