Java Arrays and Strings - Programs, Exercise & Assignements

Reverse a string

Description: Reverse the text "Java" to "avaJ". Do it with a loop and with StringBuilder.

Hint:

Go from the last index to 0 and join the characters.
StringBuilder has a reverse() method.

Solution

public class Main {
    static String reverseLoop(String s) {
        String result = "";
        for (int i = s.length() - 1; i >= 0; i--) {
            result += s.charAt(i);
        }
        return result;
    }

    public static void main(String[] args) {
        System.out.println(reverseLoop("Java"));
        System.out.println(new StringBuilder("Java").reverse());
    }
}

Output

avaJ
avaJ

How it works

charAt(i) gives one character. The loop reads them from the end.
StringBuilder.reverse() is the shortest way.

Read Java String Methods.

Check if a string is a palindrome

Description: A palindrome reads the same from both sides, like madam. Write isPalindrome(String s) which ignores the case and the spaces. Test it with “Madam”, “Nurses run” and “Java”.

Hint:

Clean the text with replace(" ", "") and toLowerCase(), then compare it with its reverse.

Solution

public class Main {
    static boolean isPalindrome(String s) {
        String clean = s.replace(" ", "").toLowerCase();
        String reversed = new StringBuilder(clean).reverse().toString();
        return clean.equals(reversed);
    }

    public static void main(String[] args) {
        for (String text : new String[]{"Madam", "Nurses run", "Java"}) {
            System.out.println(text + " -> " + isPalindrome(text));
        }
    }
}

Output

Madam -> true
Nurses run -> true
Java -> false

How it works

We remove the spaces and use one case, so only the letters are compared.
equals() compares the text. Do not use ==.

Read Java Strings.

Count the vowels in a string

Description: Count how many vowels (a, e, i, o, u) and how many consonants are in "Learn Java Programming".

Hint:

Use toCharArray(), Character.isLetter() and indexOf on the string "aeiou".

Solution

public class Main {
    public static void main(String[] args) {
        String text = "Learn Java Programming";
        int vowels = 0, consonants = 0;

        for (char c : text.toLowerCase().toCharArray()) {
            if (!Character.isLetter(c)) continue;          // skip the spaces
            if ("aeiou".indexOf(c) >= 0) {
                vowels++;
            } else {
                consonants++;
            }
        }
        System.out.println("Vowels: " + vowels + ", consonants: " + consonants);
    }
}

Output

Vowels: 7, consonants: 13

How it works

"aeiou".indexOf(c) gives -1 when the letter is not a vowel.
continue skips the spaces.

Read Java String Methods.

Find the second largest number in an array

Description: Find the second largest number in {12, 35, 1, 10, 34, 1} without sorting the array.

Hint:

Keep two variables: largest and second. Update both in one loop.

Solution

public class Main {
    static int secondLargest(int[] a) {
        int largest = Integer.MIN_VALUE, second = Integer.MIN_VALUE;
        for (int n : a) {
            if (n > largest) {
                second = largest;
                largest = n;
            } else if (n > second && n != largest) {
                second = n;
            }
        }
        return second;
    }

    public static void main(String[] args) {
        System.out.println(secondLargest(new int[]{12, 35, 1, 10, 34, 1}));
        System.out.println(secondLargest(new int[]{5, 5, 5, 2}));
    }
}

Output

34
2

How it works

When a new biggest value comes, the old biggest becomes the second.
n != largest ignores the duplicates of the biggest value.

Read Java Arrays.

Sort an array with bubble sort

Description: Sort {64, 34, 25, 12, 22, 11, 90} in ascending order without using Arrays.sort(). Then check your answer with Arrays.sort().

Hint:

Compare each pair of neighbours and swap them if they are in the wrong order. Repeat.

Solution

import java.util.Arrays;

public class Main {
    static void bubbleSort(int[] a) {
        for (int i = 0; i < a.length - 1; i++) {
            boolean swapped = false;
            for (int j = 0; j < a.length - 1 - i; j++) {
                if (a[j] > a[j + 1]) {
                    int temp = a[j];
                    a[j] = a[j + 1];
                    a[j + 1] = temp;
                    swapped = true;
                }
            }
            if (!swapped) break;           // already sorted
        }
    }

    public static void main(String[] args) {
        int[] data = {64, 34, 25, 12, 22, 11, 90};
        int[] check = data.clone();

        bubbleSort(data);
        Arrays.sort(check);
        System.out.println(Arrays.toString(data));
        System.out.println(Arrays.equals(data, check));
    }
}

Output

[11, 12, 22, 25, 34, 64, 90]
true

How it works

After each round of the outer loop, the biggest remaining value goes to its place at the end.
If a full round makes no swap, the array is sorted and we stop.
Bubble sort is good to learn, but slow (O(n squared)). Real code uses Arrays.sort().

Read Java Arrays.

Transpose a matrix

Description: Transpose the matrix {{1,2,3},{4,5,6}}. The result must be {{1,4},{2,5},{3,6}}.

Hint:

The new matrix has columns rows and rows columns. Copy m[i][j] to t[j][i].

Solution

import java.util.Arrays;

public class Main {
    static int[][] transpose(int[][] m) {
        int rows = m.length, cols = m[0].length;
        int[][] t = new int[cols][rows];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                t[j][i] = m[i][j];
            }
        }
        return t;
    }

    public static void main(String[] args) {
        int[][] m = {{1, 2, 3}, {4, 5, 6}};
        System.out.println(Arrays.deepToString(transpose(m)));
    }
}

Output

[[1, 4], [2, 5], [3, 6]]

How it works

A 2 x 3 matrix becomes a 3 x 2 matrix, so a new array with the sizes swapped is needed.

Read Multidimensional Arrays.