Coloured Triangle

Computer Science Level 3

A colored triangle is created from a row of colors ( B , G , or R ).

Successive rows, each containing one fewer color than the last, are generated by considering the two touching colors in the previous row. If these colors are identical, the same color is used in the new row; if they are different, the missing color is used in the new row.

This is continued until the final row with only a single color is generated, as illustrated below. 

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R R G B R G
 R B R G B
  G G B R
   G R G
    B B
     B

What will be the last color with this 1000 color row?

Note: Want to try a much harder version of this problem? Check it out on CodeWars .

B G R

Antoine Crbs by Antoine Crbs , 19, France

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3 solutions

Antoine Crbs
Nov 16, 2018

Simple Python Solution : 

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def triangle(s):
    while len(s) > 1: s= ''.join(({'R', 'G', 'B'} - {a, b}).pop() if a != b else a for a, b in zip(s, s[1:]))
    return s


output: 'R'
9 Helpful 0 Interesting 1 Brilliant 0 Confused

java solution: 

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public class ColoredTriangle {

    //replace lastRow value with first row.
    static String lastRow="RRGBRG";
    public static void main(String[] args) {
        StringBuilder nextRow = new StringBuilder();
        while(lastRow.length() > 1) {
            for(int i = 0; i < lastRow.length() - 1; i++) {
                nextRow.append(color(lastRow.charAt(i), lastRow.charAt(i+1)));

            }
            lastRow = nextRow.toString();
            nextRow = new StringBuilder();
        }
        System.out.println("lastRow is " + lastRow);
    }

    public static char color(char first, char second) {
        //if same - return the color. else, return the missing one.
        return first == second ? first : "RGB".replace(first, ' ').replace(second, ' ').trim().charAt(0);
    }

}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

Python solution similar to @Antoine Crbs, but without zip: 

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colors = {'R', 'G', 'B'}
row = list(input())
while len(row) != 1:
    row = [(colors-{row[i], row[i+1]}).pop() if row[i] != row[i+1] else row[i] for i in range(len(row)-1)]
print(row[0])

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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