# LeetCode - Compare Version Numbers

## Problem statement

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.

Problem statement taken from: https://leetcode.com/problems/compare-version-numbers

Example 1:

``````Input: version1 = '1.01', version2 = '1.001'
Output: 0
Explanation: Ignoring leading zeroes, both '01' and '001' represent the same integer '1'.
``````

Example 2:

``````Input: version1 = '1.0', version2 = '1.0.0'
Output: 0
Explanation: version1 does not specify revision 2, which means it is treated as '0'.
``````

Example 3:

``````Input: version1 = '0.1', version2 = '1.1'
Output: -1
Explanation: version1's revision 0 is '0', while version2's revision 0 is '1'. 0 < 1, so version1 < version2.
``````

Constraints:

``````- 1 <= version1.length, version2.length <= 500
- version1 and version2 only contain digits and '.'
- version1 and version2 are valid version numbers.
- All the given revisions in version1 and version2 can be stored in a 32-bit integer.
``````

### Explanation

We cannot compare the strings because of the dots. We traverse through the strings and separate numeric parts. If the numeric parts are equal, we compare the next numeric part.

Let's check the algorithm now.

``````- set vnum1 = 0, vnum2 = 0

- loop for i = 0, j = 0 && i < version1.length && j < version2.length
- loop while i < version1.length && version1[i] != '.'
- set vnum1 = vnum1 * 10 + (version1[i] - '0')
- increment i++

- loop while j < version2.length && version2[j] != '.'
- set vnum2 = vnum2 * 10 + (version2[j] - '0')
- increment j++

- if vnum1 > vnum2
- return 1
- else if vnum2 > vnum1
- return -1

- set vnum1 = 0, vnum2 = 0
- increment i++
- increment j++

- for end

- return 0
``````

This function's time complexity is O(n), and the space complexity is O(1).

Let's check our solutions in C++, Golang, and Javascript.

#### C++ solution

``````class Solution {
public:
int compareVersion(string version1, string version2) {
int vnum1 = 0, vnum2 = 0;

for(int i = 0, j = 0; i < version1.length() || j < version2.length();) {
while(i < version1.length() && version1[i] != '.') {
vnum1 = vnum1*10 + (version1[i] - '0');
i++;
}

while(j < version2.length() && version2[j] != '.') {
vnum2 = vnum2*10 + (version2[j] - '0');
j++;
}

if(vnum1 > vnum2) {
return 1;
} else if(vnum2 > vnum1) {
return -1;
}

vnum1 = 0;
vnum2 = 0;
i++;
j++;
}

return 0;
}
};``````

#### Golang solution

``````func compareVersion(version1 string, version2 string) int {
vnum1, vnum2 := 0, 0
i, j := 0, 0

for i < len(version1) || j < len(version2) {
for i < len(version1) && version1[i] != '.' {
vnum1 = vnum1 * 10 + int(version1[i] - '0')
i++
}

for j < len(version2) && version2[j] != '.' {
vnum2 = vnum2 * 10 + int(version2[j] - '0')
j++
}

if vnum1 > vnum2 {
return 1
} else if vnum2 > vnum1 {
return -1
}

vnum1, vnum2 = 0, 0
i++
j++
}

return 0;
}``````

#### Javascript solution

``````var compareVersion = function(version1, version2) {
let vnum1 = 0, vnum2 = 0;

for(let i = 0, j = 0; i < version1.length || j < version2.length;) {
while(i < version1.length && version1[i] != '.') {
vnum1 = vnum1 * 10 + parseInt(version1[i] - '0');
i++;
}

while(j < version2.length && version2[j] != '.') {
vnum2 = vnum2 * 10 + parseInt(version2[j] - '0');
j++;
}

if(vnum1 > vnum2) {
return 1;
} else if(vnum2 > vnum1) {
return -1;
}

vnum1 = 0, vnum2 = 0;
i++;
j++;
}

return 0;
};``````

#### Dry Run

Let's dry-run our algorithm to see how the solution works.

``````Input: version1 = '1.01'
version2 = '1.001'

Step 1: vnum1 = 0, vnum2 = 0

Step 2: loop for i = 0, j = 0; i < version1.length() || j < version2.length()
i < version1.length() || j < version2.length()
0 < 4 || 0 < 5
true

loop while i < version1.length() && version1[i] != '.'
0 < 4 && version1[0] != '.'
true && '1' != '.'
true

vnum1 = vnum1 * 10 + (version1[i] - '0')
= 0 * 10 + version1[0] - '0'
= 0 + 1
= 1

i++
i = 1

i < version1.length() && version1[i] != '.'
1 < 4 && version1[1] != '.'
true && '.' != '.'
false

loop while j < version2.length() && version2[j] != '.'
0 < 5 && version2[0] != '.'
true && '1' != '.'
true

vnum1 = vnum1 * 10 + (version2[j] - '0')
= 0 * 10 + version2[j] - '0'
= 0 + 1
= 1

j++
j = 1

if vnum1 > vnum2
1 > 1
false
else if vnum2 > vnum1
1 > 1
false

vnum1 = 0
vnum2 = 0

i++
i = 2

j++
j = 2

Step 3: loop for i < version1.length() || j < version2.length()
i < version1.length() || j < version2.length()
2 < 4 || 2 < 5
true

loop while i < version1.length() && version1[i] != '.'
2 < 4 && version1[2] != '.'
true && '0' != '.'
true

vnum1 = vnum1 * 10 + (version1[i] - '0')
= 0 * 10 + version1[2] - '0'
= 0 + 0
= 0

i++
i = 3

i < version1.length() && version1[i] != '.'
3 < 4 && version1[3] != '.'
true && '1' != '.'
true

vnum1 = vnum1 * 10 + (version1[i] - '0')
= 0 * 10 + version1[3] - '0'
= 0 + 1
= 0

i++
i = 4

i < version1.length()
4 < 4
false

loop while j < version2.length() && version2[j] != '.'
2 < 5 && version2[2] != '.'
true && '0' != '.'
true

vnum2 = vnum2 * 10 + (version2[j] - '0')
= 0 * 10 + version2[2] - '0'
= 0 + 0
= 0

j++
j = 3

j < version2.length() && version2[j] != '.'
3 < 5 && version2[3] != '.'
true && '0' != '.'
true

vnum2 = vnum2 * 10 + (version2[j] - '0')
= 0 * 10 + version2[3] - '0'
= 0 + 1
= 0

j++
j = 4

j < version2.length() && version2[j] != '.'
4 < 5 && version2[4] != '.'
true && '1' != '.'
true

vnum2 = vnum2 * 10 + (version2[j] - '0')
= 0 * 10 + version2[4] - '0'
= 0 + 1
= 1

j++
j = 5

j < version2.length()
5 < 5
false

if vnum1 > vnum2
1 > 1
false
else if vnum2 > vnum1
1 > 1
false

vnum1 = 0
vnum2 = 0

i++
i = 5

j++
j = 6

Step 4: loop for i < version1.length() || j < version2.length()
5 < 4 || 6 < 5
false

Step 5:  return 0

We return the answer as 0.
``````