Compare the two versions, version1, version2.
If version1 > version2 return 1
If version1 < version2 return -1
if version1 = version2 return 0
version strings are non-empty and contain only digits and the ‘.’ character. The ‘.’ character does not represent a decimal point and is used to separate number sequences.
Example of version ordering.
0.1 < 1.1 < 1.2 < 1.13 < 1.13.4
Note : Here the numbers present inside the string can be huge so don’t try to convert these numbers
to unsigned long long. Eg. version1 = 1.234565434523423423523423423423434432.23.0
Input : version1 : 002.0005.12.3 version2 : 126.96.36.199 Output : 0 Input : version1 : 451231654684151546847799885544662 version2 : 1.2188.8.131.52 Output : 1 Input : version1 : 1.21.20 version2 : 1.21.25 Output : -1 Input : version1 : 1.2 version2 : 184.108.40.206.0 Output : 0 Input : version1 : 1.2 version2 : 1.0.1 Output : -1
We have discussed a solution in below post.
Compare two Version numbers
The previous solution discussed above has problems like, it does not handle leading zeros and does not work for large numbers as individual parts of version numbers are stored as int.
In this solution, above issues are addressed. We traverse the both versions at same time and process them until both of them gets fully traversed. Store the numbers from version1 and version 2 in different strings i.e substr_version1 and substr_version2. Compare these substrings,
if length of substr_version1 > substr_version2 the clearly substr_version1 is greater in value so return +1. Similar is case when substr_version2 > substr_version1, we will return -1. But if both substrings are similar in length then
we will have to check each character from both substrings and then compare those characters and then return the result appropriately.
Time Complexity : O(2 * N) –> O(N)
Here worst case is when both the versions are equal so after extracting both substrings each of them will again be compared in compareSubstr() function which will take the complexity upto (2 * N).
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