Compare Version Numbers

To solve this coding challenge, the main task is to compare two version strings (

version1

and

version2

). Each version string consists of revisions separated by dots

'.'

. To compare the version strings, consider each revision value as an integer, ignoring any leading zeros. If one version string has fewer revisions than the other, treat the missing revisions as

0

.

Explanation

1. Input Parsing :
We begin by parsing the input version strings
version1
and
version2
. This involves splitting each version string by the
'.'
character. This will give us lists of revision components for both version strings.
2. Normalization of Revisions :
Since the two version strings can have different numbers of revisions, we need to pad the shorter list with zeros to make both lists of equal length. This ensures we can easily compare corresponding revisions without running into index errors.
3. Comparison of Revisions :
Starting from the leftmost revision, compare the corresponding revisions of both version strings as integers. The comparison rules are:
• If revision from
  version1
  is less than the corresponding revision from
  version2
  , return
  -1
  .
• If revision from
  version1
  is greater than the corresponding revision from
  version2
  , return
  1
  .
• If they are equal, move to the next revision and continue the comparison.
4. Final Decision :
If all corresponding revisions are equal after iterating through both lists, the version strings are considered equal, and we return
0
.
These steps ensure a thorough and accurate comparison between the two version strings as per the problem requirements.

Detailed Steps in Pseudocode

5. Split the Input Strings :
• Split
  version1
  by
  '.'
  to get
  revisions1
  .
• Split
  version2
  by
  '.'
  to get
  revisions2
  .
6. Determine Maximum Length :
• Find the maximum length of
  revisions1
  and
  revisions2
  .
7. Normalize Revision Lists :
• Iterate through a range from
  0
  to the maximum length.
• For each index, if the index exists in
  revisions1
  , use its value, otherwise use
  0
  .
• Similarly, if the index exists in
  revisions2
  , use its value, otherwise use
  0
  .
8. Compare Corresponding Revisions :
• Convert revision strings to integers and compare.
• Return
  -1
  ,
  1
  , or continue based on comparison results.
9. Return
   0
   if All Revisions are Equal .

Pseudocode

                                            
# Split the version strings into lists of revision components
revisions1 = version1.split('.')  # Split version1 by '.'
revisions2 = version2.split('.')  # Split version2 by '.'

# Determine the maximum length between the two lists of revisions
max_length = max(len(revisions1), len(revisions2))  # Length to standardize both lists

# Iterate through each revision component up to the maximum length
for i in range(max_length):
# Fetch the revision for version1, or default to '0' if beyond list length
revision1_value = int(revisions1[i]) if i < len(revisions1) else 0

# Fetch the revision for version2, or default to '0' if beyond list length
revision2_value = int(revisions2[i]) if i < len(revisions2) else 0

# Compare the two revisions
if revision1_value < revision2_value:
return -1  # version1 is less than version2
elif revision1_value > revision2_value:
return 1  # version1 is greater than version2

# If all revisions are equal, the versions are considered equal
return 0

                                        

In this pseudocode:
10. The
    split
    function call splits the version strings into lists of their respective revision components.
11. The
    max
    function call determines the maximum length of the lists to handle cases where versions may have different numbers of revisions.
12. A
    for
    loop iterates through each revision, and the
    if
    statements ensure appropriate comparison and handling of missing revisions by treating them as
    0
    .
13. Finally, based on the comparisons, the function returns
    -1
    ,
    1
    , or
    0
    as required by the problem statement.