Create Perfect Keyboard Explanation.txt

Author: MathProgrammer

2020/Educational Rounds/Educational Round 82/Explanations/Perfect Keyboard Explanation.txt

@@ -0,0 +1,143 @@
+1. Draw a graph 
+The vertices are the alphabets and the edges are drawn between consecutive characters. 
+We have to be careful about not adding the same edge twice. This is very important.
+
+    for(int i = 1; i < password.size(); i++)
+    {
+        if(adjacency[password[i] - 'a'][password[i - 1] - 'a'] == false)
+        {
+            adjacency[password[i] - 'a'][password[i - 1] - 'a'] = true;
+            adjacency[password[i - 1] - 'a'][password[i] - 'a'] = true;
+                                         
+            graph[password[i] - 'a'].push_back(password[i - 1] - 'a');
+            graph[password[i - 1] - 'a'].push_back(password[i] - 'a');
+            
+            degree[password[i] - 'a']++;
+            degree[password[i - 1] - 'a']++;
+        }
+        
+    }
+
+----
+
+2. If any vertex has degree > 2, it is not possible as any key can only have 2 neighbours
+
+    for(int alpha = 0; alpha < NO_OF_ALPHABETS; alpha++)
+    {
+        if(degree[alpha] > 2)
+        {
+            cout << "NO\n";
+            return;
+        }
+    }
+
+-----
+
+3. Now, we must look at each connected component of the graph seperately. 
+We must find the component number of each vertex. We can do this by performing a series of DFS 
+
+    int component = 1;
+    
+    for(int alpha = 0; alpha < NO_OF_ALPHABETS; alpha++)
+    {
+        if(component_no[alpha] == 0)
+        {
+            dfs(alpha, -1, component++);
+        }
+    }
+    
+-----
+
+Here is the DFS itself 
+
+void dfs(int v, int parent_v, int c)
+{
+    //cout << "V = " << v << " C = " << c << "\n";
+    component_no[v] = c;
+    
+    for(int i = 0; i < graph[v].size(); i++)
+    {
+        int child_v = graph[v][i];
+        
+        if(child_v == parent_v || component_no[child_v] == c)
+        {
+            continue;
+        }
+        
+        dfs(child_v, v, c);
+    }
+}
+
+-----
+
+4. Every component will be printed together. 
+So, first we will print characters of component 1, then characters of component 2, and so on. 
+
+-----
+
+5. How to print characters of a given component ? 
+There have to be 2 vertices in this component of degree 1 and every other vertex must have degree 2. 
+We will start from one of the vertices of degree 1 and keep going till we reach the vertex of degree 2
+
+If the component is of size 1, then we can just print it
+Otherwise, we will look for the first border and then DFS till we reach the second border
+
+for(int c = 1; c < component; c++)
+    {
+        vector <int> this_component;
+        for(int alpha = 0; alpha < NO_OF_ALPHABETS; alpha++)
+        {
+            if(component_no[alpha] == c)
+            {
+                this_component.push_back(alpha);
+            }
+        }
+        
+        if(this_component.size() == 1)
+        {
+            keyboard += (char)('a' + this_component[0]);
+            used[this_component[0]] = true;
+            continue;
+        }
+        
+        int border_1 = -1, border_2 = -1;
+        for(int i = 0; i < this_component.size(); i++)
+        {
+            if(degree[this_component[i]] == 1)
+            {
+                if(border_1 == -1)
+                    border_1 = this_component[i];
+                else
+                    border_2 = this_component[i];
+            }
+        }
+        
+        if(border_1 == -1 || border_2 == -1)
+        {
+            cout << "NO\n";
+            
+            return;
+        }
+        
+        keyboard += (char)('a' + border_1);
+        used[border_1] = true;
+        
+        while(keyboard.back() - 'a' != border_2)
+        {
+            for(int i = 0; i < graph[keyboard.back() - 'a'].size(); i++)
+            {
+                if(!used[graph[keyboard.back() - 'a'][i]])
+                {
+                    int next = graph[keyboard.back() - 'a'][i];
+                    
+                    keyboard += (char)('a' + next);
+                    
+                    used[next] = true;
+                    
+                    break;
+                }
+            }
+        }
+        
+    }
+