<p>Given a positive integer <code>n</code>,&nbsp;you can apply one of the following&nbsp;operations:</p>

<ol>
	<li>If <code>n</code> is even, replace <code>n</code> with <code>n / 2</code>.</li>
	<li>If <code>n</code> is odd, replace <code>n</code> with either <code>n + 1</code> or <code>n - 1</code>.</li>
</ol>

<p>Return <em>the minimum number of operations needed for</em> <code>n</code> <em>to become</em> <code>1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 8
<strong>Output:</strong> 3
<strong>Explanation:</strong> 8 -&gt; 4 -&gt; 2 -&gt; 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 7
<strong>Output:</strong> 4
<strong>Explanation: </strong>7 -&gt; 8 -&gt; 4 -&gt; 2 -&gt; 1
or 7 -&gt; 6 -&gt; 3 -&gt; 2 -&gt; 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 4
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    unordered_map<long long, long long> memo;

long long solve(long long n) {
    if (n == 1) return 0; // Base case: If n is 1, no replacements are needed.

    if (memo.find(n) != memo.end()) return memo[n]; // Check if result is already computed.

    long long cnt = 0;
    while (n % 2 == 0) {
        cnt++;
        n /= 2;
    }

    if (n == 1) {
        memo[n] = cnt;
        return cnt;
    } else {
        long long result = cnt + 1 + min(solve(n - 1), solve(n + 1)); // Recursive calls with memoization.
        memo[n] = result;
        return result;
    }
}

int integerReplacement(int n) {
    return solve(n);
}
};
9000