String Programs in Java
String Programs in Java: In Java, a String is an immutable object that represents a sequence of characters. For example, “Tutorial” is a string that consists of 8 characters: ‘T’, ‘u’, ‘t’, ‘o’, ‘r’, ‘i’, ‘a’, ‘l’. Immutable object means once a string is created, it cannot be changed, and if we try to change the string, we end up creating a new string. In this section, we will create different String programs.
The Java String class is used for the creating of a string object. However, we do not need to import the class String. This is because the Java String class is part of the java.lang package, and java.lang package is imported by default in every Java program. In Java, a string is represented by characters enclosed in double quotes (“”). For characters, single quotes are used (‘’);
Creating a String in Java
There are three ways of creating a string/ string object in Java:
- Using string literal
- Using the new keyword
- Using a character array
Using string literal
FileName: StringExample.java
public class StringExample
{
public static void main(String argvs[])
{
// creating strings
String country = "India";
String country1 = "France";
String country2 = "Mexico";
// printing strings
System.out.println( country );
System.out.println( country1 );
System.out.println( country2 );
}
}
Output:
India France Mexico
Explanation: In the above program, we have created three strings: country, country1 and country2 and assigned the strings: “India”, “France”, and “Mexico”. Note that we have not create any object of the class String. However, we know that a string is an object in Java. Hence, under the hood, JVM creates three objects for the three strings and assigns the reference to the string variables. Memory allocation for the three strings is done in the string pool ( a place in the heap memory used by JVM for avoiding string redundancy), whereas the string variables get placed in stack memory.
Using new Keyword
FileName: StringExample1.java
public class StringExample1
{
public static void main(String argvs[])
{
// creating strings using the new Keyword
String country = new String("India");
String country1 = new String("France");
String country2 = new String("Mexico");
// printing strings
System.out.println( country );
System.out.println( country1 );
System.out.println( country2 );
}
}
Output:
India France Mexico
Explanation: In the above program, we have used the new keyword for creating the string object. The string objects are created in heap memory but not in the string pool. Memory for the string variables is allocated in the stack memory.
Using Character Array
FileName: StringExample2.java
public class StringExample2
{
public static void main(String argvs[])
{
// a character array
char arrChar[]={'A', 'm', 'e', 'r', 'i', 'c', 'a'};
// creating a string using the character array
String str = new String(arrChar);
// printing the string
System.out.println(str);
}
}
Output:
America
Explanation: Using the given character array, we are creating string str. This time also, the new keyword comes in handy. Eventually, we are displaying the result.
The length() method
To calculate the size of a string, we use the length() method of String class. The signature of the method length() is:
public int length()
Since the keyword static is not used, the length() method cannot be invoked using the class name, i.e., String.length(). We have to use objects to call the length() method. It does not accept any parameter. It returns the length of the string. The following example illustrates the same.
FileName: StringExample3.java
public class StringExample3
{
public static void main(String argvs[])
{
int size; // for storing size of the input string
String s = "Apple"; // input string
// invoking the method length()
size = s.length();
// displaying the outcome
System.out.println("The size of the string \"" + s + "\" is " + size);
}
}
Output:
The size of the string "Apple" is 5
Explanation: The string “Apple” internally gets converted to a character array. Then, the length field of the array is used to find its size, which is 5 in our case. The value of the length field is returned by the method length().
Manipulating Strings
Suppose we have a string s = “Apple”. The task is to change it to “Rpple”, i.e., instead of the letter ‘A’; we have to use the letter ‘R’. In c++, we can easily achieve the task. At the index 0, put the letter R, i.e., S[0] = ‘R’. But it is not possible in Java. Because strings are immutable. Therefore, we need to find a workaround to do the manipulation. One way is to use a character array. Observe the following Java program.
Using Character Array
FileName: StringExample4.java
public class StringExample4
{
public static void main(String argvs[])
{
String s = "Apple"; // input string
int size = s.length(); // calculating size of the input string
// creating a character array of the same size as the input string
char ch[] = new char[size];
// assigning characters from the input string
// to the character array
for(int i = 0; i < size; i++)
{
ch[i] = s.charAt(i);
}
ch[0] = 'R'; // updating character at the index 0
// creating a new string using the character array
s = new String(ch);
// printing the new string
System.out.println("The new string is " + s);
}
}
Output:
The new string is Rpple
Explanation: We know that string is immutable in Java. However, the same is not true for arrays. This is why we need to generate a character array from the given input string. The above program does the same. First, the program allocates the size of the character array and then fills the array by scanning each and every character of the given string. In order to retrieve characters from the input array, the method charAt() is used. The method is defined in the class String. It accepts an int value as in its argument and returns the character present at that particular index. The signature of the method is:
public char charAt(int index)
If the value of the index is negative or greater than the size of the array, the method throws StringIndexOutOfBoundsException.
After assigning all the values to the character array, we update the value present at the index 0. Then, using the character array, we have created a string and assigned its reference to the same variable that was holding the input string. It means that the input string has no reference. Thus, the input string is ready for garbage collection. Eventually, we are displaying the result. Note that, at no place, we have manipulated the input string. We have to create a completely new string to get our result.
Using the replace() Method
Another approach is to use the predefined method replace(). The method is present in the class String. Observe the following Java program.
FileName: StringExample5.java
public class StringExample5
{
public static void main(String argvs[])
{
String str = "Apple"; // input string
// invoking the method replace()
String strFinal = str.replace('A', 'R');
// printing the new string
System.out.println("The new string is " + strFinal);
}
}
Output:
The new string is Rpple
Explanation: Instead of creating a character array, we have used the method replace() to accomplish our task. The replace() method, used in our code, takes two arguments: one is the character we want to replace (oldChar), and another is that character that replaces the oldChar (newChar). The signature of the replace() method is:
public String replace(char oldChar, char newChar)
From the signature, it is obvious that the replace() method returns a string. The replace() method creates a character array on the basis of the string upon which the method is called. In our code, str is calling the method. Thus, a character array is created of size equal to the length of string referred by str, which is 5. Then, all the occurrence of letter ‘A’ is replaced by ‘R’. From the character array a new sting a generated (similar to the previous example). The newly generated string is then returned, whose reference is being stored in the variable strFinal. Eventually, we are displaying the result. Thus, we see that even though we are using an in-built method, the immutability of the string remains intact.
Pattern Searching Program
In pattern searching, a word/ pattern and a text are given. The task is to find whether the given pattern is present in the text or not. For example, text = “llalalaland” and pattern = “land”. The pattern “land” is present in the text “llalalaland” at index 7. Observe the following code.
FileName: StringExample7.java
public class StringExample7
{
public static void main(String argvs[])
{
// input text
String text = "llalalaland";
// finding text size
int textSize = text.length();
// pattern to be found in the input text
String pattern = "land";
// finding pattern size
int patternSize = pattern.length();
// for storing starting index of the pattern present in the text
int index = -1;
// flag for indicating whether the pattern in found or not
// in the input text
Boolean isPatternFound = false;
// We only iterate till that character of the input string,
// from where we can accommodate the pattern. This is because a
// pattern size can never be greater than input text size
for(int i = 0; i < textSize - patternSize + 1; i++)
{
// from temp, we iterate to match the characters of text with pattern
int temp = i;
for(int j = 0; j < patternSize; j++)
{
if(text.charAt(temp) == pattern.charAt(j))
{
// character matched!!
// check the next character of the pattern
temp++;
// assuming the pattern is found!!
isPatternFound = true;
}
else
{
// character mismatch!!
// assumption made in if-block is not true.
isPatternFound = false;
// We have encountered a mismatch.
// No need to check further
break;
}
}
if(isPatternFound)
{
// pattern found!
// updating the index
index = i;
// Since pattern is found,
// there is no need to iterate further
break;
}
}
// printing the outcome
if(isPatternFound)
{
System.out.println("The pattern \"" + pattern + "\" starts at index " + index + " in the text \"" + text + "\"");
}
else
{
System.out.println("The pattern " + pattern + "is not found in the text " + text);
}
}
}
Output:
The pattern "land" starts at index 7 in the text "llalalaland"
Explanation: In the above program, we have used nested Java for-loop for finding the pattern. Inside the for loop, we have used the two pointers approach. One pointer is the outer loop variable i, for the given text, and another pointer is the inner loop variable j, for the given pattern. In each iteration of the outer loop, we assign the value of i to the variable temp. Because i marks the index, where we have to start the search of pattern in the text. Till this point, temp and i both point to the same index. Then, we enter the inner loop, where j is pointing to the first index of the pattern (j = 0).
Now, the comparison of characters of the text and the pattern starts. If the character pointed by temp matches with the character pointed by the j if-block of the inner loop gets executed. The variables j and temp are incremented by 1, and the flag isPatternFound is true. In the case of mismatch, the else block is executed, where the flag isPatternFound is assigned false value, and the inner loop is terminated.
Beneath the inner loop, there is an if block. The if block only gets executed if isPatternFound is true, which is only possible when the else block of the inner for loop is not executed at all. It means that the inner loop only leaves true value to isPatternFound when there is no mismatch. Inside the if-block, the index stores the value of the outer loop variable i, as i is the beginning index of the pattern matching in the text.
Let’s understand each iteration of the outer for-loop with the help of following steps.
For the 1st iteration of the outer loop,
i = 0, temp = 0, j = 0
llalalaland
land
The first character of the text matches with the first character of the pattern. Hence, the inner loop iterates once more.
i = 0, temp = 1, j = 1
llalalaland
land
The second character of the text does not match with the second character of the pattern. Hence, the else block of the inner block is executed, and the inner loop is terminated.
For the 2nd iteration of the outer loop
i = 1, temp = 1, j = 0
llalalaland
land
The second character of the text matches with the first character of the pattern. Hence, the iteration continues of the inner loop.
i = 1, temp = 2, j = 1
llalalaland
land
The second character of the pattern also matches. Therefore, temp and j again get incremented by 1.
i = 1, temp = 3, j = 2
llalalaland
land
Now, we encounter a mismatch. Hence, the inner loop is terminated. Still, we did not find the pattern in the text. Therefore, the search continues.
For the 3rd iteration of the outer loop,
i = 2, temp = 2, j = 0
llalalaland
land
In the first iteration of the inner loop, we get a mismatch. Hence, the inner loop terminates.
For the 4th iteration of the outer loop,
i = 3, temp = 3, j = 0
llalalaland
land
The fourth character of the text matches with the first character of the pattern. Therefore, iteration of the inner loop continues.
i = 3, temp = 4, j = 1
llalalaland
land
Again, the characters match. The inner loop iterates one more time.
i = 3, temp = 5, j = 2
llalalaland
land
l ? n. Therefore, because of the mismatch, the inner loop terminates.
For the 5th iteration of the outer loop,
i = 4, temp = 4, j = 0
llalalaland
land
a ? l. Hence, the inner loop does not iterate further.
For the 6th iteration of the outer loop,
i = 5, temp = 5, j = 0
llalalaland
land
l = l. Therefore, temp and j are incremented by 1.
i = 5, temp = 6, j = 1
llalalaland
land
Again, there is a match. Thus, we move to the next iteration of the inner for-loop.
i = 5, temp = 7, j = 2
llalalaland
land
l ? n. Because of the mismatch, the inner loop terminates.
For the 7th iteration of the outer loop,
i = 6, temp = 6, j = 0
llalalaland
land
There is a mismatch. The control moves to the outer loop, again.
For the 8th iteration of the outer loop,
i = 7, temp = 7, j = 0
llalalaland
land
The characters match. The inner loop iterates.
i = 7, temp = 8, j = 1
llalalaland
land
Again, there is a match. The inner loop comes into the picture again.
i = 7, temp = 9, j = 2
llalalaland
land
n = n. Still, there is a match. The inner loop iterates one more time
i = 7, temp = 10, j = 3
llalalaland
land
The last character of the pattern matches and the inner and outer loop terminates. We see at i = 7; the successful search has occurred. Therefore, “index 7” is printed on the console. Therefore, we have successfully searched our pattern in the given text.