3602. Hexadecimal and Hexatrigesimal Conversion

Problem Link

Description

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You are given an integer n.

Return the concatenation of the hexadecimal representation of n2 and the hexatrigesimal representation of n3.

A hexadecimal number is defined as a base-16 numeral system that uses the digits 0 – 9 and the uppercase letters A - F to represent values from 0 to 15.

A hexatrigesimal number is defined as a base-36 numeral system that uses the digits 0 – 9 and the uppercase letters A - Z to represent values from 0 to 35.

 

Example 1:

Input: n = 13

Output: "A91P1"

Explanation:

• n2 = 13 * 13 = 169. In hexadecimal, it converts to (10 * 16) + 9 = 169, which corresponds to "A9".
• n3 = 13 * 13 * 13 = 2197. In hexatrigesimal, it converts to (1 * 362) + (25 * 36) + 1 = 2197, which corresponds to "1P1".
• Concatenating both results gives "A9" + "1P1" = "A91P1".

Example 2:

Input: n = 36

Output: "5101000"

Explanation:

• n2 = 36 * 36 = 1296. In hexadecimal, it converts to (5 * 162) + (1 * 16) + 0 = 1296, which corresponds to "510".
• n3 = 36 * 36 * 36 = 46656. In hexatrigesimal, it converts to (1 * 363) + (0 * 362) + (0 * 36) + 0 = 46656, which corresponds to "1000".
• Concatenating both results gives "510" + "1000" = "5101000".

 

Constraints:

• 1 <= n <= 1000

Solution

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Python3

class Solution:
    def concatHex36(self, n: int) -> str:
        base16 = pow(n, 2)
        base36 = pow(n, 3)
        b2 = ''
        
        while base36 > 0:
            mod = base36 % 36
            if mod <= 9:
                b2 += str(mod)
            else:
                mod -= 10
                b2 += chr(ord('A') + mod)
                
            base36 //= 36
 
        b1 = ''
 
        while base16 > 0:
            mod = base16 % 16
            if mod <= 9:
                b1 += str(mod)
            else:
                mod -= 10
                b1 += chr(ord('A') + mod)
                
            base16 //= 16
        
 
        return b1[::-1] + b2[::-1]