LeetCode 0800. Similar RGB Color Solution in Java, Python & Go | Explanation + Code

Description

The red-green-blue color "#AABBCC" can be written as "#ABC" in shorthand.

• For example, "#15c" is shorthand for the color "#1155cc".

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

Given a string color that follows the format "#ABCDEF", return a string represents the color that is most similar to the given color and has a shorthand (i.e., it can be represented as some "#XYZ").

Any answer which has the same highest similarity as the best answer will be accepted.

 

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.

Example 2:

Input: color = "#4e3fe1"
Output: "#5544dd"

 

Constraints:

• color.length == 7
• color[0] == '#'
• color[i] is either digit or character in the range ['a', 'f'] for i > 0.

Solutions

Solution 1

PythonJavaGo

class Solution: def similarRGB(self, color: str) -> str: def f(x): y, z = divmod(int(x, 16), 17) if z > 8: y += 1 return '{:02x}'.format(17 * y) a, b, c = color[1:3], color[3:5], color[5:7] return f'#{f(a)}{f(b)}{f(c)}'(code-box)

class Solution { public String similarRGB(String color) { String a = color.substring(1, 3), b = color.substring(3, 5), c = color.substring(5, 7); return "#" + f(a) + f(b) + f(c); } private String f(String x) { int q = Integer.parseInt(x, 16); q = q / 17 + (q % 17 > 8 ? 1 : 0); return String.format("%02x", 17 * q); } }(code-box)

func similarRGB(color string) string { f := func(x string) string { q, _ := strconv.ParseInt(x, 16, 64) if q%17 > 8 { q = q/17 + 1 } else { q = q / 17 } return fmt.Sprintf("%02x", 17*q) } a, b, c := color[1:3], color[3:5], color[5:7] return "#" + f(a) + f(b) + f(c) }(code-box)