Leetcode 800. Similar RGB Color

In the following, every capital letter represents some hexadecimal digit from 0 to f.

The red-green-blue color "#AABBCC" can be written as "#ABC" in shorthand.  For example, "#15c" is shorthand for the color "#1155cc".

Now, say the similarity between two colors "#ABCDEF" and "#UVWXYZ" is -(AB - UV)^2 - (CD - WX)^2 - (EF - YZ)^2.

Given the color "#ABCDEF", return a 7 character color that is most similar to #ABCDEF, and has a shorthand (that is, it can be represented as some "#XYZ"

Example 1:
Input: color = "#09f166"
Output: "#11ee66"
Explanation:  
The similarity is -(0x09 - 0x11)^2 -(0xf1 - 0xee)^2 - (0x66 - 0x66)^2 = -64 -9 -0 = -73.
This is the highest among any shorthand color.

Note:

• color is a string of length 7.
• color is a valid RGB color: for i > 0, color[i] is a hexadecimal digit from 0 to f
• Any answer which has the same (highest) similarity as the best answer will be accepted.
• All inputs and outputs should use lowercase letters, and the output is 7 characters.

Code (Java):

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class Solution {     public String similarRGB(String color) {         if (color == null || color.length() == 0) {             return "";         }           String[] dict = new String[]{"00", "11", "22", "33", "44", "55", "66", "77", "88", "99",                                      "aa", "bb", "cc", "dd", "ee", "ff"};         int minDistance = Integer.MAX_VALUE;         String ans = "";           for (String R : dict) {             for (String G: dict) {                 for (String B : dict) {                     String s = "#" + R + G + B;                     int distance = getDistance(s, color);                     if (distance < minDistance) {                         ans = s;                         minDistance = distance;                     }                 }             }         }           return ans;     }       private int getDistance(String a, String b) {         int distance = 0;         for (int i = 1; i < 7; i += 2) {             int n1 = Integer.parseInt(a.substring(i, i + 2), 16);             int n2 = Integer.parseInt(b.substring(i, i + 2), 16);             distance += (n1 - n2) * (n1 - n2);         }           return distance;     } }