Question: 8. What is the time complexity of the following dynamic programming implementation of the minimum number of insertions to form a palindrome problem?<div class="hk1_style-wrap4"><div class="hk1_style-wrap3"><div class="hk1_style-wrap2"><div class="hk1_style-wrap"><div class="hk1_style"><div class="c"><pre class="de1"><span class="co2">#include&lt;stdio.h&gt;</span> <span class="co2">#include&lt;string.h&gt;</span> <span class="kw4">int</span> max<span class="br0">(</span><span class="kw4">int</span> a<span class="sy0">,</span> <span class="kw4">int</span> b<span class="br0">)</span> <span class="br0">{</span> <span class="kw1">if</span><span class="br0">(</span>a <span class="sy0">&gt;</span> b<span class="br0">)</span> <span class="kw1">return</span> a<span class="sy0">;</span> <span class="kw1">return</span> b<span class="sy0">;</span> <span class="br0">}</span> <span class="kw4">int</span> min_ins<span class="br0">(</span><span class="kw4">char</span> <span class="sy0">*</span>s<span class="br0">)</span> <span class="br0">{</span> <span class="kw4">int</span> len <span class="sy0">=</span> <span class="kw3">strlen</span><span class="br0">(</span>s<span class="br0">)</span><span class="sy0">,</span> i<span class="sy0">,</span> j<span class="sy0">;</span> <span class="kw4">int</span> arr<span class="br0">[</span>len <span class="sy0">+</span> <span class="nu0">1</span><span class="br0">]</span><span class="br0">[</span>len <span class="sy0">+</span> <span class="nu0">1</span><span class="br0">]</span><span class="sy0">;</span> <span class="kw4">char</span> rev<span class="br0">[</span>len <span class="sy0">+</span> <span class="nu0">1</span><span class="br0">]</span><span class="sy0">;</span> <span class="kw3">strcpy</span><span class="br0">(</span>rev<span class="sy0">,</span> s<span class="br0">)</span><span class="sy0">;</span> strrev<span class="br0">(</span>rev<span class="br0">)</span><span class="sy0">;</span> <span class="kw1">for</span><span class="br0">(</span>i <span class="sy0">=</span> <span class="nu0">0</span><span class="sy0">;</span>i <span class="sy0">&lt;=</span> len<span class="sy0">;</span> i<span class="sy0">++</span><span class="br0">)</span> arr<span class="br0">[</span>i<span class="br0">]</span><span class="br0">[</span><span class="nu0">0</span><span class="br0">]</span> <span class="sy0">=</span> <span class="nu0">0</span><span class="sy0">;</span> <span class="kw1">for</span><span class="br0">(</span>i <span class="sy0">=</span> <span class="nu0">0</span><span class="sy0">;</span> i <span class="sy0">&lt;=</span> len<span class="sy0">;</span> i<span class="sy0">++</span><span class="br0">)</span> arr<span class="br0">[</span><span class="nu0">0</span><span class="br0">]</span><span class="br0">[</span>i<span class="br0">]</span> <span class="sy0">=</span> <span class="nu0">0</span><span class="sy0">;</span> <span class="kw1">for</span><span class="br0">(</span>i <span class="sy0">=</span> <span class="nu0">1</span><span class="sy0">;</span> i <span class="sy0">&lt;=</span> len<span class="sy0">;</span> i<span class="sy0">++</span><span class="br0">)</span> <span class="br0">{</span> <span class="kw1">for</span><span class="br0">(</span>j <span class="sy0">=</span> <span class="nu0">1</span><span class="sy0">;</span> j <span class="sy0">&lt;=</span> len<span class="sy0">;</span> j<span class="sy0">++</span><span class="br0">)</span> <span class="br0">{</span> <span class="kw1">if</span><span class="br0">(</span>s<span class="br0">[</span>i <span class="sy0">-</span> <span class="nu0">1</span><span class="br0">]</span> <span class="sy0">==</span> rev<span class="br0">[</span>j <span class="sy0">-</span> <span class="nu0">1</span><span class="br0">]</span><span class="br0">)</span> arr<span class="br0">[</span>i<span class="br0">]</span><span class="br0">[</span>j<span class="br0">]</span> <span class="sy0">=</span> arr<span class="br0">[</span>i <span class="sy0">-</span> <span class="nu0">1</span><span class="br0">]</span><span class="br0">[</span>j <span class="sy0">-</span> <span class="nu0">1</span><span class="br0">]</span> <span class="sy0">+</span> <span class="nu0">1</span><span class="sy0">;</span> <span class="kw1">else</span> arr<span class="br0">[</span>i<span class="br0">]</span><span class="br0">[</span>j<span class="br0">]</span> <span class="sy0">=</span> max<span class="br0">(</span>arr<span class="br0">[</span>i <span class="sy0">-</span> <span class="nu0">1</span><span class="br0">]</span><span class="br0">[</span>j<span class="br0">]</span><span class="sy0">,</span> arr<span class="br0">[</span>i<span class="br0">]</span><span class="br0">[</span>j <span class="sy0">-</span> <span class="nu0">1</span><span class="br0">]</span><span class="br0">)</span><span class="sy0">;</span> <span class="br0">}</span> <span class="br0">}</span> <span class="kw1">return</span> len <span class="sy0">-</span> arr<span class="br0">[</span>len<span class="br0">]</span><span class="br0">[</span>len<span class="br0">]</span><span class="sy0">;</span> <span class="br0">}</span> <span class="kw4">int</span> main<span class="br0">(</span><span class="br0">)</span> <span class="br0">{</span> <span class="kw4">char</span> s<span class="br0">[</span><span class="br0">]</span> <span class="sy0">=</span> <span class="st0">"abcda"</span><span class="sy0">;</span> <span class="kw4">int</span> ans <span class="sy0">=</span> min_ins<span class="br0">(</span>s<span class="br0">)</span><span class="sy0">;</span> <span class="kw3">printf</span><span class="br0">(</span><span class="st0">"%d"</span><span class="sy0">,</span>ans<span class="br0">)</span><span class="sy0">;</span> <span class="kw1">return</span> <span class="nu0">0</span><span class="sy0">;</span> <span class="br0">}</span></pre></div></div></div></div></div></div>
Options:
O(1)
O(n)
O(n2)
O(mn)
Answer: D