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Formatted question description: https://leetcode.ca/all/1271.html

1271. Hexspeak

Level

Easy

Description

A decimal number can be converted to its Hexspeak representation by first converting it to an uppercase hexadecimal string, then replacing all occurrences of the digit 0 with the letter O, and the digit 1 with the letter I. Such a representation is valid if and only if it consists only of the letters in the set {"A", "B", "C", "D", "E", "F", "I", "O"}.

Given a string num representing a decimal integer N, return the Hexspeak representation of N if it is valid, otherwise return "ERROR".

Example 1:

Input: num = “257”

Output: “IOI”

Explanation: 257 is 101 in hexadecimal.

Example 2:

Input: num = “3”

Output: “ERROR”

Constraints:

• 1 <= N <= 10^12
• There are no leading zeros in the given string.
• All answers must be in uppercase letters.

Solution

The solution is quite straightforward. Convert the decimal number to the hexadecimal string, change the string to uppercase, and replace 0 with O and 1 with I. Then check each character in the string. If a character out of the set {"A", "B", "C", "D", "E", "F", "I", "O"} occurs, return "ERROR". If the string is valid, return the string.

• Java
• C++
• Python

• class Solution {
      public String toHexspeak(String num) {
          long decimal = Long.parseLong(num);
          String hex = toHex(decimal);
          hex = hex.toUpperCase();
          hex = hex.replaceAll("0", "O");
          hex = hex.replaceAll("1", "I");
          String validStr = "ABCDEFIO";
          char[] hexArray = hex.toCharArray();
          for (char c : hexArray) {
              if (validStr.indexOf(c) < 0)
                  return "ERROR";
          }
          return hex;
      }
  
      public String toHex(long num) {
          String hex = "";
          while (num > 0) {
              int remainder = (int) (num % 16);
              String digit = remainder < 10 ? String.valueOf(remainder) : String.valueOf((char) ('A' + (remainder - 10)));
              hex = digit + hex;
              num /= 16;
          }
          return hex;
      }
  }
  

• // OJ: https://leetcode.com/problems/hexspeak/
  // Time: O(N)
  // Space: O(N)
  class Solution {
  public:
      string toHexspeak(string n) {
          stringstream ss;
          ss << uppercase << hex << stoll(n);
          string ans = ss.str();
          for (char &c : ans) {
              if (c == '0') c = 'O';
              else if (c == '1') c = 'I';
              else if (isdigit(c)) return "ERROR";
          }
          return ans;
      }
  };
  

• class Solution:
      def toHexspeak(self, num: str) -> str:
          s = set('ABCDEFIO')
          t = hex(int(num))[2:].upper().replace('0', 'O').replace('1', 'I')
          return t if all(c in s for c in t) else 'ERROR'
  
  
  

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