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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.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  };
  

• 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
  
  

• 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;
  }
  
  

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