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<meta charset="utf-8">
<meta name="viewport" content="width=device-width,initial-scale=1">
<meta name="description" content="Data Structures and Algorithms Crash Course with Animated Illustrations and Off-the-Shelf Code">
<meta name="description" content="Data structures and algorithms tutorial with animated illustrations and ready-to-run code">
<meta name="author" content="krahets">
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<span class="md-ellipsis">
Chapter 1. Encounter With Algorithms
Chapter 1. Encounter with Algorithms
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Chapter 1. Encounter With Algorithms
Chapter 1. Encounter with Algorithms
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Chapter 4. Array and Linked List
Chapter 4. Arrays and Linked Lists
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Chapter 4. Array and Linked List
Chapter 4. Arrays and Linked Lists
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4.4 Memory and Cache *
4.4 Random-Access Memory and Cache *
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Chapter 5. Stack and Queue
Chapter 5. Stacks and Queues
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Chapter 5. Stack and Queue
Chapter 5. Stacks and Queues
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5.3 Double-Ended Queue
5.3 Deque
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Chapter 6. Hashing
Chapter 6. Hash Table
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Chapter 6. Hashing
Chapter 6. Hash Table
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7.3 Array Representation of Tree
7.3 Array Representation of Binary Trees
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8.2 Building a Heap
8.2 Heap Construction Operation
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8.3 Top-K Problem
8.3 Top-k Problem
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10.2 Binary Search Insertion
10.2 Binary Search Insertion Point
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10.3 Binary Search Edge Cases
10.3 Binary Search Boundaries
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10.5 Search Algorithms Revisited
10.5 Searching Algorithms Revisited
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11.1 Sorting Algorithms
11.1 Sorting Algorithm
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12.4 Hanoi Tower Problem
12.4 Hanota Problem
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16.3 Terminology Table
16.3 Glossary
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<!-- Page content -->
<h1 id="101-binary-search">10.1 &nbsp; Binary Search<a class="headerlink" href="#101-binary-search" title="Permanent link">&para;</a></h1>
<p><u>Binary search</u> is an efficient searching algorithm based on the divide-and-conquer strategy. It leverages the orderliness of data to reduce the search range by half in each round until the target element is found or the search interval becomes empty.</p>
<p><u>Binary search</u> is an efficient search algorithm based on the divide-and-conquer strategy. It leverages the sorted order of the data to reduce the search range by half in each round until the target element is found or the search interval becomes empty.</p>
<div class="admonition question">
<p class="admonition-title">Question</p>
<p>Given an array <code>nums</code> of length <span class="arithmatex">\(n\)</span> with elements arranged in ascending order and no duplicates, search for and return the index of element <code>target</code> in the array. If the array does not contain the element, return <span class="arithmatex">\(-1\)</span>. An example is shown in Figure 10-1.</p>
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<p>If the array does not contain the target element, the search interval will eventually shrink to empty. In this case, return <span class="arithmatex">\(-1\)</span>.</p>
<p>If the array does not contain the target element, the search interval will eventually become empty. In this case, return <span class="arithmatex">\(-1\)</span>.</p>
<div class="tabbed-set tabbed-alternate" data-tabs="1:7"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><input id="__tabbed_1_6" name="__tabbed_1" type="radio" /><input id="__tabbed_1_7" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">&lt;1&gt;</label><label for="__tabbed_1_2">&lt;2&gt;</label><label for="__tabbed_1_3">&lt;3&gt;</label><label for="__tabbed_1_4">&lt;4&gt;</label><label for="__tabbed_1_5">&lt;5&gt;</label><label for="__tabbed_1_6">&lt;6&gt;</label><label for="__tabbed_1_7">&lt;7&gt;</label></div>
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<p align="center"> Figure 10-2 &nbsp; Binary search process </p>
<p>It's worth noting that since both <span class="arithmatex">\(i\)</span> and <span class="arithmatex">\(j\)</span> are of <code>int</code> type, <strong><span class="arithmatex">\(i + j\)</span> may exceed the range of the <code>int</code> type</strong>. To avoid large number overflow, we typically use the formula <span class="arithmatex">\(m = \lfloor {i + (j - i) / 2} \rfloor\)</span> to calculate the midpoint.</p>
<p>It's worth noting that since both <span class="arithmatex">\(i\)</span> and <span class="arithmatex">\(j\)</span> are of <code>int</code> type, <strong><span class="arithmatex">\(i + j\)</span> may exceed the range of the <code>int</code> type</strong>. To avoid integer overflow, we typically use the formula <span class="arithmatex">\(m = \lfloor {i + (j - i) / 2} \rfloor\)</span> to calculate the midpoint.</p>
<p>The code is shown below:</p>
<div class="tabbed-set tabbed-alternate" data-tabs="2:13"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><input id="__tabbed_2_5" name="__tabbed_2" type="radio" /><input id="__tabbed_2_6" name="__tabbed_2" type="radio" /><input id="__tabbed_2_7" name="__tabbed_2" type="radio" /><input id="__tabbed_2_8" name="__tabbed_2" type="radio" /><input id="__tabbed_2_9" name="__tabbed_2" type="radio" /><input id="__tabbed_2_10" name="__tabbed_2" type="radio" /><input id="__tabbed_2_11" name="__tabbed_2" type="radio" /><input id="__tabbed_2_12" name="__tabbed_2" type="radio" /><input id="__tabbed_2_13" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">Python</label><label for="__tabbed_2_2">C++</label><label for="__tabbed_2_3">Java</label><label for="__tabbed_2_4">C#</label><label for="__tabbed_2_5">Go</label><label for="__tabbed_2_6">Swift</label><label for="__tabbed_2_7">JS</label><label for="__tabbed_2_8">TS</label><label for="__tabbed_2_9">Dart</label><label for="__tabbed_2_10">Rust</label><label for="__tabbed_2_11">C</label><label for="__tabbed_2_12">Kotlin</label><label for="__tabbed_2_13">Ruby</label></div>
<div class="tabbed-content">
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</div>
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<p><strong>Time complexity is <span class="arithmatex">\(O(\log n)\)</span></strong>: In the binary loop, the interval is reduced by half each round, so the number of loops is <span class="arithmatex">\(\log_2 n\)</span>.</p>
<p><strong>Time complexity is <span class="arithmatex">\(O(\log n)\)</span></strong>: In the binary search loop, the interval is reduced by half each round, so the number of iterations is <span class="arithmatex">\(\log_2 n\)</span>.</p>
<p><strong>Space complexity is <span class="arithmatex">\(O(1)\)</span></strong>: Pointers <span class="arithmatex">\(i\)</span> and <span class="arithmatex">\(j\)</span> use constant-size space.</p>
<h2 id="1011-interval-representation-methods">10.1.1 &nbsp; Interval Representation Methods<a class="headerlink" href="#1011-interval-representation-methods" title="Permanent link">&para;</a></h2>
<p>In addition to the closed interval mentioned above, another common interval representation is the "left-closed right-open" interval, defined as <span class="arithmatex">\([0, n)\)</span>, meaning the left boundary includes itself while the right boundary does not. Under this representation, the interval <span class="arithmatex">\([i, j)\)</span> is empty when <span class="arithmatex">\(i = j\)</span>.</p>
<p>In addition to the closed interval mentioned above, another common interval representation is the "left-closed right-open" interval, defined as <span class="arithmatex">\([0, n)\)</span>, meaning that the left boundary is inclusive while the right boundary is exclusive. Under this representation, the interval <span class="arithmatex">\([i, j)\)</span> is empty when <span class="arithmatex">\(i = j\)</span>.</p>
<p>We can implement a binary search algorithm with the same functionality based on this representation:</p>
<div class="tabbed-set tabbed-alternate" data-tabs="3:13"><input checked="checked" id="__tabbed_3_1" name="__tabbed_3" type="radio" /><input id="__tabbed_3_2" name="__tabbed_3" type="radio" /><input id="__tabbed_3_3" name="__tabbed_3" type="radio" /><input id="__tabbed_3_4" name="__tabbed_3" type="radio" /><input id="__tabbed_3_5" name="__tabbed_3" type="radio" /><input id="__tabbed_3_6" name="__tabbed_3" type="radio" /><input id="__tabbed_3_7" name="__tabbed_3" type="radio" /><input id="__tabbed_3_8" name="__tabbed_3" type="radio" /><input id="__tabbed_3_9" name="__tabbed_3" type="radio" /><input id="__tabbed_3_10" name="__tabbed_3" type="radio" /><input id="__tabbed_3_11" name="__tabbed_3" type="radio" /><input id="__tabbed_3_12" name="__tabbed_3" type="radio" /><input id="__tabbed_3_13" name="__tabbed_3" type="radio" /><div class="tabbed-labels"><label for="__tabbed_3_1">Python</label><label for="__tabbed_3_2">C++</label><label for="__tabbed_3_3">Java</label><label for="__tabbed_3_4">C#</label><label for="__tabbed_3_5">Go</label><label for="__tabbed_3_6">Swift</label><label for="__tabbed_3_7">JS</label><label for="__tabbed_3_8">TS</label><label for="__tabbed_3_9">Dart</label><label for="__tabbed_3_10">Rust</label><label for="__tabbed_3_11">C</label><label for="__tabbed_3_12">Kotlin</label><label for="__tabbed_3_13">Ruby</label></div>
<div class="tabbed-content">
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<p align="center"> Figure 10-3 &nbsp; Two interval definitions </p>
<h2 id="1012-advantages-and-limitations">10.1.2 &nbsp; Advantages and Limitations<a class="headerlink" href="#1012-advantages-and-limitations" title="Permanent link">&para;</a></h2>
<p>Binary search performs well in both time and space aspects.</p>
<p>Binary search offers good performance in both time and space.</p>
<ul>
<li>Binary search has high time efficiency. With large data volumes, the logarithmic time complexity has significant advantages. For example, when the data size <span class="arithmatex">\(n = 2^{20}\)</span>, linear search requires <span class="arithmatex">\(2^{20} = 1048576\)</span> loop rounds, while binary search only needs <span class="arithmatex">\(\log_2 2^{20} = 20\)</span> rounds.</li>
<li>Binary search has high time efficiency. With large data volumes, the logarithmic time complexity has significant advantages. For example, when the data size <span class="arithmatex">\(n = 2^{20}\)</span>, linear search requires <span class="arithmatex">\(2^{20} = 1048576\)</span> iterations, while binary search only needs <span class="arithmatex">\(\log_2 2^{20} = 20\)</span> iterations.</li>
<li>Binary search requires no extra space. Compared to searching algorithms that require additional space (such as hash-based search), binary search is more space-efficient.</li>
</ul>
<p>However, binary search is not suitable for all situations, mainly for the following reasons:</p>
<ul>
<li>Binary search is only applicable to sorted data. If the input data is unsorted, sorting specifically to use binary search would be counterproductive, as sorting algorithms typically have a time complexity of <span class="arithmatex">\(O(n \log n)\)</span>, which is higher than both linear search and binary search. For scenarios with frequent element insertions, maintaining array orderliness requires inserting elements at specific positions with a time complexity of <span class="arithmatex">\(O(n)\)</span>, which is also very expensive.</li>
<li>Binary search is only applicable to arrays. Binary search requires jump-style (non-contiguous) element access, and jump-style access has low efficiency in linked lists, making it unsuitable for linked lists or data structures based on linked list implementations.</li>
<li>Binary search is only applicable to sorted data. If the input data is unsorted, sorting specifically to use binary search would be counterproductive, as sorting algorithms typically have a time complexity of <span class="arithmatex">\(O(n \log n)\)</span>, which is higher than both linear search and binary search. For scenarios with frequent element insertions, keeping the array sorted requires inserting elements at specific positions with a time complexity of <span class="arithmatex">\(O(n)\)</span>, which is also very expensive.</li>
<li>Binary search is only applicable to arrays. Binary search requires non-contiguous, jump-style access to elements, and this kind of access is inefficient in linked lists, making it unsuitable for linked lists or linked-list-based data structures.</li>
<li>For small data volumes, linear search performs better. In linear search, each round requires only 1 comparison operation; while in binary search, it requires 1 addition, 1 division, 1-3 comparison operations, and 1 addition (subtraction), totaling 4-6 unit operations. Therefore, when the data volume <span class="arithmatex">\(n\)</span> is small, linear search is actually faster than binary search.</li>
</ul>
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10.2 Binary Search Insertion
10.2 Binary Search Insertion Point
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