How to Alphabetically Compare Two Strings in Java During Binary Search?

How can I compare two strings in Java to determine which is smaller alphabetically?

String userInput = "abcda";
String fileString = "abcza";
int comparisonResult = userInput.compareTo(fileString);

In Java, comparing two strings alphabetically is straightforward using the built-in compareTo() method. This method allows you to determine the lexicographical order of the strings, which is crucial for operations like binary search in sorted collections or files.

public class BinarySearchExample {
    public static void main(String[] args) {
        String userInput = "abcda";
        String[] sortedArray = {"abcaa", "abcab", "abcac", "abcad", "abcza"};
        int result = binarySearch(sortedArray, userInput);
        if (result >= 0) {
            System.out.println("String found at index: " + result);
        } else {
            System.out.println("String not found.");
        }
    }

    public static int binarySearch(String[] array, String key) {
        int low = 0;
        int high = array.length - 1;
        while (low <= high) {
            int mid = (low + high) / 2;
            int comparison = key.compareTo(array[mid]);
            if (comparison < 0) {
                high = mid - 1;
            } else if (comparison > 0) {
                low = mid + 1;
            } else {
                return mid; // Key found
            }
        }
        return -(low + 1); // Key not found
    }
}

Causes

• Strings are compared based on Unicode values of each character.
• Using compareTo() handles the logic for alphabetical order automatically.

Solutions

• Use String's compareTo() method to compare two strings: `string1.compareTo(string2)` returns a negative integer, zero, or a positive integer if the first string is less than, equal to, or greater than the second string, respectively.
• In a binary search within a sorted file, retrieve the string from the middle and compare it with the user input string using compareTo(). Based on the result, decide whether to continue searching in the left or right half.

Mistake: Not using compareTo() method, instead trying to compare strings with relational operators (e.g., <, >).

Solution: Always use compareTo() for string comparison to avoid errors.

Mistake: Assuming strings are in the same encoding without checking.

Solution: Ensure input strings are in the same character encoding before comparison.