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<ul class="md-nav__list">
<li class="md-nav__item">
<a href="#1-reusing-the-search-for-the-left-boundary" class="md-nav__link">
<a href="#1-reuse-the-left-boundary-search" class="md-nav__link">
<span class="md-ellipsis">
1. &nbsp; Reusing the search for the left boundary
1. &nbsp; Reuse the left boundary search
</span>
</a>
</li>
<li class="md-nav__item">
<a href="#2-transforming-into-an-element-search" class="md-nav__link">
<a href="#2-transform-into-an-element-search" class="md-nav__link">
<span class="md-ellipsis">
2. &nbsp; Transforming into an element search
2. &nbsp; Transform into an element search
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<ul class="md-nav__list">
<li class="md-nav__item">
<a href="#1-reusing-the-search-for-the-left-boundary" class="md-nav__link">
<a href="#1-reuse-the-left-boundary-search" class="md-nav__link">
<span class="md-ellipsis">
1. &nbsp; Reusing the search for the left boundary
1. &nbsp; Reuse the left boundary search
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<a href="#2-transforming-into-an-element-search" class="md-nav__link">
<a href="#2-transform-into-an-element-search" class="md-nav__link">
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2. &nbsp; Transforming into an element search
2. &nbsp; Transform into an element search
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<p class="admonition-title">Question</p>
<p>Given a sorted array <code>nums</code> of length <span class="arithmatex">\(n\)</span>, which may contain duplicate elements, return the index of the leftmost element <code>target</code>. If the element is not present in the array, return <span class="arithmatex">\(-1\)</span>.</p>
</div>
<p>Recall the method of binary search for an insertion point, after the search is completed, <span class="arithmatex">\(i\)</span> points to the leftmost <code>target</code>, <strong>thus searching for the insertion point is essentially searching for the index of the leftmost <code>target</code></strong>.</p>
<p>Consider implementing the search for the left boundary using the function for finding an insertion point. Note that the array might not contain <code>target</code>, which could lead to the following two results:</p>
<p>Recalling the method of binary search for an insertion point, after the search is completed, the index <span class="arithmatex">\(i\)</span> will point to the leftmost occurrence of <code>target</code>. Therefore, <strong>searching for the insertion point is essentially the same as finding the index of the leftmost <code>target</code></strong>.</p>
<p>We can use the function for finding an insertion point to find the left boundary of <code>target</code>. Note that the array might not contain <code>target</code>, which could lead to the following two results:</p>
<ul>
<li>The index <span class="arithmatex">\(i\)</span> of the insertion point is out of bounds.</li>
<li>The element <code>nums[i]</code> is not equal to <code>target</code>.</li>
@@ -3733,15 +3733,15 @@
</div>
</div>
<h2 id="1032-find-the-right-boundary">10.3.2 &nbsp; Find the right boundary<a class="headerlink" href="#1032-find-the-right-boundary" title="Permanent link">&para;</a></h2>
<p>So how do we find the rightmost <code>target</code>? The most straightforward way is to modify the code, replacing the pointer contraction operation in the case of <code>nums[m] == target</code>. The code is omitted here, but interested readers can implement it on their own.</p>
<p>Below we introduce two more cunning methods.</p>
<h3 id="1-reusing-the-search-for-the-left-boundary">1. &nbsp; Reusing the search for the left boundary<a class="headerlink" href="#1-reusing-the-search-for-the-left-boundary" title="Permanent link">&para;</a></h3>
<p>In fact, we can use the function for finding the leftmost element to find the rightmost element, specifically by <strong>transforming the search for the rightmost <code>target</code> into a search for the leftmost <code>target + 1</code></strong>.</p>
<p>As shown in Figure 10-7, after the search is completed, the pointer <span class="arithmatex">\(i\)</span> points to the leftmost <code>target + 1</code> (if it exists), while <span class="arithmatex">\(j\)</span> points to the rightmost <code>target</code>, <strong>thus returning <span class="arithmatex">\(j\)</span> is sufficient</strong>.</p>
<p>How do we find the rightmost occurrence of <code>target</code>? The most straightforward way is to modify the traditional binary search logic by changing how we adjust the search boundaries in the case of <code>nums[m] == target</code>. The code is omitted here. If you are interested, try to implement the code on your own.</p>
<p>Below we are going to introduce two more ingenious methods.</p>
<h3 id="1-reuse-the-left-boundary-search">1. &nbsp; Reuse the left boundary search<a class="headerlink" href="#1-reuse-the-left-boundary-search" title="Permanent link">&para;</a></h3>
<p>To find the rightmost occurrence of <code>target</code>, we can reuse the function used for locating the leftmost <code>target</code>. Specifically, we transform the search for the rightmost target into a search for the leftmost target + 1.</p>
<p>As shown in Figure 10-7, after the search is complete, pointer <span class="arithmatex">\(i\)</span> will point to the leftmost <code>target + 1</code> (if exists), while pointer <span class="arithmatex">\(j\)</span> will point to the rightmost occurrence of <code>target</code>. Therefore, returning <span class="arithmatex">\(j\)</span> will give us the right boundary.</p>
<p><a class="glightbox" href="../binary_search_edge.assets/binary_search_right_edge_by_left_edge.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="Transforming the search for the right boundary into the search for the left boundary" class="animation-figure" src="../binary_search_edge.assets/binary_search_right_edge_by_left_edge.png" /></a></p>
<p align="center"> Figure 10-7 &nbsp; Transforming the search for the right boundary into the search for the left boundary </p>
<p>Please note, the insertion point returned is <span class="arithmatex">\(i\)</span>, therefore, it should be subtracted by <span class="arithmatex">\(1\)</span> to obtain <span class="arithmatex">\(j\)</span>:</p>
<p>Note that the insertion point returned is <span class="arithmatex">\(i\)</span>, therefore, it should be subtracted by <span class="arithmatex">\(1\)</span> to obtain <span class="arithmatex">\(j\)</span>:</p>
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</div>
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</div>
<h3 id="2-transforming-into-an-element-search">2. &nbsp; Transforming into an element search<a class="headerlink" href="#2-transforming-into-an-element-search" title="Permanent link">&para;</a></h3>
<p>We know that when the array does not contain <code>target</code>, <span class="arithmatex">\(i\)</span> and <span class="arithmatex">\(j\)</span> will eventually point to the first element greater and smaller than <code>target</code> respectively.</p>
<h3 id="2-transform-into-an-element-search">2. &nbsp; Transform into an element search<a class="headerlink" href="#2-transform-into-an-element-search" title="Permanent link">&para;</a></h3>
<p>When the array does not contain <code>target</code>, <span class="arithmatex">\(i\)</span> and <span class="arithmatex">\(j\)</span> will eventually point to the first element greater and smaller than <code>target</code> respectively.</p>
<p>Thus, as shown in Figure 10-8, we can construct an element that does not exist in the array, to search for the left and right boundaries.</p>
<ul>
<li>To find the leftmost <code>target</code>: it can be transformed into searching for <code>target - 0.5</code>, and return the pointer <span class="arithmatex">\(i\)</span>.</li>
@@ -3846,10 +3846,10 @@
<p><a class="glightbox" href="../binary_search_edge.assets/binary_search_edge_by_element.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="Transforming the search for boundaries into the search for an element" class="animation-figure" src="../binary_search_edge.assets/binary_search_edge_by_element.png" /></a></p>
<p align="center"> Figure 10-8 &nbsp; Transforming the search for boundaries into the search for an element </p>
<p>The code is omitted here, but two points are worth noting.</p>
<p>The code is omitted here, but here are two important points to note about this approach.</p>
<ul>
<li>The given array does not contain decimals, meaning we do not need to worry about how to handle equal situations.</li>
<li>Since this method introduces decimals, the variable <code>target</code> in the function needs to be changed to a floating point type (no change needed in Python).</li>
<li>The given array <code>nums</code> does not contain decimal, so handling equal cases is not a concern.</li>
<li>However, introducing decimals in this approach requires modifying the <code>target</code> variable to a floating-point type (no change needed in Python).</li>
</ul>
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