187 lines
7.3 KiB
HTML
187 lines
7.3 KiB
HTML
<!doctype html>
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<html lang="en">
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defer
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src="/public/katex/katex-render.js"
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<title>Barrett Ruth</title>
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</head>
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<body class="graph-background">
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<header>
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<a
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href="/"
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style="text-decoration: none; color: inherit"
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onclick="goHome(event)"
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>
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<div class="terminal-container">
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<span class="terminal-prompt">barrett@ruth:~$ /algorithms</span>
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<span class="terminal-cursor"></span>
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</div>
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</a>
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</header>
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<main class="main">
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<div class="post-container">
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<header class="post-header">
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<h1 class="post-title">Leetcode Daily</h1>
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</header>
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<article class="post-article">
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<div class="fold">
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<h2>
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<a
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target="blank"
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href="https://leetcode.com/problems/minimum-array-end/"
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>minimum array end</a
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>
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— 9/11/24
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</h2>
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</div>
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<div class="problem-content">
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<h3>problem statement</h3>
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<p>
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Given some \(x\) and \(n\), construct a strictly increasing array
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(say
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<code>nums</code>
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) of length \(n\) such that
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<code>nums[0] & nums[1] ... & nums[n - 1] == x</code>
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, where
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<code>&</code>
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denotes the bitwise AND operator.
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</p>
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<p>
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Finally, return the minimum possible value of
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<code>nums[n - 1]</code>.
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</p>
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</div>
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<h3>understanding the problem</h3>
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<p>
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The main difficulty in this problem lies in understanding what is
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being asked (intentionally or not, the phrasing is terrible). Some
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initial notes:
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</p>
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<ul>
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<li>The final array need not be constructed</li>
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<li>
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If the element-wise bitwise AND of an array equals
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<code>x</code> if and only if each element has
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<code>x</code>'s bits set—and no other bit it set by
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all elements
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</li>
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<li>
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It makes sense to set <code>nums[0] == x</code> to ensure
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<code>nums[n - 1]</code> is minimal
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</li>
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</ul>
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<h3>developing an approach</h3>
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<p>
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An inductive approach is helpful. Consider the natural question:
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“If I had correctly generated <code>nums[:i]</code>”,
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how could I find <code>nums[i]</code>? In other words,
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<i
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>how can I find the next smallest number such that
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<code>nums</code>
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's element-wise bitwise AND is still \(x\)?</i
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>
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</p>
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<p>
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Hmm... this is tricky. Let's think of a similar problem to
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glean some insight: “Given some \(x\), how can I find the next
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smallest number?”. The answer is, of course, add one (bear
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with me here).
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</p>
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<p>
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We also know that all of <code>nums[i]</code> must have at least
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\(x\)'s bits set. Therefore, we need to alter the unset bits of
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<code>nums[i]</code>.
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</p>
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<p>
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The key insight of this problem is combining these two ideas to
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answer our question:
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<i
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>Just “add one” to <code>nums[i - 1]</code>'s
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unset bits</i
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>. Repeat this to find <code>nums[n - 1]</code>.
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</p>
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<p>
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One last piece is missing—how do we know the element-wise
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bitwise AND is <i>exactly</i> \(x\)? Because
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<code>nums[i > 0]</code> only sets \(x\)'s unset bits, every
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number in <code>nums</code> will have at least \(x\)'s bits
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set. Further, no other bits will be set because \(x\) has them
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unset.
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</p>
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<h3>carrying out the plan</h3>
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<p>Let's flesh out the remaining parts of the algorithm:</p>
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<ul>
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<li>
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<code>len(nums) == n</code> and we initialize
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<code>nums[0] == x</code>. So, we need to “add one”
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<code>n - 1</code> times
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</li>
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<li>
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How do we carry out the additions? We could iterate \(n - 1\)
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times and simulate them. However, we already know how we want to
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alter the unset bits of <code>nums[0]</code> inductively—
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(add one) <i>and</i> how many times we want to do this (\(n -
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1\)). Because we're adding one \(n-1\) times to \(x\)'s
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unset bits (right to left, of course), we simply set its unset
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bits to those of \(n - 1\).
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</li>
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</ul>
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<p>
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The implementation is relatively straightfoward. Traverse \(x\) from
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least-to-most significant bit, setting its \(i\)th unset bit to \(n
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- 1\)'s \(i\)th bit. Use a bitwise mask <code>mask</code> to
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traverse \(x\).
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</p>
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<div class="post-code">
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<pre><code class="language-cpp">long long minEnd(int n, long long x) {
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int bits_to_distribute = n - 1;
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long long mask = 1;
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while (bits_to_distribute > 0) {
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if ((x & mask) == 0) {
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// if the bit should be set, set it-otherwise, leave it alone
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if ((bits_to_distribute & 1) == 1)
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x |= mask;
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bits_to_distribute >>= 1;
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}
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mask <<= 1;
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}
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return x;
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}</code></pre>
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</div>
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<h3>asymptotic complexity</h3>
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<p>
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<u>Space Complexity</u>: \(\Theta(1)\)—a constant amount of
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numeric variables are allocated regardless of \(n\) and \(x\).
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</p>
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<p>
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<u>Time Complexity</u>: in the worst case, may need to traverse the
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entirety of \(x\) to distribute every bit of \(n - 1\) to \(x\).
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This occurs if and only if \(x\) is all ones (\(\exists k\gt 0 :
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2^k-1=x\))). \(x\) and \(n\) have \(lg(x)\) and \(lg(n)\) bits
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respectively, so the solution is \(O(lg(x) + lg(n))\in O(log(xn))\).
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\(1\leq x,n\leq 1e8\), so this runtime is bounded by
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\(O(log(1e8^2))=O(log(1e16))=O(16)\).
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</p>
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</article>
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