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  1. Number Line Jumps
  2. Discussions

Number Line Jumps

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recency

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125 Discussions

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  • jimiwills
     + 3 comments

    I'm confused by this one. 0 3 4 2 So, the steps should be as follows... 0 3 6 9 for kangaroo 1 4 6 8 10 for kangaroo 2

    The expected answer is YES, but the steps don't line up. The question reads like the kangaroos must jump at the same time.

  • dharan_ec22
     + 0 comments
    public static String kangaroo(int x1, int v1, int x2, int v2) {
        if (v1 == v2) 
        {
                return (x1 == x2) ? "YES" : "NO";
        }
        if ((x1 - x2) % (v2 - v1) == 0 && (x1 - x2) / (v2 - v1) >= 0) 
        {
                return "YES";
        }

		return "NO";
}




``}

  • dharan_ec22
     + 0 comments

    public static String kangaroo(int x1, int v1, int x2, int v2) { if (v1 == v2) { return (x1 == x2) ? "YES" : "NO"; } if ((x1 - x2) % (v2 - v1) == 0 && (x1 - x2) / (v2 - v1) >= 0) { return "YES"; }

        return "NO";

    }

    }

  • mmaffezzoli01
     + 0 comments

    Thought about as just intersecting lines.

    def kangaroo(x1, v1, x2, v2):
    if v2 >= v1:
        return "NO"
    xint = (x2 - x1) / (v1 - v2)
    if xint % 1 != 0:
        return "NO"
    return "YES"

  • vdaons
     + 0 comments

    Python3 1 liner:

    def kangaroo(x1, v1, x2, v2):
    return 'YES' if v1 > v2 and (x2-x1) % (v1-v2) == 0 else 'NO'
    
    #general case (no constain x1 < x2)
    # if v1 == v2:
    #     return 'YES' if x1 == x2 else 'NO'
    # return 'YES' if (x2-x1) / (v1-v2) >= 0 and (x2-x1) % (v1-v2) == 0 else 'NO'