ยง Sliding window implementation style

I usually implement sliding window as:
// [l, r)
int l = r = 0;
while (r < n) {
 assert(l <= r);
 if (extend_window) { r++; }
 else { 
    l--; //contract window
 }
}
However, there are cases where we have complicated invariants on the sliding window, such as a maximum length. An example is codeforces 676c , where we must maintain a sliding window which contains at most k >= 0 "illegal" elements. My flawed implementation using a while loop was:
int best = 0;
for(int c = 'a'; c <= 'b'; ++c) {
    // window: [l, r)
    int l = 0, r = 0;
    // number of illegal letters changed. <= k
    int changed = 0; 
    while(r < n) {
        assert(changed <= k);
        assert(l <= r);
        if (s[r] == c) { r++; } // legal, extend.
        else {
            // need to change a letter to extend, s[r] != c.
            if (changed == k) {
                // cannot extend, contract from left.
                if (s[l] != c) { changed--; }
                l++; 
            } else {
                // extend, spending a change.
                r++;
                changed++;
            }
        }
        // keep track of best window size.
        best = max(best, r-l);
    }
}
Unfortunately, the above code is flawed. It does not work when the window size is zero. (TODO: explain) on the other hand, the implementation where we always stride forward with the r value in a for loop, and only deciding what happens with l does not suffer from this ( link to implementation ):
int best = 0;
for(int c = 'a'; c <= 'b'; ++c) {
    int l = 0;
    // number of illegal letters changed. <= k
    int changed = 0;
    // [l, r]
    for(int r = 0; r < n; ++r) {
        // change to 'a'.
        if (s[r] != c) { changed++; }
        // maintain invariants: must have changed <= k,
        // and at the end of a loop trip, we must have l <= r.
        while(changed > k && l < r) { 
            if (s[l] != c) { changed--; }
            l++;
        }
        assert(l <= r);
        // keep track of best window size.
        best = max(best, r-l+1);
    }
}