I am doing the leetcode problem Container with most water. I am wondering why one method goes over the time limit

Question

I solved the leetcode problem https://leetcode.com/problems/container-with-most-water/ I had to write two different solutions because one of them would go over the time limit. I would expect behavior for both of them to be O(n) time and I was wondering what might be causing one of them to have a longer runtime. I am not looking for a cleaner solution this is more for my understanding of why there is a significant difference in runtime for what I perceive to be very similar solutions. My guess is maybe erase() copies over the vector and takes O(n) but I really do not understand how it works. I have attached a link to the problem for context: https://leetcode.com/problems/container-with-most-water/ and my solutions below. Thank you in advance!

class Solution {
public:
    int maxArea(vector<int>& height) {
        int largest = 0;
        int first = 0;
        int last = 0;
        first = height.front(); 
        last = height.back();
       
         while(height.size() != 1) {
        first = height.front(); 
        last = height.back();
        int temp = min(first,last)*(height.size()-1);
        largest = max(largest, temp);
        if(first <= last){
            height.erase(height.begin());
        }
        else
        {
            height.pop_back();
        }
       
    }
        return largest;
        
    }
};

The solution above did not pass as it took too much time on the longer examples, which makes me think I am getting a runtime that is not O(n).

class Solution {
public:
    int maxArea(vector<int>& height) {
        unsigned int largest = 0;
        unsigned int first = 0;
        unsigned int last = 0;
        unsigned int i = 0;
        unsigned int j = height.size() - 1;
        
         while(i!=j) {
       first = height[i];
        last = height[j];
        unsigned int temp = min(first,last)*(j - i);
        largest = max(largest, temp);
        if(first <= last){
           i++;
        }
        else
        {
           j--;
        }
       
    }
        return largest;
        
    }
};

This solution passed fine. I do not understand why my first solution would have a significant runtime difference in comparison to my second solution.