How to Compare Version Numbers in Go – Interview Solution

Version numbers consist of multiple revision numbers separated by "."; each segment may have leading zeros that should be ignored, and the length of version strings can differ (e.g., "1.0" vs "1.0.0"). Therefore the comparison must proceed from left to right, comparing numeric values of each segment.

The solution first splits both version strings by "." using strings.Split, then iterates over the corresponding segments, converting each to an integer with strconv.Atoi. If a segment in version1 is greater than the counterpart, the function returns 1; if smaller, it returns -1.

If the loop finishes without a decisive difference, the algorithm checks the remaining segments of the longer version. Any non‑zero segment in the longer version means that version is greater (treated as compared against an implicit 0 in the shorter version). The function finally returns 0 when all compared segments are equal.

func compareVersion(version1 string, version2 string) int {
    // 1. split by "."
    slicev1 := strings.Split(version1, ".")
    slicev2 := strings.Split(version2, ".")
    // 2. compare corresponding positions
    i, j := 0, 0
    for ; i < len(slicev1) && j < len(slicev2); i, j = i+1, j+1 {
        val1, _ := strconv.Atoi(slicev1[i])
        val2, _ := strconv.Atoi(slicev2[j])
        if val1 > val2 {
            // version1 larger
            return 1
        } else if val1 < val2 {
            // version2 larger
            return -1
        }
    }
    // 3. remaining parts of version1 (treated as >0)
    for i < len(slicev1) {
        val1, _ := strconv.Atoi(slicev1[i])
        i++
        if val1 > 0 {
            return 1
        } else if val1 < 0 {
            return -1
        }
    }
    // 4. remaining parts of version2 (treated as >0)
    for j < len(slicev2) {
        val2, _ := strconv.Atoi(slicev2[j])
        j++
        if val2 > 0 {
            return -1
        } else if val2 < 0 {
            return 1
        }
    }
    // 5. all equal
    return 0
}