LeetCode #420 — HARD

Strong Password Checker

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

A password is considered strong if the below conditions are all met:

  • It has at least 6 characters and at most 20 characters.
  • It contains at least one lowercase letter, at least one uppercase letter, and at least one digit.
  • It does not contain three repeating characters in a row (i.e., "Baaabb0" is weak, but "Baaba0" is strong).

Given a string password, return the minimum number of steps required to make password strong. if password is already strong, return 0.

In one step, you can:

  • Insert one character to password,
  • Delete one character from password, or
  • Replace one character of password with another character.

Example 1:

Input: password = "a"
Output: 5

Example 2:

Input: password = "aA1"
Output: 3

Example 3:

Input: password = "1337C0d3"
Output: 0

Constraints:

  • 1 <= password.length <= 50
  • password consists of letters, digits, dot '.' or exclamation mark '!'.
Patterns Used
🎯Greedy→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: A password is considered strong if the below conditions are all met: It has at least 6 characters and at most 20 characters. It contains at least one lowercase letter, at least one uppercase letter, and at least one digit. It does not contain three repeating characters in a row (i.e., "Baaabb0" is weak, but "Baaba0" is strong). Given a string password, return the minimum number of steps required to make password strong. if password is already strong, return 0. In one step, you can: Insert one character to password, Delete one character from password, or Replace one character of password with another character.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy

Example 1

"a"

Example 2

"aA1"

Example 3

"1337C0d3"

Related Problems

  • Strong Password Checker II (strong-password-checker-ii)
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #420: Strong Password Checker
class Solution {
    public int strongPasswordChecker(String password) {
        int types = countTypes(password);
        int n = password.length();
        if (n < 6) {
            return Math.max(6 - n, 3 - types);
        }
        char[] chars = password.toCharArray();
        if (n <= 20) {
            int replace = 0;
            int cnt = 0;
            char prev = '~';
            for (char curr : chars) {
                if (curr == prev) {
                    ++cnt;
                } else {
                    replace += cnt / 3;
                    cnt = 1;
                    prev = curr;
                }
            }
            replace += cnt / 3;
            return Math.max(replace, 3 - types);
        }
        int replace = 0, remove = n - 20;
        int remove2 = 0;
        int cnt = 0;
        char prev = '~';
        for (char curr : chars) {
            if (curr == prev) {
                ++cnt;
            } else {
                if (remove > 0 && cnt >= 3) {
                    if (cnt % 3 == 0) {
                        --remove;
                        --replace;
                    } else if (cnt % 3 == 1) {
                        ++remove2;
                    }
                }
                replace += cnt / 3;
                cnt = 1;
                prev = curr;
            }
        }
        if (remove > 0 && cnt >= 3) {
            if (cnt % 3 == 0) {
                --remove;
                --replace;
            } else if (cnt % 3 == 1) {
                ++remove2;
            }
        }
        replace += cnt / 3;

        int use2 = Math.min(Math.min(replace, remove2), remove / 2);
        replace -= use2;
        remove -= use2 * 2;

        int use3 = Math.min(replace, remove / 3);
        replace -= use3;
        remove -= use3 * 3;
        return (n - 20) + Math.max(replace, 3 - types);
    }

    private int countTypes(String s) {
        int a = 0, b = 0, c = 0;
        for (char ch : s.toCharArray()) {
            if (Character.isLowerCase(ch)) {
                a = 1;
            } else if (Character.isUpperCase(ch)) {
                b = 1;
            } else if (Character.isDigit(ch)) {
                c = 1;
            }
        }
        return a + b + c;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n log n)
Space
O(1)

Approach Breakdown

EXHAUSTIVE
O(2ⁿ) time
O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY
O(n log n) time
O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

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