在我的算法中，从最多K个数字替换中找到最大的回文数，我遗漏了什么极端情况

我已经为此工作了几个小时，我的解决方案符合性能基准，并且通过了几乎所有的测试用例。它是0 (n)在评论中有很好的描述。不幸的是，测试用例是如此巨大，以至于不可能逐一完成它们，所以我希望你们中的一个人可以提供一组新的眼睛，看看我是否可能在这里遗漏了一个角落用例。

static string LargestPalindromeAfterKReplacements(string number, int k)
{
    // Finds the largest palindromic number (digits are the same in reverse order)
    // that can be formed by at most k replacements of digits.
    // e.g. "001", k=2 --> "909"
    //      "001", k=1 --> "101"
    //      "001", k=0 --> "-1" (not possible)
    // Get number as a StringBuilder element
    var sb = new StringBuilder(number);
    // For the first pass through the string, replace any of the unequal
    // characters on the left and right side by the larger of the two. For
    // later use, keep track of the left index of each of these replacements.
    var replacementIndices = new Queue<int>(); 
    for(int i = 0, j = sb.Length - 1; i < j && k > 0; ++i, --j)
    {
        if(sb[i] < sb[j])
        {
            sb[i] = sb[j];
            replacementIndices.Enqueue(i);
            k -= 1;
        }
        else if(sb[i] > sb[j])
        {
            sb[j] = sb[i];
            replacementIndices.Enqueue(i);
            k -= 1;
        }
    } 
    // If sb isn't a palindrome at this point, then it was never possible to
    // make it one. 
    if(!IsPalindrome(sb.ToString()))
        return "-1";
    // If here, sb is a palindrome. If we have any k left over, then for any of
    // the indices where we made a replacement, if the pair isn't already both 9, 
    // we coudld've made them both 9 during the first pass through the string, at
    // a cost decreasing k by 2 rather than by 1. "Replay" the original pass like this.
    while(k > 0 && replacementIndices.Count > 0) 
    {
        int i = replacementIndices.Dequeue(), j = sb.Length - i - 1;
        if(sb[i] != '9')
        {
            sb[i] = '9';
            sb[j] = '9';
            k -= 1;
        }
    }
    // In case we still have k > 0, that means we ran out of replacementIndices.
    // Make a third pass through the string and make any non-equal characters
    // on opposite ends become 9. 
    for(int i = 0, j = sb.Length - 1; i <= j && k > 0; ++i, --j)
    {
        if(sb[i] != '9')
        {
            if(k > 1)
            {
              sb[i] = '9';
              sb[j] = '9'; 
              k -= 2;
            }
            else if(i == j) // k = 1 and i = j
            {
                sb[i] = '9';
            }
        }
    }
    // In case we ran out of replacement indices, 
    return sb.ToString();
}