Create Meximum Array Explanation.txt

Author: MathProgrammer

2022/Contests/Div 2/767/Explanations/Meximum Array Explanation.txt

@@ -0,0 +1,74 @@
+We need the lexicographically maximum sequence. 
+This means that we must be choosing the largest element whenever possible. 
+
+We will choose the Mex of the entire array at every given point in time. 
+If there are multiple MEX, we will look at the 
+
+-----
+
+void solve()
+{
+    int no_of_elements;
+    cin >> no_of_elements;
+
+    vector <int> A(no_of_elements + 1);
+    for(int i = 1; i <= no_of_elements; i++)
+    {
+        cin >> A[i];
+    }
+
+    vector <int> frequency(no_of_elements + 1, 0);
+    for(int i = 1; i <= no_of_elements; i++)
+    {
+        frequency[A[i]]++;
+    }
+
+    int array_mex = 0;
+    for(int i = 0; i <= no_of_elements; i++)
+    {
+        if(frequency[i] == 0)
+        {
+            array_mex = i;
+            break;
+        }
+    }
+
+    set <int> all_elements, unseen;
+    for(int i = 0; i <= array_mex; i++)
+    {
+        all_elements.insert(i);
+    }
+    unseen = all_elements;
+
+    vector <int> answer;
+    int mex_after_here = array_mex;
+    for(int i = 1; i <= no_of_elements; i++)
+    {
+        frequency[A[i]]--;
+
+        if(unseen.find(A[i]) != unseen.end())
+        {
+            unseen.erase(A[i]);
+        }
+
+        if(frequency[A[i]] == 0)
+        {
+            mex_after_here = min(mex_after_here, A[i]);
+        }
+
+        if(*(unseen.begin()) == array_mex)
+        {
+            answer.push_back(array_mex);
+            array_mex = mex_after_here;
+
+            unseen = all_elements;
+        }
+    }
+
+    cout << answer.size() << "\n";
+    for(int i = 0; i < answer.size(); i++)
+    {
+        cout << answer[i] << " ";
+    }
+    cout << "\n";
+}