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[LeetCode Sync] Runtime - 11 ms (8.12%), Memory - 17.6 MB (89.32%)
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<p>You are given an integer <code>num</code>. You will apply the following steps to <code>num</code> <strong>two</strong> separate times:</p>
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<ul>
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<li>Pick a digit <code>x (0 &lt;= x &lt;= 9)</code>.</li>
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<li>Pick another digit <code>y (0 &lt;= y &lt;= 9)</code>. Note <code>y</code> can be equal to <code>x</code>.</li>
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<li>Replace all the occurrences of <code>x</code> in the decimal representation of <code>num</code> by <code>y</code>.</li>
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</ul>
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<p>Let <code>a</code> and <code>b</code> be the two results from applying the operation to <code>num</code> <em>independently</em>.</p>
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<p>Return <em>the max difference</em> between <code>a</code> and <code>b</code>.</p>
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<p>Note that neither <code>a</code> nor <code>b</code> may have any leading zeros, and <strong>must not</strong> be 0.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> num = 555
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<strong>Output:</strong> 888
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<strong>Explanation:</strong> The first time pick x = 5 and y = 9 and store the new integer in a.
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The second time pick x = 5 and y = 1 and store the new integer in b.
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We have now a = 999 and b = 111 and max difference = 888
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> num = 9
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<strong>Output:</strong> 8
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<strong>Explanation:</strong> The first time pick x = 9 and y = 9 and store the new integer in a.
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The second time pick x = 9 and y = 1 and store the new integer in b.
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We have now a = 9 and b = 1 and max difference = 8
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= num &lt;= 10<sup>8</sup></code></li>
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</ul>
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class Solution:
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def maxDiff(self, num: int) -> int:
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def swap(x,y):
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return str(num).replace(str(x), str(y)) # change all x to y in num.
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mi = ma = num
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for x in range(10):
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for y in range(10):
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res = swap(x,y) # this is a string!
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# leading zeros check
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if res[0] != "0":
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mi = min(mi, int(res))
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ma = max(ma, int(res))
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return ma - mi

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