feat(leetcode-daily): today

posts/algorithms/leetcode-daily.html

@@ -35,6 +35,41 @@
           <h1 class="post-title">Leetcode Daily</h1>
         </header>
         <article class="post-article">
+          <h2>
+            <a
+              target="blank"
+              href="https://leetcode.com/problems/minimum-number-of-operations-to-make-elements-in-array-distinc"
+              >minimum number of operations to make array distinct</a
+            >
+            &mdash; 9/13/24
+          </h2>
+          <div class="problem-content">
+            <h3>understanding the problem</h3>
+            <p>
+              You can remove elements in groups of 3 <i>solely</i> from the
+              beginning of the array. Perform this operation until there are no
+              more duplicates left, returning the number of times you had to
+              perform the operation.
+            </p>
+            <h3>solution: rephrase the question</h3>
+            <p>
+              Definitionally, you remove the <i>last</i> duplicate. If such
+              duplicate is at 0-indexed <code>i</code>, it belongs to the \(\lceil
+              \frac{i + 1}{3}\rceil\)th chunk of 3 (i.e. operation). Find the last
+              duplicate by leveraging a frequency map and iterating backwards
+              through the input.
+            </p>
+            <div class="code" data-file="mnootmad.cpp"></div>
+            <h3>asymptotic complexity</h3>
+            <p>
+              The solution is optimal, considering the least amount of elements
+              possible in:
+            </p>
+            <ul>
+              <li><u>Time Complexity</u>: \(\O(n)\)</li>
+              <li><u>Space Complexity</u>: \(\Theta(1)\)</li>
+            </ul>
+          </div>
           <h2>
             <a
               target="blank"

public/code/algorithms/leetcode-daily/mnootmad.cpp

@@ -0,0 +1,13 @@
+class Solution {
+public:
+    int minimumOperations(vector<int>& nums) {
+        vector<int> freq(101, 0);
+        int i;
+        for (i = nums.size() - 1; i >= 0; --i) {
+            if (++freq[nums[i]] == 2) {
+                return ceil((i + 1) / 3.0);
+            }
+        }
+        return 0;
+    }
+};