Merge branch 'master' of https://github.com/smv1999/CompetitiveProgrammingQuestionBank into Devincept13

Author: Karnika06

Codechef Problems/Red_alert.c

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+/* PROBLEM STATEMENT 
+
+
+Finally a monsoon has come. According to the Meteorological Department, there will be rain in the upcoming N days in the city. Initially, the water level of the city is zero millimetres. The amount of rain on the i-th day can be described by an integer Ai as follows:
+
+If Ai>0, the water level of the city increases by Ai millimetres on the i-th day.
+If Ai=0, there is no rain on the i-th day. The water level of the city decreases by D millimetres on such a day. However, if the water level is less than D millimetres before the i-th day, then it becomes zero instead.
+There will be a red alert in the city if the water level becomes strictly greater than H millimetres on at least one of the N days. Determine if there will be a red alert.
+
+Input Format
+The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
+The first line of each test case contains three space-separated integers N, D and H.
+The second line contains N space-separated integers A1,A2,…,AN.
+Output Format
+For each test case, print a single line containing the string "YES" if there will be a red alert or "NO" otherwise.
+
+Constraints
+1≤T≤103
+1≤N,D≤102
+0≤Ai≤102 for each valid i
+1≤H≤104
+Subtasks
+Subtask #1 (100 points): original constraints
+
+Sample Input 1 
+4
+4 2 6
+1 3 0 2
+2 1 100
+1 100
+4 2 3
+1 2 0 2
+3 7 10
+5 3 9 
+Sample Output 1 
+NO
+YES
+NO
+YES
+Explanation
+Example case 1:
+
+On the first day, the water level of the city increases to 1 millimtre.
+On the second day, the water level increases by 3 millimeters and becomes 1+3=4 millimtres.
+On the third day, there is no rain in the city, so the water level decreases by D=2 millimtres and becomes 4−2=2 millimtres.
+On the fourth day, the water level increases by 2 millimtres and becomes 2+2=4 millimtres.
+There will be no red alert in the city because the water level does not exceed H=6 millimtres on any of the four days.
+
+Example case 2: The water level of the city on the 2-nd day is equal to 101 millimtres, which is greater than H=100 millimtres, so there will be a red alert in the city.
+
+Example case 3: The water levels of the city on the four days are [1,3,1,3]. The water level is equal to H=3 millimtres on the second and fourth day, but it does not exceed the threshold.
+
+Example case 4: There will be a red alert in the city on the 3-rd day.  */
+
+// SOLUTION
+
+#include <stdio.h>
+
+int main(void) {
+	
+	int t,n,d,h;
+	
+	scanf("%d",&t);
+	// t represents the number of test cases
+
+	 while(t--) //the loop runs for t times 
+	 {
+	     	scanf("%d %d %d",&n,&d,&h);
+		 // n, d and h respectively stores the input values
+	     	int sum=0;int val; int ret=0;
+	     	for(int i=0;i<n;i++) // this loop take n values as user input
+	     	{
+			
+	     	    scanf("%d",&val); // the values inputted are stored each time in the variable val
+			//note: here a single variable is used instead of an array beacuse the operations are done on the variable here itself i.e. within this loop.
+	     	    if(val>0)
+	     	    {
+			    //if val( the level of water on ith day > 0
+	     	        sum+=val; // total level = previous(stored in sum itsef) + val;
+	     	       
+	     	    }
+	     	    if(val==0)
+	     	    {
+			    //if val( the level of water on ith day = 0
+	     	        sum=(sum<d)?0 : (sum-d);
+	     	       // the ternary operator above checks if sum<d for true: sum =0 for false: sum-d
+	     	    }
+			
+	     	     if(sum>h) //checks the level each day (level is calculated by carrying out arithmetic operations on variable val which is finally stored in variable sum
+	     	        {
+	     	            ret=1; //if red alert  i.e. sum > h , the variable ret is initialised with value 1.
+			     break; 
+	     	        }
+	     	}
+	     	if(ret==0) 
+	     	printf("NO\n");
+	     	else 
+	     	printf("YES\n");
+	     	// Time Complexity : O(N)
+		 // Space Complexity: 0(1)
+	     	
+	 }
+	 
+	
+	return 0;
+}

Data Structures/Stacks/redundant_braces.cpp

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+// Given a string A of length n denoting an expression. It contains the following operators ’+’, ‘-‘, ‘*’, ‘/’.
+// Check whether A has redundant braces or not.
+
+#include <bits/stdc++.h> 
+using namespace std; 
+
+int redundantBraces(string A) {
+ 
+    int n= A.length();
+    stack <char> s;
+    
+    for(int i=0; i< n; i++)
+    {
+        //we'll push the operators and opening braces in the stack
+        if(A[i]=='+' || A[i]=='-'|| A[i]=='/'|| A[i]=='*' || A[i]=='(' )
+            s.push(A[i]);
+        
+     //when we encounter closing braces, we need to start popping
+        if(A[i]==')')
+        { 
+         //if there are no operators, it means we have redundant braces for eg. ()
+            if(s.top()=='(')
+                return 1;
+         
+         //else we start popping till we encounter opening braces
+         //count variable will keep count of number of operators enclosed within the braces
+            int count=0;
+            while(!s.empty() && s.top()!='(')
+            {
+                s.pop();
+                count++;
+            }
+         //count=0 means we have something like (a) which has redundant braces
+            if(count==0) 
+                return 1;
+            s.pop();
+        }
+    }
+ 
+        return 0;
+} 
+
+int main() 
+{
+    string str;
+    cin>>str;
+    if(redundantBraces(str))
+        cout<<str+" has redundant braces";
+    else 
+        cout<<str+" doesn't have any redundant braces";
+return 0;
+
+}  
+
+// Test Cases
+// Input 1:
+//     A = "((a + b))"
+// Output 1:
+//     1
+//     Explanation 1:
+//         ((a + b)) has redundant braces so answer will be 1.
+// Input 2:
+//     A = "(a + (a + b))"
+// Output 2:
+//     0
+//     Explanation 2:
+//         (a + (a + b)) doesn't have have any redundant braces so answer will be 0.
+
+// Time complexity: O(n)

Hackerrank solutions/30 days Of Code/Day8-Dictionaries_and_Maps.c

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+/* PROBLEM STATEMENT
+Given n names and phone numbers, assemble a phone book that maps friends' names to their respective phone numbers. 
+You will then be given an unknown number of names to query your phone book for. 
+For each  name queried, print the associated entry from your phone book on a new line in the form name=phoneNumber; 
+if an entry for name  is not found, print Not found instead.
+
+Note: Your phone book should be a Dictionary/Map/HashMap data structure.
+
+Input Format
+
+The first line contains an integer,n , denoting the number of entries in the phone book.
+Each of the n subsequent lines describes an entry in the form of 2 space-separated values on a single line. 
+The first value is a friend's name, and the second value is an 8-digit phone number.
+
+After the n lines of phone book entries, there are an unknown number of lines of queries. 
+Each line (query) contains a name to look up, and you must continue reading lines until there is no more input.
+
+Note: Names consist of lowercase English alphabetic letters and are first names only.
+
+Output Format
+
+On a new line for each query, print Not found if the name has no corresponding entry in the phone book; otherwise,
+print the full  name and  phone number in the format name=phoneNumber.
+
+Sample Input
+
+3
+sam 99912222
+tom 11122222
+harry 12299933
+sam
+edward
+harry
+ 
+ Sample Output
+
+sam=99912222
+Not found
+harry=12299933
+Explanation
+
+We add the following n=3 (Key,Value) pairs to our map so it looks like this:
+phonebook={ (sam ,99912222),(tom, 11122222),(harry, 12299933)}
+
+We then process each query and print key=value if the queried  is found in the map; otherwise, we print Not found.
+
+Query 0: sam
+Sam is one of the keys in our dictionary, so we print sam=99912222.
+
+Query 1: edward
+Edward is not one of the keys in our dictionary, so we print Not found.
+
+Query 2: harry
+Harry is one of the keys in our dictionary, so we print harry=12299933.     */
+
+
+#include <cmath>
+#include <cstdio>
+#include <vector>
+#include <iostream>
+#include <algorithm>
+#include <map>
+using namespace std;
+
+
+int main() {
+   
+    int n;
+string name;
+long num;
+cin >> n; 
+cin.ignore();
+map <string, long> pBook;
+for (int i = 0; i < n; i++) {
+    cin >> name;
+    cin >> num;
+    pBook[name] = num;
+}
+while(cin >> name) {
+    if (pBook.find(name) != pBook.end())
+     //pBook.end means that it has reached to the last element of pBook. 
+        //so if we use != , it means that it has not reached to the end. Thus it means that the name is found somewhere in the middle of pBook.
+    {
+      
+        cout << name << "=" << pBook.find(name)->second << endl;
+      //map.find() iterates through your map structure. if it finds the element then it returns an iterator to that element else it returns end. 
+      //End being an element past the last element of that container, essentially telling you it wasn't found.
+      //Remember containers in c++: arrays, vectors, maps, queues, stacks, etc. start at 0. 
+      //So if you have a map container with 10 elements and an index of 10 is returned this is outside the bounds of the container so in this function end is returned.
+     
+    } else {
+        cout << "Not found" << endl;
+    }
+}  
+    return 0;
+}
+
+// Time Comlexity: O(n)
+// Space Complexity: O(1)

Trie/Trie_add_Search_using_dict.py

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+"""
+Purpose:
+
+Implementation of Trie has been asked in my interviews like in Google
+Store word in Dictionary
+Search word in the dictionary returns True if exist
+Stores words in a compressed manner
+"""
+"""
+Inputs:
+1nd line no of word you want to store in Trie (int)
+2rd line to (2+1st line input)th line (string)
+Next line enter number of words want to search
+then enter strings in each line
+
+"""
+class Trie:
+    root = {} #main dictionary of Trie
+    
+    def add(self, word): # This funtion adds new word to Trie
+        cur = self.root  # cur is iterator to check whether new key/paths are required to add new word and root is main dictionary
+        
+        for ch in word: # ch is all the charactors in the word 
+            if ch not in cur: # if charactor is not present add new key/path initiate it empty
+                cur[ch] = {}  
+            cur = cur[ch]    # now go to the key that was created or was present
+        cur['*'] = True   # set * key to true at the end of each word
+        #so larger the word the dictionary nesting will take place and * key represant the end state of the word
+        
+    def search(self,word): # This function searches whether the word is presant or not presant
+        cur = self.root  #cur is iterator to check whether new key/paths are presant or not
+        
+        for ch in word:    # ch is all the charactors in the word 
+            if ch not in cur:  # if ch is not found in cur that means word not presant so result is false
+                return False
+            cur = cur[ch] #if ch present then go till the end unless it becomes false
+        if '*' in cur: # if * key not present that means words is prefix of a word that is present but word is not Present
+            return True
+        else:
+            return False
+# dynamic input 
+n2 = int(input("No of words to be Added:\n")) 
+print("enter words:")
+words = []
+checks = []
+for j in range(n2):
+    words.append(input())
+n3 = int(input("No of words to be searched:\n"))
+print("enter words:")
+for j in range(n3):
+    checks.append(input())
+
+        
+
+"""
+Time complexity :
+Insert: O(word Length)
+Search: O(word Length)
+
+Space complexity :
+Insert: O(1)
+Search: O(1)
+
+Storages required for storing words:
+if no word has same prefix the required (total word length+1)*(char size)
+
+"""
+dictionary = Trie() # Trie object created
+
+
+for k in words:
+    dictionary.add(k) # adding word in Trie from words list
+for k in checks:
+    print(dictionary.search(k)) # printing Searched word results if present then prints True if not present then returns False