Compare Version Numbers in Python

Suppose we have to compare two version numbers version1 and version2. If the version1 > version2 then return 1; otherwise when version1 < version2 return -1; otherwise return 0. We can assume that the version strings are non-empty and contain only digits and the dot (.) characters. The dot character does not represent a decimal point and is used to separate number sequences. So for example, 2.5 is not "two and a half" or "halfway to version three", it is the fifth second-level revision of the second first-level revision.

We can assume the default revision number for each level of a version number to be 0. For example, the version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

So if the input is like version1 = “1.0.1” and version2 = “1”, then it will return +1.

To solve this, we will follow these steps −

  • version1_arr = an array of numbers separated by dot for version1

  • version2_arr = an array of numbers separated by dot for version2

  • for i in range 0 to max of size of version1_arr and size of version2_arr −

    • v1 := version1_arr[i] if i < size of version1_arr, otherwise 0

    • v2 := version2_arr[i] if i < size of version2_arr, otherwise 0

    • if v1 > v2, then return 1, otherwise when v1 < v2, return -1

  • return 0

Example (Python)

Let us see the following implementation to get better understanding −

 Live Demo

class Solution:
   def compareVersion(self, version1, version2):
      versions1 = [int(v) for v in version1.split(".")]
      versions2 = [int(v) for v in version2.split(".")]
      for i in range(max(len(versions1),len(versions2))):
         v1 = versions1[i] if i < len(versions1) else 0
         v2 = versions2[i] if i < len(versions2) else 0
         if v1 > v2:
            return 1
         elif v1 <v2:
            return -1
      return 0
ob1 = Solution()





Updated on: 02-May-2020

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