<filename>medium/165-Compare Version Numbers.py<gh_stars>1-10
"""
https://leetcode.com/problems/compare-version-numbers/
Compare two version numbers version1 and version2.
If version1 > version2 return 1; if version1 < version2 return -1;otherwise return 0.

You may assume that the 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.

For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

You may assume the default revision number for each level of a version number to be 0. For example, 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.

 

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1
Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1
Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1
Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”
Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"
 

Note:

Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
Version strings do not start or end with dots, and they will not be two consecutive dots.
"""

# time complexity: O(n+m), space complexity: O(n)

class Solution:
    def compareVersion(self, version1: str, version2: str) -> int:
        first = version1.split('.')
        second = version2.split('.')
        import itertools
        for a, b in itertools.zip_longest(first, second, fillvalue=0):
            if int(a) > int(b):
                return 1
            elif int(a) < int(b):
                return -1
        return 0

        # second solution that doesn't use itertools.zip_longest
        first = [0]
        second = [0]
        for letter in version1:
            if '0' <= letter <= '9':
                first[-1] = first[-1]*10+int(letter)
            else:
                first.append(0)
        
        for letter in version2:
            if '0' <= letter <= '9':
                second[-1] = second[-1] * 10 + int(letter)
            else:
                second.append(0)
        
        i = 0
        while i < max(len(first), len(second)):
            num1 = 0 if i >= len(first) else first[i]
            num2 = 0 if i >= len(second) else second[i]
            if num1 < num2:
                return -1
            elif num1 > num2:
                return 1
            i += 1
        return 0

<filename>medium/165-Compare Version Numbers.py<gh_stars>1-10
"""
https://leetcode.com/problems/compare-version-numbers/
Compare two version numbers version1 and version2.
If version1 > version2 return 1; if version1 < version2 return -1;otherwise return 0.

You may assume that the 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.

For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

You may assume the default revision number for each level of a version number to be 0. For example, 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.

 

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1
Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1
Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1
Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”
Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"
 

Note:

Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
Version strings do not start or end with dots, and they will not be two consecutive dots.
"""

# time complexity: O(n+m), space complexity: O(n)

class Solution:
    def compareVersion(self, version1: str, version2: str) -> int:
        first = version1.split('.')
        second = version2.split('.')
        import itertools
        for a, b in itertools.zip_longest(first, second, fillvalue=0):
            if int(a) > int(b):
                return 1
            elif int(a) < int(b):
                return -1
        return 0

        # second solution that doesn't use itertools.zip_longest
        first = [0]
        second = [0]
        for letter in version1:
            if '0' <= letter <= '9':
                first[-1] = first[-1]*10+int(letter)
            else:
                first.append(0)
        
        for letter in version2:
            if '0' <= letter <= '9':
                second[-1] = second[-1] * 10 + int(letter)
            else:
                second.append(0)
        
        i = 0
        while i < max(len(first), len(second)):
            num1 = 0 if i >= len(first) else first[i]
            num2 = 0 if i >= len(second) else second[i]
            if num1 < num2:
                return -1
            elif num1 > num2:
                return 1
            i += 1
        return 0

https://leetcode.com/problems/compare-version-numbers/ Compare two version numbers version1 and version2. If version1 > version2 return 1; if version1 < version2 return -1;otherwise return 0.  You may assume that the 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.  For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.  You may assume the default revision number for each level of a version number to be 0. For example, 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.     Example 1:  Input: version1 = "0.1", version2 = "1.1" Output: -1 Example 2:  Input: version1 = "1.0.1", version2 = "1" Output: 1 Example 3:  Input: version1 = "7.5.2.4", version2 = "7.5.3" Output: -1 Example 4:  Input: version1 = "1.01", version2 = "1.001" Output: 0 Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1” Example 5:  Input: version1 = "1.0", version2 = "1.0.0" Output: 0 Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"    Note:  Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes. Version strings do not start or end with dots, and they will not be two consecutive dots. # time complexity: O(n+m), space complexity: O(n) # second solution that doesn't use itertools.zip_longest