Check whether the structure of given Graph is same as the game of Rock-Paper-Scissor
Given an array arr[] of N pairs of strings representing a directed graph, where each pair of the strings represents an edge between the nodes represented by the pair of strings, the task is to check whether the structure of the given directed graph is the same as that of the game of Rock-Paper-Scissor. If found true, then print “Yes“. Otherwise, print “No“.
Note:
In the real game of Rock-Paper-Scissor, the rock dominates over the scissor, scissor dominates over the paper and paper dominates over the rock.The game can be modelled using a directed graph as below, in which the directed edge (a, b) shows that a dominates over b.
Examples:
Input: arr[] = {{“Snake”, “Water”}, {“Water”, “Gun”}, {“Gun”, “Snake”}}
Output: Yes
Explanation:Therefore, from the above image it can be seen that the given graph structure is similar to Rock-Paper-Scissor.
Input: arr[]={{“Snake”, “Water”}, {“Water”, “Gun”}, {“Gun”, “Gun”}}
Output: No
Approach: The problem can be solved by using the indegree and outdegree of the graphs based on the observation that in the directed graph Rock-Paper-Scissor there are only three nodes and indegree and outdegree of each node is one. Follow the steps below to solve the problem.
- Initialize two maps of {strings, int} say inDegree and outDegree, to store the indegree and outdegree of each node in these maps.
- Initialize a set of strings say st and insert all the nodes i.e. strings in this set.
- Check if the size of any map is not equal to 3 or the size of the set st of nodes is not equal to 3, then print “No“.
- Iterate over both the maps and check if the second value of any entry map is not 1, then, print “No” and break.
- Finally, if none of the above cases satisfy, then print “Yes“.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if the given directed // graph has same structure to Rock-Paper-Scissor string similarGame(vector<vector<string> > arr) { // Stores indegree of each node unordered_map<string, int > indegree; // Stores outdegree of each node unordered_map<string, int > outdegree; // Stores all nodes set<string> st; // Traverse through the edges for ( int i = 0; i < arr.size(); i++) { // Check presence of self loop if (arr[i][0] == arr[i][1]) return "No" ; // Insert the node arr[i][0] and arr[i][1] // into the set st.insert(arr[i][0]); st.insert(arr[i][1]); // Increment the outdegree and indegree of // nodes arr[i][0] and arr[i][1] outdegree[arr[i][0]]++; indegree[arr[i][1]]++; } // Check base conditions if (outdegree.size() != 3 || indegree.size() != 3 || st.size() != 3) return "No" ; // Traverse the array outdegree for ( auto it : outdegree) { if (it.second != 1) return "No" ; } // Traverse the array indegree for ( auto it : indegree) { if (it.second != 1) return "No" ; } // Return return "Yes" ; } // Driver Code int main() { // Given Input vector<vector<string> > arr = { { "Snake" , "Water" }, { "Water" , "Gun" }, { "Gun" , "Snake" } }; // Function Call cout << similarGame(arr); return 0; } |
Python3
# Python 3 program for the above approach # Function to check if the given directed # graph has same structure to Rock-Paper-Scissor def similarGame(arr): # Stores indegree of each node indegree = {} # Stores outdegree of each node outdegree = {} # Stores all nodes st = set () # Traverse through the edges for i in range ( len (arr)): # Check presence of self loop if (arr[i][ 0 ] = = arr[i][ 1 ]): return "No" # Insert the node arr[i][0] and arr[i][1] # into the set st.add(arr[i][ 0 ]) st.add(arr[i][ 1 ]) # Increment the outdegree and indegree of # nodes arr[i][0] and arr[i][1] if arr[i][ 0 ] in outdegree: outdegree[arr[i][ 0 ]] + = 1 else : outdegree[arr[i][ 0 ]] = 0 if arr[i][ 1 ] in indegree: indegree[arr[i][ 1 ]] + = 1 else : indegree[arr[i][ 1 ]] = 0 # Check base conditions if ( len (outdegree) ! = 3 and len (indegree) ! = 3 and len (st) ! = 3 ): return "No" ; # Traverse the array outdegree for key,value in outdegree.items(): if (value = = 1 ): return "No" # Traverse the array indegree for key,value in indegree.items(): if (value = = 1 ): return "No" # Return return "Yes" # Driver Code if __name__ = = '__main__' : # Given Input arr = [[ "Snake" , "Water" ],[ "Water" , "Gun" ],[ "Gun" , "Snake" ]] # Function Call print (similarGame(arr)) # This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script> // JavaScript program for the above approach // Function to check if the given directed // graph has same structure to Rock-Paper-Scissor function similarGame(arr) { // Stores indegree of each node let indegree = new Map(); // Stores outdegree of each node let outdegree = new Map(); // Stores all nodes let st = new Set(); // Traverse through the edges for (let i = 0; i < arr.length; i++) { // Check presence of self loop if (arr[i][0] == arr[i][1]) return "No" ; // add the node arr[i][0] and arr[i][1] // into the set st.add(arr[i][0]); st.add(arr[i][1]); // Increment the outdegree and indegree of // nodes arr[i][0] and arr[i][1] if (outdegree.has(arr[i][0]) == true )outdegree.set(arr[i][0],outdegree.get(arr[i][0])+1); else outdegree.set(arr[i][0],1); if (indegree.has(arr[i][1]) == true )indegree.set(arr[i][1],indegree.get(arr[i][1])+1); else indegree.set(arr[i][1],1); } // Check base conditions if (outdegree.size != 3 || indegree.size != 3 || st.size != 3) return "No" ; // Traverse the array outdegree for (let [key,value] in outdegree) { if (value != 1) return "No" ; } // Traverse the array indegree for (let [key,value] in indegree) { if (value != 1) return "No" ; } // Return return "Yes" ; } // Driver Code // Given Input let arr = [ [ "Snake" , "Water" ], [ "Water" , "Gun" ], [ "Gun" , "Snake" ] ]; // Function Call document.write(similarGame(arr), "</br>" ); // This code is contributed by shinjanpatra. </script> |
Java
/*package whatever //do not write package name here */ import java.io.*; import java.util.*; class GFG { // Java program for the above approach // Function to check if the given directed // graph has same structure to Rock-Paper-Scissor static String similarGame(String[][] arr) { // Stores indegree of each node HashMap<String,Integer> indegree = new HashMap<>(); // Stores outdegree of each node HashMap<String,Integer> outdegree = new HashMap<>(); // Stores all nodes TreeSet<String> st = new TreeSet<>(); // Traverse through the edges for ( int i = 0 ; i < arr.length; i++) { // Check presence of self loop if (arr[i][ 0 ] == arr[i][ 1 ]) return "No" ; // Insert the node arr[i][0] and arr[i][1] // into the set st.add(arr[i][ 0 ]); st.add(arr[i][ 1 ]); // Increment the outdegree and indegree of // nodes arr[i][0] and arr[i][1] if (outdegree.containsKey(arr[i][ 0 ])){ outdegree.put(arr[i][ 0 ],outdegree.get(arr[i][ 0 ])+ 1 ); } else outdegree.put(arr[i][ 0 ], 1 ); if (indegree.containsKey(arr[i][ 0 ])){ indegree.put(arr[i][ 0 ],indegree.get(arr[i][ 0 ])+ 1 ); } else indegree.put(arr[i][ 0 ], 1 ); } // Check base conditions if (outdegree.size() != 3 || indegree.size() != 3 || st.size() != 3 ) return "No" ; // Traverse the array outdegree for (Map.Entry it : outdegree.entrySet()) { if (( int )it.getValue() != 1 ) return "No" ; } // Traverse the array indegree for (Map.Entry it : indegree.entrySet()) { if (( int )it.getValue() != 1 ) return "No" ; } // Return return "Yes" ; } /* Driver program to test above function*/ public static void main(String args[]) { // Given Input String[][] arr = { { "Snake" , "Water" },{ "Water" , "Gun" },{ "Gun" , "Snake" } }; // Function Call System.out.println(similarGame(arr)); } } // This Solution is contributed by shinjanpatra. |
C#
// C# program for the equivalent code using System; using System.Collections.Generic; class GFG { // Function to check if the given directed // graph has same structure to Rock-Paper-Scissor static string SimilarGame( string [, ] arr) { // Stores indegree of each node Dictionary< string , int > indegree = new Dictionary< string , int >(); // Stores outdegree of each node Dictionary< string , int > outdegree = new Dictionary< string , int >(); // Stores all nodes SortedSet< string > st = new SortedSet< string >(); // Traverse through the edges for ( int i = 0; i < arr.GetLength(0); i++) { // Check presence of self loop if (arr[i, 0] == arr[i, 1]) return "No" ; // Insert the node arr[i][0] and arr[i][1] // into the set st.Add(arr[i, 0]); st.Add(arr[i, 1]); // Increment the outdegree and indegree of // nodes arr[i][0] and arr[i][1] if (outdegree.ContainsKey(arr[i, 0])) { outdegree[arr[i, 0]]++; } else outdegree.Add(arr[i, 0], 1); if (indegree.ContainsKey(arr[i, 0])) { indegree[arr[i, 0]]++; } else indegree.Add(arr[i, 0], 1); } // Check base conditions if (outdegree.Count != 3 || indegree.Count != 3 || st.Count != 3) return "No" ; // Traverse the array outdegree foreach (KeyValuePair< string , int > it in outdegree) { if (it.Value != 1) return "No" ; } // Traverse the array indegree foreach (KeyValuePair< string , int > it in indegree) { if (it.Value != 1) return "No" ; } // Return return "Yes" ; } // Driver program to test above function public static void Main( string [] args) { // Given Input string [, ] arr = { { "Snake" , "Water" }, { "Water" , "Gun" }, { "Gun" , "Snake" } }; // Function Call Console.WriteLine(SimilarGame(arr)); } } |
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)