In this article, we will discuss how we can compare two strings lexicographically in Java. One solution is to use Java compareTo() method. The method compareTo() is used for comparing two strings lexicographically in Java. Each character of both the strings is converted into a Unicode value for comparison.
int compareTo(String str) :
It returns the following values:
- if (string1 > string2) it returns a positive value.
- if both the strings are equal lexicographically i.e.(string1 == string2) it returns 0.
- if (string1 < string2) it returns a negative value.
Java
// Java program to show how to compare Strings // using library function public class Test { public static void main(String[] args) { String s1 = "Ram"; String s2 = "Ram"; String s3 = "Shyam"; String s4 = "ABC"; System.out.println(" Comparing strings with compareTo:"); System.out.println(s1.compareTo(s2)); System.out.println(s1.compareTo(s3)); System.out.println(s1.compareTo(s4)); } } |
Comparing strings with compareTo: 0 -1 17
Time Complexity: O(N)
Auxiliary Space: O(1)
Refer How to Initialize and Compare Strings in Java? for more details. How to compare two strings without using library function?
1. Input two strings string 1 and string 2.
2. for (int i = 0; i < str1.length() &&
i < str2.length(); i ++)
(Loop through each character of both
strings comparing them until one
of the string terminates):
a. If unicode value of both the characters
is same then continue;
b. If unicode value of character of
string 1 and unicode value of string 2
is different then return (str1[i]-str2[i])
3. if length of string 1 is less than string2
return str2[str1.length()]
else
return str1[str2.length()]
Below is the implementation of the above algorithm.
Java
// Java program to Compare two strings // lexicographically class Compare { // This method compares two strings // lexicographically without using // library functions public static int stringCompare(String str1, String str2) { for (int i = 0; i < str1.length() && i < str2.length(); i++) { if ((int)str1.charAt(i) == (int)str2.charAt(i)) { continue; } else { return (int)str1.charAt(i) - (int)str2.charAt(i); } } // Edge case for strings like // String 1="Geeky" and String 2="Geekyguy" if (str1.length() < str2.length()) { return (str1.length()-str2.length()); } else if (str1.length() > str2.length()) { return (str1.length()-str2.length()); } // If none of the above conditions is true, // it implies both the strings are equal else { return 0; } } // Driver function to test the above program public static void main(String args[]) { String string1 = new String("Geeks"); String string2 = new String("Practice"); String string3 = new String("Geeks"); String string4 = new String("Geeksforneveropen"); System.out.println(stringCompare(string1, string2)); System.out.println(stringCompare(string1, string3)); System.out.println(stringCompare(string2, string1)); // To show for edge case // In these cases, the output is the difference of // length of the string System.out.println(stringCompare(string1, string4)); System.out.println(stringCompare(string4, string1)); } } |
-9 0 9 -8 8
Time Complexity: O(n)
Auxiliary Space: O(1)
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