Here is problem solution in Python, java, c++, c and javascript - https://programmingoneonone.com/hackerrank-day-8-dictionaries-and-maps-30-days-of-code-solution.html

n=int(input()) d={} for _ in range(n): details=input() f_name,phone_no=map(str,details.split()) d[f_name]=phone_no

while True: try: name=input() if name in d: print(f'{name}={d[name]}') else: print("Not found") except EOFError: break

public static void main(String []argh){ Scanner in = new Scanner(System.in); int n = in.nextInt(); Map phonebook=new HashMap<>(); for(int i = 0; i < n; i++){ String name = in.next(); int phone = in.nextInt(); phonebook.put(name,phone); // Write code here } while(in.hasNext()){ String s = in.next(); //write your code here if(phonebook.containsKey(s)) System.out.println(s+"="+phonebook.get(s)); else System.out.println("Not found"); } in.close(); }

x = int(input())
dic = {}

Reading phonebook entries

for i in range(x):
    a = input().split(' ', 1)
    dic[a[0]] = a[1]

Handling multiple queries safely until EOF

try: 
    while True:
        a = input()
        if len(a) > 0:
            if a in dic:
                print(f'{a}={dic[a]}')
            else:
                print('Not found')
except EOFError:
    pass

This is python solution can any one fix the error it failed test case1 n= int(input()) phonebook = {} for i in range(n):

entry = input().split()
name = entry[0]
phone = entry [1]
phonebook[name] = phone

queries = []

for i in range(n): query = input() queries.append(query)

for query in queries: if query in phonebook: print(f"{query}={phonebook[query]}")

  else:
        print("Not found")