Strong Password Checker

1. Model the problem as a shortest path problem on a graph where nodes are password strings and edges represent one operation (insert, delete, replace).
2. Use Breadth-First Search (BFS) to find the shortest path from the initial password to any strong password.
3. Initialize a queue with the starting password and a visited set to avoid cycles.
4. In each level of the BFS, explore all possible passwords that can be generated by one operation from the passwords in the current level.
5. For each password dequeued, check if it meets the strong password criteria.
6. If a strong password is found, the current level number (number of steps) is the minimum required, so return it.
7. If the password is not strong, generate all valid next states (by inserting, deleting, or replacing a character) and add them to the queue if they haven't been visited.

The core idea is to build a state graph where each node is a unique password string. An edge connects two nodes if one password can be transformed into the other via a single operation (insertion, deletion, or replacement). The problem then becomes finding the shortest path from the initial password node to any node representing a strong password.

BFS is the natural algorithm for this, as it explores the graph level by level, guaranteeing that the first time we reach a goal state, it will be via a shortest path.

We start with a queue containing the initial password. We also use a Set to keep track of visited passwords to prevent redundant computations and infinite loops. The search proceeds in levels, where each level corresponds to one additional operation. For every password processed, we check if it's strong. If not, we generate all possible new passwords by applying one of the three operations at every possible position. These new, unvisited passwords are then added to the queue for the next level of the search.

1// Conceptual code for BFS approach2// Note: This is not a practical solution due to performance issues.3class Solution {4    public int strongPasswordChecker(String password) {5        Queue<String> queue = new LinkedList<>();6        Set<String> visited = new HashSet<>();7        queue.offer(password);8        visited.add(password);9        int steps = 0;10 11        while (!queue.isEmpty()) {12            int levelSize = queue.size();13            for (int i = 0; i < levelSize; i++) {14                String current = queue.poll();15                if (isStrong(current)) {16                    return steps;17                }18                // Generate all next possible passwords via one operation19                // (insert, delete, replace) and add to queue if not visited.20            }21            steps++;22        }23        return -1; // Should not be reached24    }25 26    private boolean isStrong(String s) {27        if (s.length() < 6 || s.length() > 20) return false;28        boolean hasLower = false, hasUpper = false, hasDigit = false;29        for (char c : s.toCharArray()) {30            if (Character.isLowerCase(c)) hasLower = true;31            if (Character.isUpperCase(c)) hasUpper = true;32            if (Character.isDigit(c)) hasDigit = true;33        }34        if (!hasLower || !hasUpper || !hasDigit) return false;35        for (int i = 0; i <= s.length() - 3; i++) {36            if (s.charAt(i) == s.charAt(i + 1) && s.charAt(i + 1) == s.charAt(i + 2)) {37                return false;38            }39        }40        return true;41    }42}

1. First, perform a single pass over the password to gather key metrics:
   • n: the length of the password.
   • missing_types: the number of missing character types (lowercase, uppercase, digit), from 0 to 3.
   • repeating_changes: the number of replacements needed to fix all repeating sequences. A sequence of length L needs L/3 replacements.
   • For the n > 20 case, also count how many repeating sequences have lengths L such that L % 3 == 0, L % 3 == 1, and L % 3 == 2.
2. Handle the problem in three distinct cases based on the password length n:
3. Case 1: n < 6
   • The password is too short. We need at least 6 - n insertions. These insertions can also fix missing character types. The total steps will be the maximum of the changes needed for length and for character types.
   • Return max(missing_types, 6 - n).
4. Case 2: 6 <= n <= 20
   • The password length is acceptable. We only need to fix missing types and repeating characters using replacements. A single replacement can address both issues simultaneously.
   • Return max(missing_types, repeating_changes).
5. Case 3: n > 20
   • The password is too long. We must perform deletions_needed = n - 20 deletions.
   • Greedily use these deletions to reduce repeating_changes as efficiently as possible:
     • First, use 1 deletion on sequences where len % 3 == 0 (saves 1 replacement).
     • Second, use 2 deletions on sequences where len % 3 == 1 (saves 1 replacement).
     • Finally, use any remaining 3 deletions on any repeating sequence (saves 1 replacement).
   • After deletions, the remaining changes are max(missing_types, remaining_repeating_changes).
   • Return deletions_needed + max(missing_types, remaining_repeating_changes).

The algorithm's strength lies in its ability to correctly determine the minimum number of operations by understanding how they interact. Instead of exploring a vast search space, it directly calculates the result.

Initial Analysis: First, we iterate through the password to find:

• missing_types: The number of required character types (lowercase, uppercase, digit) that are not present.
• repeating_changes: The number of replacements needed to break up sequences of three or more identical characters. For each such sequence of length L, we need L/3 replacements. We also track the counts of these sequences based on their length modulo 3, which is crucial for the n > 20 case.

Case-based Logic:

• n < 6: The primary issue is length. We need 6 - n insertions. These insertions can simultaneously satisfy the missing_types requirement. Thus, the total operations needed is the maximum of these two values, as one operation can serve two purposes. Repeating sequences are not a concern as they will be broken by the necessary insertions/replacements.

• 6 <= n <= 20: Length is not an issue. We only need to use replacements to fix missing_types and repeating_changes. Since one replacement can fix a repeating character and introduce a missing type, the total operations are the maximum of the two counts.

• n > 20: This is the most complex case. We must perform n - 20 deletions. The greedy strategy is to use these deletions to fix the most 'expensive' repeating sequences first. A deletion is most valuable when it reduces the number of required replacements. The priority is:

  1. Delete 1 character from a sequence of length L where L % 3 == 0. This saves one replacement.
  2. Delete 2 characters from a sequence of length L where L % 3 == 1. This also saves one replacement.
  3. Any remaining deletions are used in groups of 3 to reduce the total length, which effectively saves one replacement per 3 deletions. After applying deletions, the final number of steps is the n - 20 deletions plus the number of replacements needed for any remaining issues, which is max(missing_types, remaining_repeating_changes).

1class Solution {2    public int strongPasswordChecker(String password) {3        int n = password.length();4        int missing_types = 3;5        if (password.chars().anyMatch(Character::isLowerCase)) missing_types--;6        if (password.chars().anyMatch(Character::isUpperCase)) missing_types--;7        if (password.chars().anyMatch(Character::isDigit)) missing_types--;8 9        int repeating_changes = 0;10        // repeats_mod[i] will store the count of repeating sequences with length % 3 == i11        int[] repeats_mod = new int[3];12        for (int i = 0; i < n; ) {13            char c = password.charAt(i);14            int j = i;15            while (j < n && password.charAt(j) == c) {16                j++;17            }18            int len = j - i;19            if (len >= 3) {20                repeating_changes += len / 3;21                repeats_mod[len % 3]++;22            }23            i = j;24        }25 26        if (n < 6) {27            return Math.max(missing_types, 6 - n);28        } else if (n <= 20) {29            return Math.max(missing_types, repeating_changes);30        } else { // n > 2031            int deletions_needed = n - 20;32            int total_deletions = deletions_needed;33 34            // Use 1 deletion to fix a L%3==0 sequence (saves 1 replacement)35            int use_del = Math.min(deletions_needed, repeats_mod[0]);36            deletions_needed -= use_del;37            repeating_changes -= use_del;38 39            // Use 2 deletions to fix a L%3==1 sequence (saves 1 replacement)40            use_del = Math.min(deletions_needed / 2, repeats_mod[1]);41            deletions_needed -= use_del * 2;42            repeating_changes -= use_del;43            44            // Use 3 deletions to fix any sequence (saves 1 replacement)45            use_del = Math.min(deletions_needed / 3, repeating_changes);46            repeating_changes -= use_del;47 48            return total_deletions + Math.max(missing_types, repeating_changes);49        }50    }51}