Comparing version numbers with JavaScript

Question

I have written a function that compares two version number strings (e.g. 1.2.0 and 1.2.2) and return 1 if the first string is greater, -1 if the second string is greater and 0 if the strings are equal for a code challenge that I'm attempting.

Also, it's guaranteed that both strings contain an equal number of numeric fields and all decimals are non-negative.

So,

• 1.2.0 and 1.2.2 should return -1.
• 1.05.4 and 1.5.3 should return 1.
• 1.2.6.76 and 1.2.06.0076 should return 0.

---

My logic for the function was to simply split the string at each occurrence of the period ., compare each number and then add a character (e for equal, m for more and l for less) to a temporary variable z depending on the comparison result.

I then simply use some basic regex to return 1, -1 or 0 based on the content of the z variable as can be seen in the following Code Snippet:

function checkVersion(a,b) {
    let x=a.split('.').map(e=> parseInt(e));
    let y=b.split('.').map(e=> parseInt(e));
    let z = "";

    for(i=0;i<x.length;i++) {
        if(x[i] === y[i]) {
            z+="e";
        } else
        if(x[i] > y[i]) {
            z+="m";
        } else {
            z+="l";
        }
    }
    if (!z.match(/[l|m]/g)) {
      return 0;
    } else if (!z.match(/[l]/g)) {
      return 1;
    } else {
      return -1;
    }
}

console.log(checkVersion("1.2.2","1.2.0")); // returns 1 as expected
console.log(checkVersion("1.0.5","1.1.0")); // returns -1 as expected
console.log(checkVersion("1.0.5","1.00.05")); // returns 0 as expected
console.log(checkVersion("0.9.9.9.9.9.9","1.0.0.0.0.0.0")) // returns -1 as expected;

The above function seems to be working fine with any random two version numbers that I've tried so far but when I try to submit the above function for the challenge, there is always one hidden test with two unknown version numbers that keeps failing. What logic am I missing in the above code?

---

EDIT:

Thanks to @RomanPerekhrest's comment below, I have found out that my regex is the problem. Instead of using the 2nd regex, I just remove any occurence of e from the z variable using the split() method and then just check if the first character is m or l and now the function is working correctly as seen in the following Code Snippet:

function checkVersion(a,b) {
    let x=a.split('.').map(e=> parseInt(e));
    let y=b.split('.').map(e=> parseInt(e));
    let z = "";

    for(i=0;i<x.length;i++) {
        if(x[i] === y[i]) {
            z+="e";
        } else
        if(x[i] > y[i]) {
            z+="m";
        } else {
            z+="l";
        }
    }
    if (!z.match(/[l|m]/g)) {
      return 0;
    } else if (z.split('e').join('')[0] == "m") {
      return 1;
    } else {
      return -1;
    }
}

console.log(checkVersion("2.0.5","1.0.15")); // returns 1 as expected
console.log(checkVersion("1.2.2","1.2.0")); // returns 1 as expected
console.log(checkVersion("1.0.5","1.1.0")); // returns -1 as expected
console.log(checkVersion("1.0.5","1.00.05")); // returns 0 as expected
console.log(checkVersion("0.9.9.9.9.9.9","1.0.0.0.0.0.0")) // returns -1 as expected;

---

However, I still feel like there must be a shorter, more concise and cleaner way of doing this though. Any suggestions?

Answer 1

A small review;

• Once you know that one version digit is larger than the other, you can exit immediately
• You did not declare i with const or let
• I would advise the use of a beautifier for your code, it's a bit compact in some places
• The code does not handle well versions with different counts of digits
• You should always pass the base, when you call parseInt

I wrote an alternative version with 2 extra tests;

function checkVersion(a, b) {
    const x = a.split('.').map(e => parseInt(e, 10));
    const y = b.split('.').map(e => parseInt(e, 10));

    for (const i in x) {
        y[i] = y[i] || 0;
        if (x[i] === y[i]) {
            continue;
        } else if (x[i] > y[i]) {
            return 1;
        } else {
            return -1;
        }
    }
    return y.length > x.length ? -1 : 0;
}

console.log(checkVersion("1.2.2", "1.2.0"), 1); // returns 1 as expected
console.log(checkVersion("1.0.5", "1.1.0"), -1); // returns -1 as expected
console.log(checkVersion("1.0.5", "1.00.05"), 0); // returns 0 as expected
console.log(checkVersion("0.9.9.9.9.9.9", "1.0.0.0.0.0.0"), -1) // returns -1 as expected;
console.log(checkVersion("1.0.5", "1.0"), 1); // returns 1 as expected
console.log(checkVersion("1.0", "1.0.5"), -1); // returns -1 as expected
console.log(checkVersion('2019.09', '2019.9'), 0) // returns 0