<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 cnt(self,a,b):
        A,B = len(a),len(b)
        if A < B:
            a = '0'*(B-A) + a
        elif B < A:
            b = '0'*(A-B) + b
        diff = 0
        h = [0]*10
        op = set()
        for i in range(len(a)):
            if a[i] != b[i]:
                if (a[i],b[i]) not in op and \
                    (b[i],a[i]) not in op:
                    op.add((a[i],b[i]))
                diff += 1
                h[ord(a[i])-48] += 1
                h[ord(b[i])-48] -= 1

        v = 0
        for i in range(10):
            v += abs(h[i])

        return 1 if v == 0 and (diff < 4 or (diff == 4 and len(op) <= 2)) else 0

    def gen(self,a):
        a = list(a)
        for i in range(len(a)-1):
            for j in range(i+1,len(a)):
                a[i],a[j] = a[j],a[i]
                yield ''.join(a)
                a[i],a[j] = a[j],a[i]

    def cnt2(self,a,h):
        out = h[a.lstrip('0')]
        done = set([a.lstrip('0')])
        a = '00' + a
        for i in self.gen(a):
            i = i.lstrip('0')
            if i not in done:
                out += h[i]
                done.add(i)
            for j in self.gen(i):
                j = j.lstrip('0')
                if j not in done:
                    out += h[j] 
                    done.add(j)
        return out
    def countPairs(self, nums: List[int]) -> int:
        out = 0
        h = defaultdict(int)
        for v in nums:
            v = str(v)
            
            out += self.cnt2(v,h)
            h[v] += 1

        return out

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