Merge branch 'master' of https://github.com/smv1999/CompetitiveProgrammingQuestionBank into Devincept15

Author: Karnika06

Arrays/leap.java

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+/* 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/TheLeadGame.cpp

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+/*
+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;
+}

Data Structures/Linked Lists/Singly Linked List/merge_two_sorted_linkedlist.py

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+# Class for creating nodes in linkedlist
+class SinglyLinkedListNode:
+    def __init__(self, node_data):
+        self.data = node_data
+        self.next = None
+
+#Class for building linkedlist
+class SinglyLinkedList:
+    def __init__(self):
+        self.head = None
+        self.tail = None
+    
+    #Function for inserting node
+    def insert_node(self, node_data):
+        node = SinglyLinkedListNode(node_data)
+
+        if not self.head:
+            self.head = node
+        else:
+            self.tail.next = node
+
+        self.tail = node
+
+#Function to print the linked list
+def print_singly_linked_list(node, sep):
+    while node:
+        print(str(node.data), end=' ')
+
+        node = node.next
+
+# Function for getting elements in first and second linked list in array
+def printt(headd):
+    itr = headd
+    llstr = []
+    while itr:
+        llstr.append(itr.data)
+        itr = itr.next
+    return llstr
+
+
+# This function takes two linked list and compares it and merge two list
+def mergeLists(llist1, llist2):
+    ll1 = printt(llist1)
+    ll2 = printt(llist2)
+    ll3 = (ll1 + ll2) #comparing and merging two linked list
+    ll3.sort()
+    lll = SinglyLinkedList() #creating new linkedlist
+    for ii in ll3:
+        lll.insert_node(ii) #Adding merged element to new linked list
+    return lll.head
+
+# Main funcation
+if __name__ == '__main__':
+
+    llist1_count = int(
+        input("Enter the number of element to be in linked list 1: "))
+
+    llist1 = SinglyLinkedList()
+
+    print("Enter the elements to be added in list1 line by line")
+    for _ in range(llist1_count):
+        llist1_item = int(input())
+        llist1.insert_node(llist1_item)
+
+    print("\n")
+    llist2_count = int(
+        input("Enter the number of element to be in linked list 2: "))
+
+    llist2 = SinglyLinkedList()
+
+    print("Enter the elements to be added in list2 line by line")
+    for _ in range(llist2_count):
+        llist2_item = int(input())
+        llist2.insert_node(llist2_item)
+
+    llist3 = mergeLists(llist1.head, llist2.head)
+
+    print("\n")
+    print("The merged linked list value: ")
+    print_singly_linked_list(llist3, ' ')
+
+'''
+Output window:
+
+Enter the number of element to be in linked list 1: 3
+Enter the elements to be added in list1 line by line
+1
+2
+3
+
+
+Enter the number of element to be in linked list 2: 2
+Enter the elements to be added in list2 line by line
+3
+4
+
+
+The merged linked list value:
+1 2 3 3 4
+
+Complexity:
+Time complexity: O(n)
+Space complexity: O(n)
+'''

General Questions/Polynomial_Verification.py

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+'''
+Aim: Given a polynomial P of a single indeterminate (or variable), x. Also, 
+given the values of x and k. The task is to verify if P(x) = k.
+
+'''
+
+# taking the space separated input
+x_and_k = input().split()
+# the very first number denotes 'x'
+x = int(x_and_k[0])
+# eval is used to evaluate any expression
+# here if the evaluated expression results in the value of k, it would mean that it is valid
+# the value of k is second in the entered string x_and_k, we have selected it using indicing
+print(eval(input()) == int(x_and_k[1]))
+
+'''
+
+COMPLEXITY:
+	
+	 Time Complexity -> O(1)
+	 Space Complexity -> O(1)
+     
+Sample Input:
+1 4
+x**3 + x**2 + x + 1
+
+Sample Output:
+True
+
+Explanation:
+x = 1
+k = 4
+P(x) = x**3 + x**2 + x + 1
+P(1) = 1**3 + 1**2 + 1 + 1
+     = 1 + 1 + 1 + 1
+     = 4
+     = k
+Hence, the polynomial is a valid one.
+
+'''

General Questions/possible no from word.cpp

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+/* C++ program to print a given number in words. The program
+handles numbers from 0 to 9999 
+Just take an example  if the user privide the number
+INPUT-
+1-No of test cases
+1234-the given number
+OUTPUT- 
+One thousand Two hundred and twenty four */
+#include<bits/stdc++.h>
+using namespace std;
+string one[] = { "", "one ", "two ", "three ", "four ",
+        "five ", "six ", "seven ", "eight ",
+        "nine ", "ten ", "eleven ", "twelve ",
+        "thirteen ", "fourteen ", "fifteen ",
+        "sixteen ", "seventeen ", "eighteen ",
+        "nineteen " };
+
+// strings at index 0 and 1 are not used, they is to
+// make array indexing simple
+string ten[] = { "", "", "twenty ", "thirty ", "forty ",
+        "fifty ", "sixty ", "seventy ", "eighty ",
+        "ninety " };
+/* ******* Your Functions Below ******** */
+
+
+// n is 1- or 2-digit number
+string numToWords(int n, string s)
+{
+  string str = "";
+  // if n is more than 19, divide it
+  if (n > 19)
+    str += ten[n / 10] + one[n % 10];
+  else
+    str += one[n];
+
+  // if n is non-zero
+  if (n)
+    str += s;
+
+  return str;
+}
+
+// Function to print a given number in words
+string convertToWords(long n)
+{
+  // stores word representation of given number n
+  string out;
+
+  // handles digits at ten millions and hundred
+  // millions places (if any)
+  out += numToWords((n / 10000000), "crore ");
+
+  // handles digits at hundred thousands and one
+  // millions places (if any)
+  out += numToWords(((n / 100000) % 100), "lakh ");
+
+  // handles digits at thousands and tens thousands
+  // places (if any)
+  out += numToWords(((n / 1000) % 100), "thousand ");
+
+  // handles digit at hundreds places (if any)
+  out += numToWords(((n / 100) % 10), "hundred ");
+
+  if (n > 100 && n % 100)
+    out += "and ";
+
+  // handles digits at ones and tens places (if any)
+  out += numToWords((n % 100), "");
+
+  return out;
+}
+int main()
+{
+    int t;
+    cin>>t;
+
+    while(t--)
+{
+    long n ;
+  cin>>n;
+
+  // convert given number in words
+  cout << convertToWords(n) << endl;
+
+
+}
+}
+
+/*Time complexity:
+Time complexity: O(1). 
+The loop runs for a constant amount of time.
+Auxiliary space: O(1). 
+As no extra space is required.
+Time complexity: O(1). */
+
+
+	
+