Unable to use vector.size() inside the for loop condition statement in C++

Question

I have recently come across a confusing situation. I don't know the explanation. Here is my simple code:

class Solution {
public:
    bool canAttendMeetings(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), compare);
        for (int i=0; i<intervals.size()-1; i++) {
            if (intervals[i].end > intervals[i+1].start) return false;
        }
        return true;
    }  
private:
    static bool compare(const Interval &interval1, const Interval &interval2) {
        return interval1.start < interval2.start;
    }
};

The problem is at this line:

for (int i=0; i<intervals.size()-1; i++) {

I got the runtime error for the input: []. However, I figured out the fix and it's pretty simple but hard to explain. Here is the fix, I just assign intervals.size() to a new variable before the for loop.

class Solution {
public:
    bool canAttendMeetings(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), compare);
        int n = intervals.size();
        for (int i=0; i<n-1; i++) {
            if (intervals[i].end > intervals[i+1].start) return false;
        }
        return true;
    }
private:
    static bool compare(const Interval &interval1, const Interval &interval2) {
        return interval1.start < interval2.start;
    }
};

Can anyone explain to me why the first version doesn't work but the second version works? Thanks a lot!

Answer 1

The problem is that size() returns an unsigned 0 when the container is empty.

You either want to convert that to a signed zero before subtracting one (as your working code did) or put more thought into how empty should behave. Such as:

    for (std:size_t i=1; i<intervals.size(); i++) {
        if (intervals[i-1].end > intervals[i].start) return false;

The above assumes you knew that subtracting one from an unsigned zero gives you a very big number, rather than a negative number (and you just didn't realize your code was subtracting one from an unsigned zero). But in case you thought subtracting one from an unsigned zero gave a negative result, now you know.