Given a string str, write a Java program to print all words with even length in the given string. Examples:
Input: s = "This is a java language"
Output: This
is
java
language
Input: s = "i am GFG"
Output: am
Approach:
- Take the string
- Break the string into words with the help of split() method in String class. It takes the string by which the sentence is to be broken. So here ” “(space) is passed as the parameter. As a result, the words of the string are split and returned as a string array
String[] words = str.split(" ");
- Traverse each word in the string array returned with the help of Foreach loop in Java.
for(String word : words)
{ }
- Calculate the length of each word using String.length() function.
int lengthOfWord = word.length();
- If the length is even, then print the word.
Below is the implementation of the above approach:
Java
// Java program to print// even length words in a stringclass GfG { public static void printWords(String s) { // Splits Str into all possible tokens for (String word : s.split(" ")) // if length is even if (word.length() % 2 == 0) // Print the word System.out.println(word); } // Driver Code public static void main(String[] args) { String s = "i am Geeks for Geeks and a Geek"; printWords(s); }} |
am Geek
Time complexity: O(n) where n is length of given string
Auxiliary Space: O(1)
Approach 2: Using Dynamic Programming:
In this approach, we split the string into words and store the length of each word in an array. Then, we loop through the array and check if the length of each word is even. If it is, we print the word. This approach uses dynamic programming to store the length of each word in an array, which can be reused in subsequent loops, reducing the overall number of calculations. However, in practice, the difference in performance between this approach and the original approach is likely to be negligible.
Here is the DP approach in Java:
Java
class GfG { public static void printWords(String s) { // Split string into words String[] words = s.split(" "); // Create an array to store the length of each word int[] wordLengths = new int[words.length]; // Calculate the length of each word and store in the array for (int i = 0; i < words.length; i++) { wordLengths[i] = words[i].length(); } // Check if the length of each word is even and print if true for (int i = 0; i < words.length; i++) { if (wordLengths[i] % 2 == 0) { System.out.println(words[i]); } } } // Driver code public static void main(String[] args) { String s = "i am Geeks for Geeks and a Geek"; printWords(s); }} |
Output:
am Geek
Time complexity: O(n) where n is length of given string
Auxiliary Space: O(n) where n is the total number of characters in the input string.
