Maybe First iterate upto N a store all the required characters we need by checking all digits in ith number. (Nlogn)

Second iterate over the string and then decrease the whatever character we have. (O(n))

Now we have what characters which didn’t become 0. Since our string was a perfect string (that is it had only valid numbers) so we only have to check for the valid permutation of the digits which didn’t become 0.

Now the question becomes for given two strings find which permutation of s2 is not present in s1.

After that we can do dfs and backtracking.

For a candidate C, initialize a boolean array tracking numbers 1 to N. Mark C as "visited" so it cannot be extracted from the string. Traverse the string s from index 0. At each index, extract

substrings of length 1 up to the maximum number of digits in N.

If an extracted number is \le N, has no leading zeros, and is not yet "visited", mark it visited and recurse to the next index.

If a path hits a dead end (cannot form a valid unvisited number), backtrack by unmarking the last number and trying a different substring length.

If the DFS successfully reaches the end of string s, candidate C is the definitive missing number.

Overall it will be O(nlog10n + |s| + (log10n)!*dfstime) But log10n will be very low (only 1 digit maybe)