3941. Password Strength

Description

You are given a string password.

The strength of the password is calculated based on the following rules:

1 point for each distinct lowercase letter ('a' to 'z').
2 points for each distinct uppercase letter ('A' to 'Z').
3 points for each distinct digit ('0' to '9').
5 points for each distinct special character from the set "!@#$".

Each character contributes at most once, even if it appears multiple times.

Return an integer denoting the strength of the password.

Example 1:

Input: password = "aA1!"

Output: 11

Explanation:

The distinct characters are 'a', 'A', '1' and '!'.
Thus, the strength = 1 + 2 + 3 + 5 = 11.

Example 2:

Input: password = "bbB11#"

Output: 11

Explanation:

The distinct characters are 'b', 'B', '1' and '#'.
Thus, the strength = 1 + 2 + 3 + 5 = 11.

Constraints:

1 <= password.length <= 10⁵
password consists of lowercase and uppercase English letters, digits, and special characters from "!@#$".

Solutions

Solution 1: Hash Table

We store each character in the input string in a hash set $\textit{st}$, so we can quickly ensure each distinct character is counted only once.

Then, we iterate through each character in $\textit{st}$ and compute the password strength according to the rules:

If the character is a lowercase letter ('a' to 'z'), add 1 point.
If the character is an uppercase letter ('A' to 'Z'), add 2 points.
If the character is a digit ('0' to '9'), add 3 points.
If the character is a special character (from the set "!@#$"), add 5 points.

Finally, return the computed password strength.

The time complexity is $O(n)$, where $n$ is the length of the input string. The space complexity is $O(m)$, where $m$ is the number of distinct characters in the input string.

Python3JavaC++GoTypeScript

class Solution:
    def passwordStrength(self, password: str) -> int:
        st = set(password)
        ans = 0
        for ch in st:
            if ch.islower():
                ans += 1
            elif ch.isupper():
                ans += 2
            elif ch.isdigit():
                ans += 3
            else:
                ans += 5
        return ans

class Solution {
    public int passwordStrength(String password) {
        var st = password.chars().mapToObj(c -> (char) c).collect(Collectors.toSet());

        int ans = 0;

        for (char ch : st) {
            if (Character.isLowerCase(ch)) {
                ans += 1;
            } else if (Character.isUpperCase(ch)) {
                ans += 2;
            } else if (Character.isDigit(ch)) {
                ans += 3;
            } else {
                ans += 5;
            }
        }

        return ans;
    }
}

class Solution {
public:
    int passwordStrength(string password) {
        unordered_set<char> st(password.begin(), password.end());

        int ans = 0;

        for (char ch : st) {
            if (islower(ch)) {
                ans += 1;
            } else if (isupper(ch)) {
                ans += 2;
            } else if (isdigit(ch)) {
                ans += 3;
            } else {
                ans += 5;
            }
        }

        return ans;
    }
};

func passwordStrength(password string) (ans int) {
    st := map[rune]struct{}{}

    for _, ch := range password {
        st[ch] = struct{}{}
    }

    for ch := range st {
        switch {
        case unicode.IsLower(ch):
            ans += 1
        case unicode.IsUpper(ch):
            ans += 2
        case unicode.IsDigit(ch):
            ans += 3
        default:
            ans += 5
        }
    }

    return
}

function passwordStrength(password: string): number {
    const st = new Set(password);

    let ans = 0;

    for (const ch of st) {
        if (/[a-z]/u.test(ch)) {
            ans += 1;
        } else if (/[A-Z]/u.test(ch)) {
            ans += 2;
        } else if (/\d/u.test(ch)) {
            ans += 3;
        } else {
            ans += 5;
        }
    }

    return ans;
}

3941. Password Strength

Description

Solutions

Solution 1: Hash Table

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