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800. Similar RGB Color πŸ”’

DifficultyEasy

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

Thinking

A shorthand color uses only channel values \(00,11,\ldots,ff\), i.e. multiples of \(17\). Similarity is the sum of squared channel differences, and the three channels do not interact, so enumerating all \(16^3\) shorthand colors is unnecessary.

Write a channel \(q\) as \(17y+z\). If the remainder \(z>8\), round up to the next multiple; otherwise keep the current one. Applying this independently to the three channels yields the closest shorthand color.

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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)}'

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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);
    }
}

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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)
}

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