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Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.7 MB (58.67%)
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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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# Approach 2: Greedy
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# Time: O(log(num))
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# Space: O(log(n))
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class Solution:
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def maxDiff(self, num: int) -> int:
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min_num, max_num = str(num), str(num)
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for digit in max_num:
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if digit != '9':
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max_num = max_num.replace(digit, '9')
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break
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for i, digit in enumerate(min_num):
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if i == 0:
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if digit != '1':
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min_num = min_num.replace(digit, '1')
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break
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else:
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if digit != '0' and digit != min_num[0]:
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min_num = min_num.replace(digit, '0')
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break
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return int(max_num) - int(min_num)

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