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Author: MathProgrammer

2020/Div 3/617 Div 3/Explanations/String Colouring (Easy Version) Explanation.txt

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+Let us make an observation
+
+For every character S[j] that comes after S[i],
+If S[j] > S[i], S[j] will have to meet S[i] at some point.
+
+We need to draw an edge between all pairs (i, j) such that
+(i < j) and (S[i] > S[j]) and then check for bipartite matching.
+
+We can do this in O(n^2) time.
+We will colour a point red if it is uncoloured and then
+colour all it's neighbours blue.
+
+If any of it's neighbours is forced to be red, then it is not possible.
+
+-----
+
+int main()
+{
+    int length;
+    string S;
+
+    cin >> length >> S;
+
+    vector <int> colour(length, UNCOLOURED);
+    for(int i = 0; i < length; i++)
+    {
+        if(colour[i] == UNCOLOURED)
+        {
+            colour[i] = RED;
+        }
+
+        for(int j = i + 1; j < length; j++)
+        {
+            if(S[j] < S[i] && colour[j] == UNCOLOURED)
+            {
+                colour[j] = other(colour[i]);
+
+                continue;
+            }
+
+            if(S[j] < S[i] && colour[i] == colour[j])
+            {
+                cout << "NO\n";
+
+                return 0;
+            }
+        }
+    }
+
+    cout << "YES\n";
+    for(int i = 0; i < length; i++)
+    {
+        cout << colour[i];
+    }
+    cout << "\n";
+
+    return 0;
+}

2020/Div 3/617 Div 3/Explanations/String Colouring (Hard Version) Explanation.txt

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+There is a theorem called Dilworth's Theorem. It can be applied here.
+
+We have to count the minimum number of non-decreasing sequences in this array.
+
+This is equal to the length of the longest non-increasing sequence in the array.
+
+We will calculate the length of the longest non-increasing sequence.
+
+As the size of the alphabet is small, we can do it in O(26N) time.
+
+-----
+
+int main()
+{
+    int length;
+    string S;
+    cin >> length >> S;
+
+    const int NO_OF_ALPHABETS = 26;
+    vector <int> max_till(NO_OF_ALPHABETS);
+    vector <int> sequence_no(length, 1);
+
+    int no_of_sequences = 0;
+    for(int i = 0; i < length; i++)
+    {
+        for(int alpha = S[i] - 'a' + 1; alpha < NO_OF_ALPHABETS; alpha++)
+        {
+            sequence_no[i] = max(sequence_no[i], max_till[alpha] + 1);
+        }
+
+        max_till[S[i] - 'a'] = max(max_till[S[i] - 'a'], sequence_no[i]);
+
+        no_of_sequences = max(no_of_sequences, sequence_no[i]);
+    }
+
+    cout << no_of_sequences << "\n";
+
+    for(int i = 0; i < length; i++)
+    {
+        cout << sequence_no[i] << " ";
+    }
+
+    cout << "\n";
+
+    return 0;
+}