barrett/barrettruth.com
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

feat(daily): runtime analysis

Browse source
This commit is contained in:
Barrett Ruth 2024-11-12 12:04:27 -05:00
parent 1728bae2ad
commit 7802876632
1 changed files with 56 additions and 34 deletions
Download patch file Download diff file Expand all files Collapse all files

90
posts/algorithms/leetcode-daily.html
View file

@ -97,37 +97,35 @@
<code>items</code> with a linear scan will take \(O(nlg(n))\)
time, meeting the constraints.
</p>
</div>
<h3>fleshing out the approach</h3>
<p>
A few specifics need to be understood before coding up the approach:
</p>
<ul>
<li>
Re-ordering the queries: couple <code>query[i]</code> with
<code>i</code>, then sort. When responding to queries in sorted
order, we know where to place them in an output
container&mdash;index <code>i</code>.
</li>
<li>
The linear scan: accumulate a running maximal beauty, starting at
index <code>0</code>. For some query <code>query</code>, we want
to consider all items with price less than or equal to
<code>query</code>. Therefore, loop until this condition is
<i>violated</i>&mdash; the previous index will represent the last
considered item.
</li>
<li>
Edge cases: it&apos;s perfectly possible the last considered item
is invalid (consider a query cheaper than the cheapest item).
Return <code>0</code> as specified by the problem constraints.
</li>
</ul>
<h3>
carrying out the plan
</h3>
<div class="post-code">
<pre><code class="language-cpp">vector<int> maximumBeauty(vector<vector<int>>& items, vector<int>& queries) {
<h3>carrying out the plan</h3>
<p>
A few specifics need to be understood before coding up the
approach:
</p>
<ul>
<li>
Re-ordering the queries: couple <code>query[i]</code> with
<code>i</code>, then sort. When responding to queries in sorted
order, we know where to place them in an output
container&mdash;index <code>i</code>.
</li>
<li>
The linear scan: accumulate a running maximal beauty, starting
at index <code>0</code>. For some query <code>query</code>, we
want to consider all items with price less than or equal to
<code>query</code>. Therefore, loop until this condition is
<i>violated</i>&mdash; the previous index will represent the
last considered item.
</li>
<li>
Edge cases: it&apos;s perfectly possible the last considered
item is invalid (consider a query cheaper than the cheapest
item). Return <code>0</code> as specified by the problem
constraints.
</li>
</ul>
<div class="post-code">
<pre><code class="language-cpp">vector&lt;int&gt; maximumBeauty(vector&lt;vector&lt;int&gt;&gt;& items, vector&lt;int&gt;& queries) {
std::sort(items.begin(), items.end());
std::vector&lt;pair&lt;int, int&gt;&gt; sorted_queries;
sorted_queries.reserve(queries.size());
@ -139,7 +137,7 @@
int beauty = items[0][1];
size_t i = 0;
std::vector<int> ans(queries.size());
std::vector&lt;int&gt; ans(queries.size());
for (const auto [query, index] : sorted_queries) {
while (i < items.size() && items[i][0] <= query) {
@ -152,6 +150,30 @@
return std::move(ans);</code></pre>
</div>
<h3>asymptotic complexity</h3>
<p>
Let <code>n=len(items)</code> and <code>m=len(queries)</code>.
There may be more items than queries, or vice versa. Note that a
&ldquo;looser&rdquo; upper bound can be found by analyzing the
runtime in terms of \(max\{n,m\}\).
</p>
<p>
<u>Time Complexity</u>: \(O(nlg(n)+mlg(m)+m)=O(nlg(n)+mlg(m))\).
An argument can be made that because
<code>queries[i],items[i][{0,1}]</code>\(\leq10^9\), radix sort
can be leveraged to achieve a time complexity of \(O(d \cdot (n +
k + m + k)) = O(9 \cdot (n + m + 20)) = O(9n+9m + 180) = O(9n+9m)
= O(n+m)\).
</p>
<p>
<u>Space Complexity</u>: \(\Theta(1)\), considering that \(O(m)\)
space must be allocated. If <code>queries</code>/<code
>items</code
>
cannot be modified in-place, increase the space complexity by
\(m\)/\(n\) respectively.
</p>
</div>
<div class="fold">
<h2>
<a
@ -159,7 +181,7 @@
href="https://leetcode.com/problems/shortest-subarray-with-or-at-least-k-ii/description/"
>shortest subarray with or at least k ii</a
>
&mdash; 9/12/24
&mdash; 9/11/24
</h2>
</div>
<div class="problem-content">
@ -341,7 +363,7 @@
href="https://leetcode.com/problems/minimum-array-end/"
>minimum array end</a
>
&mdash; 9/11/24
&mdash; 9/10/24
</h2>
</div>
<div class="problem-content">