Java Exercises: Count the letters, spaces, numbers and other characters of an input string
Java Basic: Exercise-38 with Solution
Write a Java program to count the letters, spaces, numbers and other characters of an input string.
Test Data:
The string is : Aa kiu, I swd skieo 236587. GH kiu: sieo?? 25.33
Pictorial Presentation:

Sample Solution:
Java Code:
import java.util.Scanner;
public class Exercise38 {
public static void main(String[] args) {
String test = "Aa kiu, I swd skieo 236587. GH kiu: sieo?? 25.33";
count(test);
}
public static void count(String x){
char[] ch = x.toCharArray();
int letter = 0;
int space = 0;
int num = 0;
int other = 0;
for(int i = 0; i < x.length(); i++){
if(Character.isLetter(ch[i])){
letter ++ ;
}
else if(Character.isDigit(ch[i])){
num ++ ;
}
else if(Character.isSpaceChar(ch[i])){
space ++ ;
}
else{
other ++;
}
}
System.out.println("The string is : Aa kiu, I swd skieo 236587. GH kiu: sieo?? 25.33");
System.out.println("letter: " + letter);
System.out.println("space: " + space);
System.out.println("number: " + num);
System.out.println("other: " + other);
}
}
Sample Output:
The string is : Aa kiu, I swd skieo 236587. GH kiu: sieo?? 25.33 letter: 23 space: 9 number: 10 other: 6
Flowchart:

Java Code Editor:
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Previous: Write a Java program to reverse a string.
Next: Write a Java program to create and display unique three-digit number using 1, 2, 3, 4. Also count how many three-digit numbers are there.
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Java: Tips of the Day
Prefer primitive classes
Wrapper classes are often slower than primitive classes. While Primitive class only has values, the wrapper class stores information about the entire class. Since the classes often deal with object values, comparing them with primitive classes does not give the desired results.
Ref: https://bit.ly/3eItLCN
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