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Formatted question description: https://leetcode.ca/all/800.html

800. Similar RGB Color (Easy)

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.

Companies:
Google

Related Topics:
Math, String

Solution 1.

  • class Solution {
        public String similarRGB(String color) {
            int red = Integer.parseInt(color.substring(1, 3), 16);
            int green = Integer.parseInt(color.substring(3, 5), 16);
            int blue = Integer.parseInt(color.substring(5, 7), 16);
            int closestRed = 0, closestGreen = 0, closestBlue = 0;
            for (int i = 0; i <= 0xff; i += 0x11) {
                int redDifference = Math.abs(red - i);
                int greenDifference = Math.abs(green - i);
                int blueDifference = Math.abs(blue - i);
                if (redDifference < Math.abs(red - closestRed))
                    closestRed = i;
                if (greenDifference < Math.abs(green - closestGreen))
                    closestGreen = i;
                if (blueDifference < Math.abs(blue - closestBlue))
                    closestBlue = i;
            }
            String similarRGB = "#" + toHex(closestRed) + toHex(closestGreen) + toHex(closestBlue);
            return similarRGB;
        }
    
        public String toHex(int num) {
            String hex = "";
            while (num > 0) {
                int remainder = num % 16;
                String curHex = remainder < 10 ? String.valueOf(remainder) : String.valueOf((char) (remainder - 10 + 'a'));
                hex = curHex + hex;
                num /= 16;
            }
            while (hex.length() < 2)
                hex = "0" + hex;
            return hex;
        }
    }
    
    ############
    
    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);
        }
    }
    
  • // OJ: https://leetcode.com/problems/similar-rgb-color/
    // Time: O(1)
    // Space: O(1)
    class Solution {
    private:
        const string digits = "0123456789abcdef";
    public:
        string similarRGB(string color) {
            string ans = "#";
            for (int i = 0; i < 3; ++i) {
                char a = color[1 + i * 2], b = color[2 + i * 2];
                int hex1 = stoi(color.substr(1 + i * 2, 2), nullptr, 16);
                int best = 0, bestSim = INT_MIN;
                for (int j = 0; j < 16; ++j) {
                    int hex2 = j * 16 + j;
                    int sim = -pow(hex1 - hex2, 2);
                    if (sim > bestSim) {
                        bestSim = sim;
                        best = j;
                    }
                }
                char c = digits[best];
                ans += c;
                ans += c;
            }
            return ans;
        }
    };
    
  • 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)}'
    
    
    
  • 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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