<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>

class Solution:
    def countPairs(self, nums: List[int]) -> int:
        dicti, n, ans = {}, len(nums), 0
        visited = [False] * n
        for index, num in enumerate(nums):
            dicti.setdefault(num, []).append(index)
            string, checked = str(num), {num}
            string = '0' * (7 - len(string)) + string
            for i in range(6):
                for j in range(i + 1, 7):
                    new = string[:i] + string[j] + string[i + 1 : j] + string[i] + string[j + 1:]
                    cur = int(new)
                    if cur not in checked:
                        dicti.setdefault(cur, []).append(index)
                        checked.add(cur)
                    for i2 in range(6):
                        for j2 in range(i2 + 1, 7):
                            cur = int(new[:i2] + new[j2] + new[i2 + 1 : j2] + new[i2] + new[j2 + 1:])
                            if cur not in checked:
                                dicti.setdefault(cur, []).append(index)
                                checked.add(cur)
        return sum(len(dicti[nums[index]]) - 1 for index in range(n)) >> 1
31291