LeetCode #2299 — EASY

Strong Password Checker II

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

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The Problem

Problem Statement

A password is said to be strong if it satisfies all the following criteria:

It has at least 8 characters.
It contains at least one lowercase letter.
It contains at least one uppercase letter.
It contains at least one digit.
It contains at least one special character. The special characters are the characters in the following string: "!@#$%^&*()-+".
It does not contain 2 of the same character in adjacent positions (i.e., "aab" violates this condition, but "aba" does not).

Given a string password, return true if it is a strong password. Otherwise, return false.

Example 1:

Input: password = "IloveLe3tcode!"
Output: true
Explanation: The password meets all the requirements. Therefore, we return true.

Example 2:

Input: password = "Me+You--IsMyDream"
Output: false
Explanation: The password does not contain a digit and also contains 2 of the same character in adjacent positions. Therefore, we return false.

Example 3:

Input: password = "1aB!"
Output: false
Explanation: The password does not meet the length requirement. Therefore, we return false.

Constraints:

1 <= password.length <= 100
password consists of letters, digits, and special characters: "!@#$%^&*()-+".

Step 01

Brute Force Baseline

Problem summary: A password is said to be strong if it satisfies all the following criteria: It has at least 8 characters. It contains at least one lowercase letter. It contains at least one uppercase letter. It contains at least one digit. It contains at least one special character. The special characters are the characters in the following string: "!@#$%^&*()-+". It does not contain 2 of the same character in adjacent positions (i.e., "aab" violates this condition, but "aba" does not). Given a string password, return true if it is a strong password. Otherwise, return false.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"IloveLe3tcode!"

Example 2

"Me+You--IsMyDream"

Example 3

"1aB!"

Core Insight

What unlocks the optimal approach

You can use a boolean flag to define certain types of characters seen in the string.
In the end, check if all boolean flags have ended up True, and do not forget to check the "adjacent" and "length" criteria.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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

Upper-end input sizes

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 #2299: Strong Password Checker II
class Solution {
    public boolean strongPasswordCheckerII(String password) {
        if (password.length() < 8) {
            return false;
        }
        int mask = 0;
        for (int i = 0; i < password.length(); ++i) {
            char c = password.charAt(i);
            if (i > 0 && c == password.charAt(i - 1)) {
                return false;
            }
            if (Character.isLowerCase(c)) {
                mask |= 1;
            } else if (Character.isUpperCase(c)) {
                mask |= 2;
            } else if (Character.isDigit(c)) {
                mask |= 4;
            } else {
                mask |= 8;
            }
        }
        return mask == 15;
    }
}

// Accepted solution for LeetCode #2299: Strong Password Checker II
func strongPasswordCheckerII(password string) bool {
	if len(password) < 8 {
		return false
	}
	mask := 0
	for i, c := range password {
		if i > 0 && byte(c) == password[i-1] {
			return false
		}
		if unicode.IsLower(c) {
			mask |= 1
		} else if unicode.IsUpper(c) {
			mask |= 2
		} else if unicode.IsDigit(c) {
			mask |= 4
		} else {
			mask |= 8
		}
	}
	return mask == 15
}

# Accepted solution for LeetCode #2299: Strong Password Checker II
class Solution:
    def strongPasswordCheckerII(self, password: str) -> bool:
        if len(password) < 8:
            return False
        mask = 0
        for i, c in enumerate(password):
            if i and c == password[i - 1]:
                return False
            if c.islower():
                mask |= 1
            elif c.isupper():
                mask |= 2
            elif c.isdigit():
                mask |= 4
            else:
                mask |= 8
        return mask == 15

// Accepted solution for LeetCode #2299: Strong Password Checker II
impl Solution {
    pub fn strong_password_checker_ii(password: String) -> bool {
        let s = password.as_bytes();
        let n = password.len();
        if n < 8 {
            return false;
        }
        let mut mask = 0;
        let mut prev = b' ';
        for &c in s.iter() {
            if c == prev {
                return false;
            }
            mask |= if c.is_ascii_uppercase() {
                0b1000
            } else if c.is_ascii_lowercase() {
                0b100
            } else if c.is_ascii_digit() {
                0b10
            } else {
                0b1
            };
            prev = c;
        }
        mask == 0b1111
    }
}

// Accepted solution for LeetCode #2299: Strong Password Checker II
function strongPasswordCheckerII(password: string): boolean {
    if (password.length < 8) {
        return false;
    }
    let mask = 0;
    for (let i = 0; i < password.length; ++i) {
        const c = password[i];
        if (i && c == password[i - 1]) {
            return false;
        }
        if (c >= 'a' && c <= 'z') {
            mask |= 1;
        } else if (c >= 'A' && c <= 'Z') {
            mask |= 2;
        } else if (c >= '0' && c <= '9') {
            mask |= 4;
        } else {
            mask |= 8;
        }
    }
    return mask == 15;
}

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.