/**
* 165. Compare Version Numbers
* this is a great question. shows that thinking thoroughly in the beginning of solving a problem is so critical.
* Given two version numbers, version1 and version2, compare them.
* Version numbers consist of one or more revisions joined by a dot ‘.’. Each revision consists of digits and may
* contain leading zeros. Every revision contains at least one character. Revisions are 0-indexed from left to right,
* with the leftmost revision being revision 0, the next revision being revision 1,
* and so on. For example 2.5.33 and 0.1 are valid version numbers.
* To compare version numbers, compare their revisions in left-to-right order. Revisions are compared using
* their integer value ignoring any leading zeros. This means that revisions 1 and 001 are considered equal.
* If a version number does not specify a revision at an index, then treat the revision as 0. For example,
* version 1.0 is less than version 1.1 because their revision 0s are the same,
* but their revision 1s are 0 and 1 respectively, and 0 < 1.
* Return the following:
* If version1 < version2, return -1.
* If version1 > version2, return 1.
* Otherwise, return 0.
* Example 1:
* Input: version1 = “1.01”, version2 = “1.001”
* Output: 0
* Explanation: Ignoring leading zeroes, both “01” and “001” represent the same integer “1”.
*/
See solution to the problem at github here: https://github.com/zcoderz/leetcode/blob/main/src/main/java/amazon/medium/CompareVersions.java
The reason to include a solution to this problem here is that this question can be very tricky to implement if we try to solve it per case basis rather than the below clean solution. The below approach is extremely simple and elegant and hence I included the question here.
Because versions could be constructed as 1.0.0.0.0.0.0.1 & 1.0.0.1 or they can be constructed as 1.0.0.0.0.10 vs 1.0.0.0.0.9 or they could be 1.1.0 and 1.1.0.0.0.0 which are same you need an algorithm as below:
1. Divide the version into chunks split on ‘.’
2. Per version chunk, do integer comparison instead of string comparison
3. Pad the version with smaller chunks with a 0 when its chunk terminates. This approach simplifies the code.
Here is the code:
public int compareVersion(String version1, String version2) {
String[] s1 = version1.split("\\.");
String[] s2 = version2.split("\\.");
int oneSz = s1.length;
int twoSz = s2.length;
int maxSz = Math.max(oneSz, twoSz);
for (int i = 0; i < maxSz; i++) {
int a = i < oneSz ? Integer.parseInt(s1[i]) : 0;
int b = i < twoSz ? Integer.parseInt(s2[i]) : 0;
if (a != b) {
return a > b ? 1 : -1;
}
}
return 0;
}