Compare Version Numbers#

Problem statement#

[1]Given two version numbers, version1 and version2, your task is to compare them.

Version numbers consist of one or more revisions joined by a dot '.'. Each revision is composed of digits and may contain leading zeros. Each revision has at least one character. Revisions are indexed from left to right, with the leftmost revision being revision 0, the next revision being revision 1, and so on.

For instance, 2.5.33 and 0.1 are valid version numbers.

To compare version numbers, you should 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 a particular index, treat that 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 is less than 1.

The function should return the following:

  • If version1 is less than version2, return -1.

  • If version1 is greater than version2, return 1.

  • If version1 and version2 are equal, return 0.

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.

Solution#

Each version can be considered as an array of revisions.

version = revisions[0].revisions[1].revisions[2]....

The problem is to compare each revisions[i] between two versions.

For example, revisions[0] of version1 is less than of version2 in Example 3. So the result is -1.

All revisions[i] of version1 and version2 are equal in Example 1. So the result is 0.

The number of revisions between the versions might not be equal (like in Example 2).

If all revisions of the shorter version are equal to the corresponding revisions of the longer one, the version having extra revisions and there exists a non-zero revision among them is the bigger one. Otherwise, the two versions are equal.

Code#

#include <iostream>
#include <vector>
#include <string>
#include <numeric>
using namespace std;
//! @return the vector of revisions of the version 
//! @example if version = "1.02.11", return {1,2,11}
vector<int> toVector(const string& version) {
    vector<int> revisions;
    string revision;
    for (auto& c : version) {
        if (c != '.') {
            // continue to build current revision 
            revision += c;
        } else {
            // current revision completes
            // uses stoi() to ignore leading zeros
            revisions.push_back(stoi(revision));

            // start a new revision
            revision = "";
        }
    }
    revisions.push_back(stoi(revision));
    return revisions;
}

int compareVersion(const string& version1, const string& version2) { 
    vector<int> r1 = toVector(version1);    
    vector<int> r2 = toVector(version2);

    int i = 0;
    // perform the comparison on the revisions
    while (i < r1.size() && i < r2.size()) {
        if (r1[i] < r2[i]) {
            return -1;
        } else if (r1[i] > r2[i]) {
            return 1;
        }
        i++;
    }
    if (i == r1.size()) {
        // if version1 is not longer than version2
        // and version2 still has some valid revisions remain
        if (accumulate(r2.begin() + i, r2.end(), 0) > 0) {
            return -1;
        }
    } else if (accumulate(r1.begin() + i, r1.end(), 0) > 0) {
        // if version2 is not longer than version1
        // and version1 still has some valid revisions remain
        return 1;
    }
    return 0;
}
int main() {
    cout << compareVersion("1.01", "1.001") << endl;
    cout << compareVersion("1.0", "1.0.0") << endl;
    cout << compareVersion("0.1", "1.1") << endl;
}

Output:
0
0
-1

Complexity#

  • Runtime: O(N) where N = max(version1.length, version2.length).

  • Extra space: O(N).

Conclusion#

This solution first converts the version strings into vectors of integers representing the individual components of the version numbers. This conversion is done by iterating through each character of the version string, accumulating digits until encountering a dot, at which point the accumulated integer is added to the revisions vector.

Once both version strings are converted into vectors, the function iterates through the vectors, comparing corresponding elements to determine the relationship between the versions. Additionally, it accounts for any remaining digits in the longer version string after the common components by summing them up and comparing the totals.

This approach simplifies the comparison process by breaking down the version strings into easily comparable components.

C++ Notes#

  • std::stoi[2] is used to convert a string to an int. It ignores the leading zeros for you.

  • std::accumulate[3] is used to compute the sum of a container.