\nInput:\n5\napple\nbanana\ngrape\norange\npear\ngrape\n\nOutput:\n\"grape\": grape\n<\/pre>\n\nInput:\n5\napple\nbanana\ngrape\norange\npear\nwatermelon\n\nOutput:\nThe word could not be found.\n<\/pre>\nProblem Solving Strategy<\/h2>\n
\n This problem requires searching for a word, so choosing an efficient data structure is key.
\n Using a HashMap allows you to search for a word with an average time complexity of O(1).
\n Therefore, you can store the dictionary in a HashMap, then search for the word entered by the user in that HashMap
\n to output its meaning.\n<\/p>\n
Java Code Implementation<\/h2>\n
\n Below is the Java code to solve the above problem. This code implements the dictionary using a HashMap and handles user input.\n<\/p>\n
\nimport java.util.HashMap;\nimport java.util.Scanner;\n\npublic class DictionarySearch {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n \n \/\/ Input the number of words in the dictionary\n int N = scanner.nextInt();\n scanner.nextLine(); \/\/ Remove the newline character\n \n \/\/ Initialize the HashMap\n HashMap dictionary = new HashMap<>();\n \n \/\/ Input words and meanings for the dictionary\n for (int i = 0; i < N; i++) {\n String word = scanner.nextLine(); \/\/ Input word\n String meaning = scanner.nextLine(); \/\/ Input meaning\n dictionary.put(word, meaning);\n }\n \n \/\/ Input the word to search\n String searchWord = scanner.nextLine();\n \n \/\/ Search for the word and output the result\n if (dictionary.containsKey(searchWord)) {\n System.out.println(\"\\\"\" + searchWord + \"\\\": \" + dictionary.get(searchWord));\n } else {\n System.out.println(\"The word could not be found.\");\n }\n\n scanner.close();\n }\n}\n<\/pre>\nCode Explanation<\/h2>\n
\n 1. The Scanner<\/code> class is used to handle user input, and HashMap<\/code> is used to store the dictionary.
\n 2. First, the size of the dictionary N<\/code> is input, and then words and their meanings are added to the HashMap for N<\/code> times.
\n 3. Next, the user inputs the word they want to search for, and it checks if that word is in the HashMap.
\n 4. If the word exists, its meaning is printed; if not, the message \"The word could not be found.\" is printed.\n<\/p>\nComplexity Analysis<\/h2>\n
\n - Time Complexity: O(1) for adding a word to the HashMap and O(1) for searching. Overall O(N).
\n - Space Complexity: O(N) as the HashMap stores N<\/code> words and their meanings.\n<\/p>\nConclusion<\/h2>\n
\n In this article, we looked at an algorithm problem to find a specific word in a dictionary using Java.
\n We learned how to efficiently search for words using a HashMap and gained insights into what approach to take
\n during coding tests through the problem-solving process.
\n I hope you can confidently solve similar problems in future coding tests.\n<\/p>\n
<\/body><\/p>\n","protected":false},"excerpt":{"rendered":"
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