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<h2class="anchor anchorWithStickyNavbar_LWe7"id="problem">Problem<ahref="#problem"class="hash-link"aria-label="Direct link to Problem"title="Direct link to Problem"></a></h2>
<p>The problem we are going to solve is one of <em>CodeWars</em> katas and is called
<p>Slide down in this case is equal to <code>1074</code>.</p>
<h2class="anchor anchorWithStickyNavbar_LWe7"id="solving-the-problem">Solving the problem<ahref="#solving-the-problem"class="hash-link"aria-label="Direct link to Solving the problem"title="Direct link to Solving the problem"></a></h2>
<divclass="theme-admonition theme-admonition-caution admonition_xJq3 alert alert--warning"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 16 16"><pathfill-rule="evenodd"d="M8.893 1.5c-.183-.31-.52-.5-.887-.5s-.703.19-.886.5L.138 13.499a.98.98 0 0 0 0 1.001c.193.31.53.501.886.501h13.964c.367 0 .704-.19.877-.5a1.03 1.03 0 0 0 .01-1.002L8.893 1.5zm.133 11.497H6.987v-2.003h2.039v2.003zm0-3.004H6.987V5.987h2.039v4.006z"></path></svg></span>caution</div><divclass="admonitionContent_BuS1"><p>I will describe the following ways you can approach this problem and implement
them in <em>Java</em><sup><ahref="#user-content-fn-1"id="user-content-fnref-1"data-footnote-ref="true"aria-describedby="footnote-label">1</a></sup>.</p></div></div>
<p>For all of the following solutions I will be using basic <code>main</code> function that
will output <code>true</code>/<code>false</code> based on the expected output of our algorithm. Any
other differences will lie only in the solutions of the problem. You can see the
<h2class="anchor anchorWithStickyNavbar_LWe7"id="naïve-solution">Naïve solution<ahref="#naïve-solution"class="hash-link"aria-label="Direct link to Naïve solution"title="Direct link to Naïve solution"></a></h2>
<p>Our naïve solution consists of trying out all the possible slides and finding
<p>First one is used as a <em>public interface</em> to the solution, you just pass in the
pyramid itself. Second one is the recursive “algorithm” that finds the slide
down.</p>
<p>It is a relatively simple solution… There's nothing to do at the bottom of the
pyramid, so we just return the value in the <em>cell</em>. Otherwise we add it and try
to slide down the available cells below the current row.</p>
<h3class="anchor anchorWithStickyNavbar_LWe7"id="time-complexity">Time complexity<ahref="#time-complexity"class="hash-link"aria-label="Direct link to Time complexity"title="Direct link to Time complexity"></a></h3>
<p>If you get the source code and run it yourself, it runs rather fine… I hope you
are wondering about the time complexity of the proposed solution and, since it
really is a naïve solution, the time complexity is pretty bad. Let's find the
<p>There's not much to do here, so we can safely say that the time complexity of
this function is bounded by <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>T</mi><mostretchy="false">(</mo><mi>n</mi><mostretchy="false">)</mo></mrow><annotationencoding="application/x-tex">T(n)</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:1em;vertical-align:-0.25em"></span><spanclass="mord mathnormal"style="margin-right:0.13889em">T</span><spanclass="mopen">(</span><spanclass="mord mathnormal">n</span><spanclass="mclose">)</span></span></span></span>, where <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>T</mi></mrow><annotationencoding="application/x-tex">T</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6833em"></span><spanclass="mord mathnormal"style="margin-right:0.13889em">T</span></span></span></span> is our second overload. This
doesn't tell us anything, so let's move on to the second overload where we are
going to define the <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>T</mi><mostretchy="false">(</mo><mi>n</mi><mostretchy="false">)</mo></mrow><annotationencoding="application/x-tex">T(n)</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:1em;vertical-align:-0.25em"></span><spanclass="mord mathnormal"style="margin-right:0.13889em">T</span><spanclass="mopen">(</span><spanclass="mord mathnormal">n</span><spanclass="mclose">)</span></span></span></span> function.</p>
<p>If you wonder why, I'll try to describe it intuitively:</p>
<ol>
<li>In each call to <code>longestSlideDown</code> we do some work in constant time,
regardless of being in the base case. Those are the <code>1</code>s in both cases.</li>
<li>If we are not in the base case, we move one row down <strong>twice</strong>. That's how we
obtained <code>2 *</code> and <code>y + 1</code> in the <em>otherwise</em> case.</li>
<li>We move row-by-row, so we move down <code>y</code>-times and each call splits to two
subtrees.</li>
<li>Overall, if we were to represent the calls as a tree, we would get a full
binary tree of height <code>y</code>, in each node we do some work in constant time,
therefore we can just sum the ones.</li>
</ol>
<divclass="theme-admonition theme-admonition-warning admonition_xJq3 alert alert--warning"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 16 16"><pathfill-rule="evenodd"d="M8.893 1.5c-.183-.31-.52-.5-.887-.5s-.703.19-.886.5L.138 13.499a.98.98 0 0 0 0 1.001c.193.31.53.501.886.501h13.964c.367 0 .704-.19.877-.5a1.03 1.03 0 0 0 .01-1.002L8.893 1.5zm.133 11.497H6.987v-2.003h2.039v2.003zm0-3.004H6.987V5.987h2.039v4.006z"></path></svg></span>warning</div><divclass="admonitionContent_BuS1"><p>It would've been more complicated to get an exact result. In the equation above
we are assuming that the width of the pyramid is bound by the height.</p></div></div>
<p>Hopefully we can agree that this is not the best we can do. <!---->😉</p>
<h2class="anchor anchorWithStickyNavbar_LWe7"id="greedy-solution">Greedy solution<ahref="#greedy-solution"class="hash-link"aria-label="Direct link to Greedy solution"title="Direct link to Greedy solution"></a></h2>
<p>We will try to optimize it a bit. Let's start with a relatively simple <em>greedy</em>
approach.</p>
<divclass="theme-admonition theme-admonition-info admonition_xJq3 alert alert--info"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 14 16"><pathfill-rule="evenodd"d="M7 2.3c3.14 0 5.7 2.56 5.7 5.7s-2.56 5.7-5.7 5.7A5.71 5.71 0 0 1 1.3 8c0-3.14 2.56-5.7 5.7-5.7zM7 1C3.14 1 0 4.14 0 8s3.14 7 7 7 7-3.14 7-7-3.14-7-7-7zm1 3H6v5h2V4zm0 6H6v2h2v-2z"></path></svg></span>Greedy algorithms</div><divclass="admonitionContent_BuS1"><p><em>Greedy algorithms</em> can be described as algorithms that decide the action on the
optimal option at the moment.</p></div></div>
<p>We can try to adjust the naïve solution. The most problematic part are the
recursive calls. Let's apply the greedy approach there:</p>
<p>OK, if we cannot go right <strong>or</strong> the right path adds smaller value to the sum,
we simply go left.</p>
<h3class="anchor anchorWithStickyNavbar_LWe7"id="time-complexity-1">Time complexity<ahref="#time-complexity-1"class="hash-link"aria-label="Direct link to Time complexity"title="Direct link to Time complexity"></a></h3>
<p>We have switched from <em>adding the maximum</em> to <em>following the “bigger” path</em>, so
we improved the time complexity tremendously. We just go down the pyramid all
the way to the bottom. Therefore we are getting:</p>
<p>We have managed to convert our exponential solution into a linear one.</p>
<h3class="anchor anchorWithStickyNavbar_LWe7"id="running-the-tests">Running the tests<ahref="#running-the-tests"class="hash-link"aria-label="Direct link to Running the tests"title="Direct link to Running the tests"></a></h3>
<p>However, if we run the tests, we notice that the second test failed:</p>
<h2class="anchor anchorWithStickyNavbar_LWe7"id="top-down-dp">Top-down DP<ahref="#top-down-dp"class="hash-link"aria-label="Direct link to Top-down DP"title="Direct link to Top-down DP"></a></h2>
<p><em>Top-down dynamic programming</em> is probably the most common approach, since (at
least looks like) is the easiest to implement. The whole point is avoiding the
unnecessary computations that we have already done.</p>
<p>In our case, we can use our naïve solution and put a <em>cache</em> on top of it that
will make sure, we don't do unnecessary calculations.</p>
immutable structure with implicitly defined <code>.equals()</code>, <code>.hashCode()</code>,
<code>.toString()</code> and getters for the attributes.</p></div></div>
<p>Because of the choice of <code>TreeMap</code>, we had to additionally define the ordering
on it.</p>
<p>In the <code>longestSlideDown</code> you can notice that the computation which used to be
at the end of the naïve version above, is now wrapped in an <code>if</code> statement that
checks for the presence of the position in the cache and computes the slide down
just when it's needed.</p>
<h3class="anchor anchorWithStickyNavbar_LWe7"id="time-complexity-2">Time complexity<ahref="#time-complexity-2"class="hash-link"aria-label="Direct link to Time complexity"title="Direct link to Time complexity"></a></h3>
<p>If you think that evaluating time complexity for this approach is a bit more
tricky, you are right. Keeping the cache in mind, it is not the easiest thing
to do. However there are some observations that might help us figure this out:</p>
<ol>
<li>Slide down from each position is calculated only once.</li>
<li>Once calculated, we use the result from the cache.</li>
</ol>
<p>Knowing this, we still cannot, at least easily, describe the time complexity of
finding the best slide down from a specific position, <strong>but</strong> we can bound it
from above for the <strong>whole</strong> run from the top. Now the question is how we can do
that!</p>
<p>Overall we are doing the same things for almost<sup><ahref="#user-content-fn-2"id="user-content-fnref-2"data-footnote-ref="true"aria-describedby="footnote-label">2</a></sup> all of the positions within
the pyramid:</p>
<ol>
<li>
<p>We calculate and store it (using the partial results stored in cache). This
is done only once.</p>
<p>For each calculation we take 2 values from the cache and insert one value.
Because we have chosen <code>TreeMap</code>, these 3 operations have logarithmic time
complexity and therefore this step is equivalent to <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>3</mn><mo>⋅</mo><msub><mrow><mi>log</mi><mo></mo></mrow><mn>2</mn></msub><mi>n</mi></mrow><annotationencoding="application/x-tex">3 \cdot \log_2{n}</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6444em"></span><spanclass="mord">3</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mbin">⋅</span><spanclass="mspace"style="margin-right:0.2222em"></span></span><spanclass="base"><spanclass="strut"style="height:0.9386em;vertical-align:-0.2441em"></span><spanclass="mop"><spanclass="mop">lo<spanstyle="margin-right:0.01389em">g</span></span><spanclass="msupsub"><spanclass="vlist-t vlist-t2"><spanclass="vlist-r"><spanclass="vlist"style="height:0.207em"><spanstyle="top:-2.4559em;margin-right:0.05em"><spanclass="pstrut"style="height:2.7em"></span><spanclass="sizing reset-size6 size3 mtight"><spanclass="mord mtight">2</span></span></span></span><spanclass="vlist-s"></span></span><spanclass="vlist-r"><spanclass="vlist"style="height:0.2441em"><span></span></span></span></span></span></span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord"><spanclass="mord mathnormal">n</span></span></span></span></span>.</p>
<p>However for the sake of simplicity, we are going to account only for the
insertion, the reason is rather simple, if we include the 2 retrievals here,
it will be interleaved with the next step, therefore it is easier to keep the
retrievals in the following point.</p>
<divclass="theme-admonition theme-admonition-caution admonition_xJq3 alert alert--warning"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 16 16"><pathfill-rule="evenodd"d="M8.893 1.5c-.183-.31-.52-.5-.887-.5s-.703.19-.886.5L.138 13.499a.98.98 0 0 0 0 1.001c.193.31.53.501.886.501h13.964c.367 0 .704-.19.877-.5a1.03 1.03 0 0 0 .01-1.002L8.893 1.5zm.133 11.497H6.987v-2.003h2.039v2.003zm0-3.004H6.987V5.987h2.039v4.006z"></path></svg></span>caution</div><divclass="admonitionContent_BuS1"><p>You might have noticed it's still not that easy, cause we're not having full
cache right from the beginning, but the sum of those logarithms cannot be
expressed in a nice way, so taking the upper bound, i.e. expecting the cache
to be full at all times, is the best option for nice and readable complexity
of the whole approach.</p></div></div>
<p>Our final upper bound of this work is therefore <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mrow><mi>log</mi><mo></mo></mrow><mn>2</mn></msub><mi>n</mi></mrow><annotationencoding="application/x-tex">\log_2{n}</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.9386em;vertical-align:-0.2441em"></span><spanclass="mop"><spanclass="mop">lo<spanstyle="margin-right:0.01389em">g</span></span><spanclass="msupsub"><spanclass="vlist-t vlist-t2"><spanclass="vlist-r"><spanclass="vlist"style="height:0.207em"><spanstyle="top:-2.4559em;margin-right:0.05em"><spanclass="pstrut"style="height:2.7em"></span><spanclass="sizing reset-size6 size3 mtight"><spanclass="mord mtight">2</span></span></span></span><spanclass="vlist-s"></span></span><spanclass="vlist-r"><spanclass="vlist"style="height:0.2441em"><span></span></span></span></span></span></span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord"><spanclass="mord mathnormal">n</span></span></span></span></span>.</p>
</li>
<li>
<p>We retrieve it from the cache. Same as in first point, but only twice, so we
get <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>2</mn><mo>⋅</mo><msub><mrow><mi>log</mi><mo></mo></mrow><mn>2</mn></msub><mi>n</mi></mrow><annotationencoding="application/x-tex">2 \cdot \log_2{n}</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6444em"></span><spanclass="mord">2</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mbin">⋅</span><spanclass="mspace"style="margin-right:0.2222em"></span></span><spanclass="base"><spanclass="strut"style="height:0.9386em;vertical-align:-0.2441em"></span><spanclass="mop"><spanclass="mop">lo<spanstyle="margin-right:0.01389em">g</span></span><spanclass="msupsub"><spanclass="vlist-t vlist-t2"><spanclass="vlist-r"><spanclass="vlist"style="height:0.207em"><spanstyle="top:-2.4559em;margin-right:0.05em"><spanclass="pstrut"style="height:2.7em"></span><spanclass="sizing reset-size6 size3 mtight"><spanclass="mord mtight">2</span></span></span></span><spanclass="vlist-s"></span></span><spanclass="vlist-r"><spanclass="vlist"style="height:0.2441em"><span></span></span></span></span></span></span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord"><spanclass="mord mathnormal">n</span></span></span></span></span>.</p>
<divclass="theme-admonition theme-admonition-caution admonition_xJq3 alert alert--warning"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 16 16"><pathfill-rule="evenodd"d="M8.893 1.5c-.183-.31-.52-.5-.887-.5s-.703.19-.886.5L.138 13.499a.98.98 0 0 0 0 1.001c.193.31.53.501.886.501h13.964c.367 0 .704-.19.877-.5a1.03 1.03 0 0 0 .01-1.002L8.893 1.5zm.133 11.497H6.987v-2.003h2.039v2.003zm0-3.004H6.987V5.987h2.039v4.006z"></path></svg></span>caution</div><divclass="admonitionContent_BuS1"><p>It's done twice because of the <code>.containsKey()</code> in the <code>if</code> condition.</p></div></div>
</li>
</ol>
<p>Okay, we have evaluated work done for each of the cells in the pyramid and now
we need to put it together.</p>
<p>Let's split the time complexity of our solution into two operands:</p>
<p><spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>r</mi></mrow><annotationencoding="application/x-tex">r</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em"></span><spanclass="mord mathnormal"style="margin-right:0.02778em">r</span></span></span></span> will represent the <em>actual</em> calculation of the cells and <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>s</mi></mrow><annotationencoding="application/x-tex">s</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em"></span><spanclass="mord mathnormal">s</span></span></span></span> will represent
the additional retrievals on top of the calculation.</p>
<p>We calculate the values only <strong>once</strong>, therefore we can safely agree on:</p>
<p>What about the <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>s</mi></mrow><annotationencoding="application/x-tex">s</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em"></span><spanclass="mord mathnormal">s</span></span></span></span> though? Key observation here is the fact that we have 2
lookups on the tree in each of them <strong>and</strong> we do it twice, cause each cell has
<divclass="theme-admonition theme-admonition-tip admonition_xJq3 alert alert--success"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 12 16"><pathfill-rule="evenodd"d="M6.5 0C3.48 0 1 2.19 1 5c0 .92.55 2.25 1 3 1.34 2.25 1.78 2.78 2 4v1h5v-1c.22-1.22.66-1.75 2-4 .45-.75 1-2.08 1-3 0-2.81-2.48-5-5.5-5zm3.64 7.48c-.25.44-.47.8-.67 1.11-.86 1.41-1.25 2.06-1.45 3.23-.02.05-.02.11-.02.17H5c0-.06 0-.13-.02-.17-.2-1.17-.59-1.83-1.45-3.23-.2-.31-.42-.67-.67-1.11C2.44 6.78 2 5.65 2 5c0-2.2 2.02-4 4.5-4 1.22 0 2.36.42 3.22 1.19C10.55 2.94 11 3.94 11 5c0 .66-.44 1.78-.86 2.48zM4 14h5c-.23 1.14-1.3 2-2.5 2s-2.27-.86-2.5-2z"></path></svg></span>tip</div><divclass="admonitionContent_BuS1"><p>You might've noticed that lookups actually take more time than the construction
of the results. This is not entirely true, since we have included the
<code>.containsKey()</code> and <code>.get()</code> from the <code>return</code> statement in the second part.</p><p>If we were to represent this more precisely, we could've gone with:</p><spanclass="katex-display"><spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mtablerowspacing="0.25em"columnalign="right left"columnspacing="0em"><mtr><mtd><mstylescriptlevel="0"displaystyle="true"><mi>r</mi></mstyle></mtd><mtd><mstylescriptlevel="0"displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mn>3</mn><mo>⋅</mo><mi>n</mi><mo>⋅</mo><mi>log</mi><mo></mo><mi>n</mi></mrow></mstyle></mtd></mtr><mtr><mtd><mstylescriptlevel="0"displaystyle="true"><mi>s</mi></mstyle></mtd><mtd><mstylescriptlevel="0"displaystyle="true"><mrow><mrow></mrow><mo>=</mo><mn>2</mn><mo>⋅</mo><mi>n</mi><mo>⋅</mo><mi>log</mi><mo></mo><mi>n</mi></mrow></mstyle></mtd></mtr></mtable><annotationencoding="application/x-tex">\begin{align*}
r &= 3 \cdot n \cdot \log{n} \\
s &= 2 \cdot n \cdot \log{n}
\end{align*}</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:3em;vertical-align:-1.25em"></span><spanclass="mord"><spanclass="mtable"><spanclass="col-align-r"><spanclass="vlist-t vlist-t2"><spanclass="vlist-r"><spanclass="vlist"style="height:1.75em"><spanstyle="top:-3.91em"><spanclass="pstrut"style="height:3em"></span><spanclass="mord"><spanclass="mord mathnormal"style="margin-right:0.02778em">r</span></span></span><spanstyle="top:-2.41em"><spanclass="pstrut"style="height:3em"></span><spanclass="mord"><spanclass="mord mathnormal">s</span></span></span></span><spanclass="vlist-s"></span></span><spanclass="vlist-r"><spanclass="vlist"style="height:1.25em"><span></span></span></span></span></span><spanclass="col-align-l"><spanclass="vlist-t vlist-t2"><spanclass="vlist-r"><spanclass="vlist"style="height:1.75em"><spanstyle="top:-3.91em"><spanclass="pstrut"style="height:3em"></span><spanclass="mord"><spanclass="mord"></span><spanclass="mspace"style="margin-right:0.2778em"></span><spanclass="mrel">=</span><spanclass="mspace"style="margin-right:0.2778em"></span><spanclass="mord">3</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mbin">⋅</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mord mathnormal">n</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mbin">⋅</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mop">lo<spanstyle="margin-right:0.01389em">g</span></span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord"><spanclass="mord mathnormal">n</span></span></span></span><spanstyle="top:-2.41em"><spanclass="pstrut"style="height:3em"></span><spanclass="mord"><spanclass="mord"></span><spanclass="mspace"style="margin-right:0.2778em"></span><spanclass="mrel">=</span><spanclass="mspace"style="margin-right:0.2778em"></span><spanclass="mord">2</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mbin">⋅</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mord mathnormal">n</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mbin">⋅</span><spanclass="mspace"style="margin-right:0.2222em"></span><spanclass="mop">lo<spanstyle="margin-right:0.01389em">g</span></span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord"><spanclass="mord mathnormal">n</span></span></span></span></span><spanclass="vlist-s"></span></span><spanclass="vlist-r"><spanclass="vlist"style="height:1.25em"><span></span></span></span></span></span></span></span></span></span></span></span><p>On the other hand we are summing both numbers together, therefore in the end it
doesn't really matter.</p><p>(<em>Feel free to compare the sums of both “splits”.</em>)</p></div></div>
<p>And so our final time complexity for the whole <em>top-down dynamic programming</em>
<p>As you can see, this is worse than our <em>greedy</em> solution that was incorrect, but
it's better than the <em>naïve</em> one.</p>
<h3class="anchor anchorWithStickyNavbar_LWe7"id="memory-complexity">Memory complexity<ahref="#memory-complexity"class="hash-link"aria-label="Direct link to Memory complexity"title="Direct link to Memory complexity"></a></h3>
<p>With this approach we need to talk about the memory complexity too, because we
have introduced cache. If you think that the memory complexity is linear to the
input, you are right. We start at the top and try to find each and every slide
down. At the end we get the final result for <code>new Position(0, 0)</code>, so we need to
<p><spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotationencoding="application/x-tex">n</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em"></span><spanclass="mord mathnormal">n</span></span></span></span> represents the total amount of cells in the pyramid, i.e.</p>
<divclass="theme-admonition theme-admonition-caution admonition_xJq3 alert alert--warning"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 16 16"><pathfill-rule="evenodd"d="M8.893 1.5c-.183-.31-.52-.5-.887-.5s-.703.19-.886.5L.138 13.499a.98.98 0 0 0 0 1.001c.193.31.53.501.886.501h13.964c.367 0 .704-.19.877-.5a1.03 1.03 0 0 0 .01-1.002L8.893 1.5zm.133 11.497H6.987v-2.003h2.039v2.003zm0-3.004H6.987V5.987h2.039v4.006z"></path></svg></span>caution</div><divclass="admonitionContent_BuS1"><p>If you're wondering whether it's correct because of the second <code>if</code> in our
function, your guess is right. However we are expressing the complexity in the
Bachmann-Landau notation, so we care about the <strong>upper bound</strong>, not the exact
number.</p></div></div>
<divclass="theme-admonition theme-admonition-tip admonition_xJq3 alert alert--success"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 12 16"><pathfill-rule="evenodd"d="M6.5 0C3.48 0 1 2.19 1 5c0 .92.55 2.25 1 3 1.34 2.25 1.78 2.78 2 4v1h5v-1c.22-1.22.66-1.75 2-4 .45-.75 1-2.08 1-3 0-2.81-2.48-5-5.5-5zm3.64 7.48c-.25.44-.47.8-.67 1.11-.86 1.41-1.25 2.06-1.45 3.23-.02.05-.02.11-.02.17H5c0-.06 0-.13-.02-.17-.2-1.17-.59-1.83-1.45-3.23-.2-.31-.42-.67-.67-1.11C2.44 6.78 2 5.65 2 5c0-2.2 2.02-4 4.5-4 1.22 0 2.36.42 3.22 1.19C10.55 2.94 11 3.94 11 5c0 .66-.44 1.78-.86 2.48zM4 14h5c-.23 1.14-1.3 2-2.5 2s-2.27-.86-2.5-2z"></path></svg></span>Can this be optimized?</div><divclass="admonitionContent_BuS1"><p>Yes, it can! Try to think about a way, how can you minimize the memory
complexity of this approach. I'll give you a hint:</p><spanclass="katex-display"><spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mimathvariant="script">O</mi><mostretchy="false">(</mo><mi>r</mi><mi>o</mi><mi>w</mi><mi>s</mi><mostretchy="false">)</mo></mrow><annotationencoding="application/x-tex">\mathcal{O}(rows)</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:1em;vertical-align:-0.25em"></span><spanclass="mord mathcal"style="margin-right:0.02778em">O</span><spanclass="mopen">(</span><spanclass="mord mathnormal">ro</span><spanclass="mord mathnormal"style="margin-right:0.02691em">w</span><spanclass="mord mathnormal">s</span><spanclass="mclose">)</span></span></span></span></span></div></div>
<h2class="anchor anchorWithStickyNavbar_LWe7"id="bottom-up-dp">Bottom-up DP<ahref="#bottom-up-dp"class="hash-link"aria-label="Direct link to Bottom-up DP"title="Direct link to Bottom-up DP"></a></h2>
<p>If you try to think in depth about the top-down DP solution, you might notice
that the <em>core</em> of it stands on caching the calculations that have been already
done on the lower “levels” of the pyramid. Our bottom-up implementation will be
using this fact!</p>
<divclass="theme-admonition theme-admonition-tip admonition_xJq3 alert alert--success"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 12 16"><pathfill-rule="evenodd"d="M6.5 0C3.48 0 1 2.19 1 5c0 .92.55 2.25 1 3 1.34 2.25 1.78 2.78 2 4v1h5v-1c.22-1.22.66-1.75 2-4 .45-.75 1-2.08 1-3 0-2.81-2.48-5-5.5-5zm3.64 7.48c-.25.44-.47.8-.67 1.11-.86 1.41-1.25 2.06-1.45 3.23-.02.05-.02.11-.02.17H5c0-.06 0-.13-.02-.17-.2-1.17-.59-1.83-1.45-3.23-.2-.31-.42-.67-.67-1.11C2.44 6.78 2 5.65 2 5c0-2.2 2.02-4 4.5-4 1.22 0 2.36.42 3.22 1.19C10.55 2.94 11 3.94 11 5c0 .66-.44 1.78-.86 2.48zM4 14h5c-.23 1.14-1.3 2-2.5 2s-2.27-.86-2.5-2z"></path></svg></span>tip</div><divclass="admonitionContent_BuS1"><p>As I have said in the <em>top-down DP</em> section, it is the easiest way to implement
DP (unless the cached function has complicated parameters, in that case it might
get messy).</p><p>Bottom-up dynamic programming can be more effective, but may be more complicated
to implement right from the beginning.</p></div></div>
<p>Let's see how we can implement it:</p>
<divclass="language-java codeBlockContainer_Ckt0 theme-code-block"style="--prism-color:#000000;--prism-background-color:#ffffff"><divclass="codeBlockContent_biex"><pretabindex="0"class="prism-code language-java codeBlock_bY9V thin-scrollbar"style="color:#000000;background-color:#ffffff"><codeclass="codeBlockLines_e6Vv"><spanclass="token-line"style="color:#000000"><spanclass="token keyword"style="color:rgb(0, 0, 255)">public</span><spanclass="token plain"></span><spanclass="token keyword"style="color:rgb(0, 0, 255)">static</span><spanclass="token plain"></span><spanclass="token keyword"style="color:rgb(0, 0, 255)">int</span><spanclass="token plain"></span><spanclass="token function"style="color:rgb(0, 0, 255)">longestSlideDown</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">(</span><spanclass="token keyword"style="color:rgb(0, 0, 255)">int</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">[</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">]</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">[</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">]</span><spanclass="token plain"> pyramid</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">)</span><spanclass="token plain"></span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">{</span><spanclass="token plain"></span><br></span><spanclass="token-line"style="color:#000000"><spanclass="token plain"></span><spanclass="token comment"style="color:rgb(0, 128, 0)">// In the beginning we declare new array. At this point it is easier to just</span><spanclass="token plain"></span><br></span><spanclass="token-line"style="color:#000000"><spanclass="token plain"></span><spanclass="token comment"style="color:rgb(0, 128, 0)">// work with the one dimension, i.e. just allocating the space for the rows.</span><spanclass="token plain"></span><br></span><spanclass="token-line"style="color:#000000"><spanclass="token plain"></span><spanclass="token keyword"style="color:rgb(0, 0, 255)">int</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">[</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">]</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">[</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">]</span><spanclass="token plain"> slideDowns </span><spanclass="token operator"style="color:rgb(0, 0, 0)">=</span><spanclass="token plain"></span><spanclass="token keyword"style="color:rgb(0, 0, 255)">new</span><spanclass="token plain"></span><spanclass="token keyword"style="color:rgb(0, 0, 255)">int</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">[</span><spanclass="token plain">pyramid</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">.</span><spanclass="token plain">length</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">]</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">[</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">]</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">;</span><spanclass="token plain"></span><br></span><spanclass="token-line"style="color:#000000"><spanclass="token plain"style="display:inline-block"></span><br></span><spanclass="token-line"style="color:#000000"><spanclass="token plain"></span><spanclass="token comment"style="color:rgb(0, 128, 0)">// Bottom row gets just copied, there's nothing else to do… It's the base</span><spanclass="token plain"></span><br></span><spanclass="token-line"style="color:#000000"><spanclass="token plain"></span><spanclass="token comment"style="color:rgb(0, 128, 0)">// case.</span><spanclass="token plain"></span><br></span><spanclass="token-line"style="color:#000000"><spanclass="token plain"> slideDowns</span><spanclass="token punctuation"style="color:rgb(4, 81, 165)">[</span><spanclass="token plain">pyramid</span><spanclass="token punctuation"
<p>I've tried to explain the code as much as possible within the comments, since it
might be more beneficial to see right next to the “offending” lines.</p>
<p>As you can see, in this approach we go from the other side<sup><ahref="#user-content-fn-3"id="user-content-fnref-3"data-footnote-ref="true"aria-describedby="footnote-label">3</a></sup>, the bottom of
the pyramid and propagate the partial results up.</p>
<divclass="theme-admonition theme-admonition-info admonition_xJq3 alert alert--info"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 14 16"><pathfill-rule="evenodd"d="M7 2.3c3.14 0 5.7 2.56 5.7 5.7s-2.56 5.7-5.7 5.7A5.71 5.71 0 0 1 1.3 8c0-3.14 2.56-5.7 5.7-5.7zM7 1C3.14 1 0 4.14 0 8s3.14 7 7 7 7-3.14 7-7-3.14-7-7-7zm1 3H6v5h2V4zm0 6H6v2h2v-2z"></path></svg></span>How is this different from the <em>greedy</em> solution???</div><divclass="admonitionContent_BuS1"><p>If you try to compare them, you might find a very noticable difference. The
greedy approach is going from the top to the bottom without <strong>any</strong> knowledge of
what's going on below. On the other hand, bottom-up DP is going from the bottom
(<em>DUH…</em>) and <strong>propagates</strong> the partial results to the top. The propagation is
what makes sure that at the top I don't choose the best <strong>local</strong> choice, but
the best <strong>overall</strong> result I can achieve.</p></div></div>
<h3class="anchor anchorWithStickyNavbar_LWe7"id="time-complexity-3">Time complexity<ahref="#time-complexity-3"class="hash-link"aria-label="Direct link to Time complexity"title="Direct link to Time complexity"></a></h3>
<p>Time complexity of this solution is rather simple. We allocate an array for the
rows and then for each row, we copy it and adjust the partial results. Doing
<p>Of course, this is an upper bound, since we iterate through the bottom row only
once.</p>
<h3class="anchor anchorWithStickyNavbar_LWe7"id="memory-complexity-1">Memory complexity<ahref="#memory-complexity-1"class="hash-link"aria-label="Direct link to Memory complexity"title="Direct link to Memory complexity"></a></h3>
<p>We're allocating an array for the pyramid <strong>again</strong> for our partial results, so
<divclass="theme-admonition theme-admonition-tip admonition_xJq3 alert alert--success"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 12 16"><pathfill-rule="evenodd"d="M6.5 0C3.48 0 1 2.19 1 5c0 .92.55 2.25 1 3 1.34 2.25 1.78 2.78 2 4v1h5v-1c.22-1.22.66-1.75 2-4 .45-.75 1-2.08 1-3 0-2.81-2.48-5-5.5-5zm3.64 7.48c-.25.44-.47.8-.67 1.11-.86 1.41-1.25 2.06-1.45 3.23-.02.05-.02.11-.02.17H5c0-.06 0-.13-.02-.17-.2-1.17-.59-1.83-1.45-3.23-.2-.31-.42-.67-.67-1.11C2.44 6.78 2 5.65 2 5c0-2.2 2.02-4 4.5-4 1.22 0 2.36.42 3.22 1.19C10.55 2.94 11 3.94 11 5c0 .66-.44 1.78-.86 2.48zM4 14h5c-.23 1.14-1.3 2-2.5 2s-2.27-.86-2.5-2z"></path></svg></span>tip</div><divclass="admonitionContent_BuS1"><p>If we were writing this in C++ or Rust, we could've avoided that, but not
really.</p><p>C++ would allow us to <strong>copy</strong> the pyramid rightaway into the parameter, so we
would be able to directly change it. However it's still a copy, even though we
don't need to allocate anything ourselves. It's just implicitly done for us.</p><p>Rust is more funny in this case. If the pyramids weren't used after the call of
<code>longest_slide_down</code>, it would simply <strong>move</strong> them into the functions. If they
were used afterwards, the compiler would force you to either borrow it, or
<em>clone-and-move</em> for the function.</p><hr><p>Since we're doing it in Java, we get a reference to the <em>original</em> array and we
can't do whatever we want with it.</p></div></div>
<h2class="anchor anchorWithStickyNavbar_LWe7"id="summary">Summary<ahref="#summary"class="hash-link"aria-label="Direct link to Summary"title="Direct link to Summary"></a></h2>
<p>And we've finally reached the end. We have seen 4 different “solutions”<sup><ahref="#user-content-fn-4"id="user-content-fnref-4"data-footnote-ref="true"aria-describedby="footnote-label">4</a></sup> of
the same problem using different approaches. Different approaches follow the
order in which you might come up with them, each approach influences its
successor and represents the way we can enhance the existing implementation.</p>
<sectiondata-footnotes="true"class="footnotes"><h2class="anchor anchorWithStickyNavbar_LWe7 sr-only"id="footnote-label">Footnotes<ahref="#footnote-label"class="hash-link"aria-label="Direct link to Footnotes"title="Direct link to Footnotes"></a></h2>
<ol>
<liid="user-content-fn-1">
<p>cause why not, right!? <ahref="#user-content-fnref-1"data-footnote-backref=""aria-label="Back to reference 1"class="data-footnote-backref">↩</a></p>
</li>
<liid="user-content-fn-2">
<p>except the bottom row <ahref="#user-content-fnref-2"data-footnote-backref=""aria-label="Back to reference 2"class="data-footnote-backref">↩</a></p>
</li>
<liid="user-content-fn-3">
<p>definitely not an RHCP reference <!---->😉 <ahref="#user-content-fnref-3"data-footnote-backref=""aria-label="Back to reference 3"class="data-footnote-backref">↩</a></p>
</li>
<liid="user-content-fn-4">
<p>one was not correct, thus the quotes <ahref="#user-content-fnref-4"data-footnote-backref=""aria-label="Back to reference 4"class="data-footnote-backref">↩</a></p>
