<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:
        count = 0
        nums.sort()
        nums = [str(num) for num in nums]
        dp = defaultdict(dict)
        def helper(num, length):
            s1 = '0' * (length - len(num)) + num
            queue = deque([s1])
            visited = set([s1])
            step = 0
            while queue:
                step += 1
                for _ in range(len(queue)):
                    curr = queue.popleft()
                    for i in range(length):
                        for j in range(i + 1, length):
                            new_s = curr[:i] + curr[j] + curr[i+1:j] + curr[i] + curr[j+1:]
                            if new_s in visited:
                                continue
                            queue.append(new_s)
                            visited.add(new_s)
                            if int(new_s) < int(s1):
                                dp[int(new_s)][int(s1)] = step
                            else:
                                dp[int(s1)][int(new_s)] = step
                if step == 2:
                    break
        maximum = 0
        for num in nums:
            maximum = max(maximum, len(num))
        for i in range(len(nums)):
            if int(nums[i]) in dp:
                continue
            num = nums[i]
            dp[int(num)][int(num)] = 0
            for length in range(len(num), maximum + 1):
                helper(num, length)

        nums = [int(num) for num in nums]
        for i in range(len(nums)):
            for j in range(i + 1, len(nums)):
                if nums[j] in dp[nums[i]] and dp[nums[i]][nums[j]] <= 2:
                    count += 1
        return count        
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