Merge branch 'smv1999:master' into master

Author: Shantanuu-eng

Arrays/leap.java

@@ -0,0 +1,58 @@
+/* Title : Check whether the year entered is leap or not .
+           If the entered year passes the conditions then it is considered to be as leap
+           otherwise not .
+
+*/
+import java.util.Scanner ;
+class   leap
+{    int leap_year;
+
+    leap (int a)   // Constructor
+    {
+        leap_year=a;
+    }
+    void check ()
+    {
+        if(leap_year%4==0)
+
+        {
+            if(leap_year%100==0)
+            {
+                if(leap_year%400==0)
+                    {   System.out.println( " Leap year .");
+                        System.out.println( " \n 366 days in " +leap_year+ " year . \n 29 days in month of Feb . \n"); }
+                else
+                    {
+                        System.out.println( " Not Leap year "); }
+            }
+            else
+            {   System.out.println( " Leap year .");
+                System.out.println( " \n 366 days in " + leap_year+ " year . \n 29 days in month of Feb . \n");}
+        }
+
+        else
+         {   System.out.println( " Not Leap year ");  }
+
+    }
+
+    public static void main (String[] args)
+    {
+        int year ;
+        Scanner in=new Scanner(System.in);
+        year=in.nextInt();
+        leap Y = new leap(year);
+        Y.check();
+
+    }
+}
+/*
+ 1900
+ Not Leap year .
+
+ 2000
+ Leap year .
+
+ 366 days in year 2000 .
+ 29 days in month of Feb .
+
+ */

Codechef Problems/MaximumArrayXOR.java

@@ -0,0 +1,77 @@
+/*
+The game of billiards involves two players knocking 3 balls around on a green baize table. Well, there is more to it, but for our purposes this is sufficient.
+
+The game consists of several rounds and in each round both players obtain a score, based on how well they played. Once all the rounds have been played, the total score of each player is determined by adding up the scores in all the rounds and the player with the higher total score is declared the winner.
+
+The Siruseri Sports Club organises an annual billiards game where the top two players of Siruseri play against each other. The Manager of Siruseri Sports Club decided to add his own twist to the game by changing the rules for determining the winner. In his version, at the end of each round, the cumulative score for each player is calculated, and the leader and her current lead are found. Once all the rounds are over the player who had the maximum lead at the end of any round in the game is declared the winner.
+
+Consider the following score sheet for a game with 5 rounds:
+
+Round	Player 1	Player 2
+1	      140	      82
+2	      89	      134
+3	      90	      110
+4	      112	      106
+5	      88	      90
+The total scores of both players, the leader and the lead after each round for this game is given below:
+
+Round	Player 1	Player 2	Leader	Lead
+1	      140	      82	   Player 1	 58
+2	      229	      216	   Player 1	 13
+3	      319	      326	   Player 2	 7
+4	      431	      432	   Player 2	 1
+5	      519	      522	   Player 2	 3
+Note that the above table contains the cumulative scores.
+
+The winner of this game is Player 1 as he had the maximum lead (58 at the end of round 1) during the game.
+
+Your task is to help the Manager find the winner and the winning lead. You may assume that the scores will be such that there will always be a single winner. That is, there are no ties.
+
+Input
+The first line of the input will contain a single integer N (N ≤ 10000) indicating the number of rounds in the game. Lines 2,3,...,N+1 describe the scores of the two players in the N rounds. Line i+1 contains two integer Si and Ti, the scores of the Player 1 and 2 respectively, in round i. You may assume that 1 ≤ Si ≤ 1000 and 1 ≤ Ti ≤ 1000.
+Output
+Your output must consist of a single line containing two integers W and L, where W is 1 or 2 and indicates the winner and L is the maximum lead attained by the winner.
+Example
+Input:
+5
+140 82
+89 134
+90 110
+112 106
+88 90
+Output:
+1 58
+*/
+#include <bits/stdc++.h>
+using namespace std;
+int main()
+{
+    int n, lead=INT_MIN,p1=0,p2=0,winner=0;
+    cin >> n;
+    for (int i = 0; i < n; i++)
+    {
+       int a,b;
+       cin>>a>>b;
+        //Adding the scores of player1 and player2 so that we can store there cumulative score sum in p1 and p2        
+        p1+=a;
+        p2+=b;
+        //If the cumulative sum of p1 is greater than p2
+        if(p1>p2){
+            //and their difference is greater than lead than the winner will be player 1
+            if(p1-p2>lead){
+                lead=p1-p2;
+                winner=1;
+            }
+        }
+        
+        else{
+            //if their difference is greater than lead than the winner will be player 2
+            if(p2-p1>lead){
+                lead=p2-p1;
+                winner=2;
+            }
+        }
+    }
+    cout<<winner<<" "<<lead;
+    return 0;
+}

Codechef Problems/TheLeadGame.cpp

@@ -0,0 +1,123 @@
+/*
+Implementation of Bridge edge in Graph problem
+
+Problem statement: 
+Given an undirected graph of V vertices and E edges and another edge (c-d), the task is to 
+find if the given edge is a bridge in graph, i.e., removing the edge disconnects the graph.
+
+Link to the problem: https://practice.geeksforgeeks.org/problems/bridge-edge-in-graph/1
+
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+
+
+class Solution
+{
+	public:
+
+    void dfs( vector<int> adj[],vector<int>&v,int i)
+    {
+        //marking visited
+        v[i]=1; 
+
+        //traversing adjacent node
+        for(auto x:adj[i]) 
+        {
+            if(!v[x])
+                dfs(adj,v,x);
+
+        }
+
+    }
+
+    bool connected(vector<int>adj[],int n,int c,int d)
+    {
+        vector<int>v(n,0);
+
+        // dfs from edge c
+        dfs(adj,v,c); 
+
+        //if edge d is not visited means not connected
+        if(v[d]==0) 
+            return false;
+
+        return true;
+
+    }
+
+    //Function to find if the given edge is a bridge in graph.
+    int isBridge(int n, vector<int>adj[], int c, int d) 
+    {
+        //if graph is not connected
+        if(!connected(adj,n,c,d)) 
+            return 0;
+
+        else
+        {
+            //removing edge c and d
+            adj[c].erase(remove(adj[c].begin(), adj[c].end(), d), adj[c].end());
+            adj[d].erase(remove(adj[d].begin(), adj[d].end(), c), adj[d].end());
+
+            //if connected means no bridge
+            if(connected(adj,n,c,d)) 
+                return 0;
+            else
+                return 1;
+        }
+
+    }
+};
+
+// Driver Code Starts
+int main()
+{
+    int t;
+    cin >> t;
+    while (t--) 
+    {
+        int V, E;
+        cin >> V >> E;
+        vector<int> adj[V];
+        int i=0;
+        while (i++<E) 
+        {
+            int u, v;
+            cin >> u >> v;
+            adj[u].push_back (v);
+            adj[v].push_back (u);
+        }
+
+        int c,d;
+        cin>>c>>d;
+
+        Solution obj;
+    	cout << obj.isBridge(V, adj, c, d) << "\n";
+    }
+
+    return 0;
+}
+
+
+/*
+Time Complexity : O(V+E)
+Space Complexity : O(V)
+
+Input:
+t=1
+V=4 E=3
+0 1
+1 2
+2 3
+c=1  d=2
+
+Output: 1
+
+Explanation:
+From the graph, we can clearly see that
+removing the edge 1-2 will result in 
+disconnection of the graph. So, it is 
+a bridge Edge and thus the Output 1.
+
+*/