Top 70 Java Program Interview Questions

Print Character And Count Of a Word: This Java program aims to count the occurrences of each character in a given input string and generate an output string that represents each character followed by its count.

The specific transformation is applied to the input string “abbcccdeee” to obtain the output string “a1b2c3d1e3”.

Check frequently asked Java programs for interviews.

Print Character And Count Of a Word

package com.softwaretestingo.interviewprograms;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Map.Entry;
public class InterviewPrograms7 
{
	/*
	 * Input= abbcccdeee 
	 * Output= a1b2c3d1e3
	 */
	public static void main(String[] args) 
	{
		String s="abbcccdeee";
		char [ ] ch= s.toCharArray();
		Map<Character, Integer> maps= new LinkedHashMap<Character,Integer>();
		for (Character c:ch)
		{
			Integer count = maps.get(c);
			if (count != null)
			{
				maps.put (c, ++count);
			} 
			else 
			{
				maps.put(c,1);
			}
		}
		StringBuffer sb = new StringBuffer();
		for (Entry< Character , Integer > entry : maps.entrySet()) 
		{
			sb.append(entry.getKey()).append(entry.getValue());
		}
		System.out.println (sb);
	}
}

Output:

a1b2c3d1e3

Here’s a step-by-step explanation of what we are achieving through this program:

• The program starts by defining a class named “InterviewPrograms7” in the package “com.softwaretestingo.interviewprograms”.
• The main method is declared as public and static, allowing it to be accessed without creating an instance of the class.
• A string variable named “s” is declared and assigned the value “abbcccdeee”.
• The input string “s” is converted into a character array using the toCharArray() method.
• A LinkedHashMap named “maps” is created to store characters as keys and their respective counts as values. LinkedHashMap is used to preserve the insertion order.
• The program iterates over each character in the character array using an enhanced for loop.
• Within the loop, it checks if the character is already present as a key in the “maps” LinkedHashMap.
• If the character is present, it retrieves the current count from the “maps” LinkedHashMap and increments it by 1.
• If the character is not present, it adds the character as a key to the “maps” LinkedHashMap and sets its count to 1.
• After iterating through all the characters in the input string, a StringBuffer named “sb” is created.
• The program iterates over each entry (key-value pair) in the “maps” LinkedHashMap using a for-each loop.
• Within the loop, it appends the character and its count to the StringBuffer “sb”.
• Finally, the program prints the StringBuffer “sb” as the compressed output string.

Through this program, we achieve the goal of transforming the input string “abbcccdeee” into its compressed representation “a1b2c3d1e3” by counting the occurrences of each character and appending their counts to the respective characters.

Explanation:

package com.softwaretestingo.interviewprograms;
public class InterviewPrograms11_1 
{
	/*
	 * Input string =abbcccdeee 
	 * Output= a1b2c3d1e3
	 */
	public static void main(String[] args) 
	{
		StringBuilder sb = new StringBuilder();
		String str = "abbcccdeee";
		String[] strArray = str.split("");
		int i = 0 ;
		while ( i < strArray.length ) 
		{
			int counter=0 ;
			int j = i ;
			while ( j < strArray . length ) 
			{
				if ( ! strArray [ i ] .equals ( strArray [ j ] ) )
					break;
				counter++ ;
				j++;
			}
			sb.append ( strArray [ i ] ) . append ( counter ) ;
			i = j ;
		}
		System.out.println (sb.toString());
	}
}

Create a StringBuilder: We create a StringBuilder object named sb to construct the final output string.

StringBuilder sb = new StringBuilder();

Initialize the input string: We assign the string value “abbcccdeee” to the variable str. This serves as the input for our program.

Split the input string into an array: We split the input string into an array of individual characters using the split(“”) method. Each character is stored as a separate element in the strArray.

String[] strArray = str.split("");

Process the input array and generate the output string: We use nested loops to iterate through the input array strArray. The outer loop iterates over each character, and the inner loop counts the consecutive occurrences of the current character. Once a different character is encountered, we break out of the inner loop. The character and its count are appended to the StringBuilder sb. The outer loop then continues from the next distinct character.

int i = 0;
while (i < strArray.length) {
    int counter = 0;
    int j = i;
    while (j < strArray.length) {
        if (!strArray[i].equals(strArray[j]))
            break;
        counter++;
        j++;
    }
    sb.append(strArray[i]).append(counter);
    i = j;
}

Print the output string: We print the generated output string by converting the StringBuilder sb to a string using the toString() method.

System.out.println(sb.toString());

In summary, this program takes an input string, counts the consecutive occurrences of each character, and generates an output string by concatenating each character with its respective count. The program demonstrates the usage of loops, string manipulation, and the StringBuilder class in Java.

package com.softwaretestingo.interviewprograms;
public class InterviewPrograms38 
{
	/*
	 * Input = AAABBBBCC 
	 * Output = A3B4C2
	 */
	public static void main(String[] args) 
	{
		String str = "AAABBBBCC";
		int length = str.length ( ) ;

		// Created an object of a StringBuffer class .
		StringBuffer sb= new StringBuffer();
		int count = 1; // counter for counting number of occurrences

		// input String
		// length of a String .
		for ( int i=0; i<length; i++)
		{
			// if i reaches at the end then append all and break the loop
			if (i == length-1)
			{
				sb.append(str.charAt(i)+""+count);
				break ;
				
			}

			// if two successive chars are equal then increase the
			if (str.charAt(i) == str.charAt(i+1))
			{
				count ++ ;
			}
			else 
			{
				// else append character with its count
				sb.append(str.charAt(i)+""+count);
				count = 1 ; // resetting the counter
			}
		}
		// String representation of a StringBuffer object
		System.out.println(sb.toString());
	}
}

Output

A3B4C2