Create Day8-Dictionaries_and_Maps.c

Author: simranquirky

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)