Hexagonal Grid Discussions | Algorithms | HackerRank
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  2. Algorithms
  3. Dynamic Programming
  4. Hexagonal Grid
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Hexagonal Grid

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  • zoyashah41306
     + 0 comments

    Show appreciation with unique corporate gifts like a "Hexagonal Grid" while fostering creativity and innovation. These gifts inspire collaboration and problem-solving skills, ideal for team-building activities or office decor. Enhance workplace dynamics with thoughtful items that encourage productivity and aesthetic appeal. Whether used for brainstorming sessions or as a decorative element, Hexagonal Grids symbolize forward-thinking and synergy within your organization. Elevate your corporate culture with meaningful gifts that reflect your commitment to excellence and teamwork.

  • venom1724
     + 0 comments

    ~10 lines Python, 100%, cool bit things

    If you figure out what the numbers mean I'll buy you a virtual coffee :D

    @cache
def f(a,b):
    x=((a&3)<<2)|(b&3)
    if a==b:
        return True
    if (a,b)==(0,1) or (a,b)==(1,0) or x in {6,11,14}:
        return False
    if x in {2,7,8,13}:
        return f(a>>1,b>>1)
    if x in {0,3,5,9,10,12,15}:
        return f(a>>2,b>>2)
    return f(a>>1,(b>>1)|1) or f((a>>1)|1,b>>1)

def hexagonalGrid(a, b):
    return "YES" if f(int(a,2),int(b,2)) else "NO"

  • thecodingsoluti2
     + 0 comments

    Here is Hexagonal Grid problem solution - https://programs.programmingoneonone.com/2021/07/hackerrank-hexagonal-grid-problem-solution.html

  • puustmorgan
     + 0 comments

    My python3 example with O(n) complexity:

    def hexagonalGrid(a, b):
    if ((a.count('1') + b.count('1')) % 2 != 0):
        return 'NO'
    white = 0
    block = False
    for i in range(len(a)):
        if (a[i] == '1'):
            if block and white % 2 != 0:
                return 'NO'
            block = True
        elif (a[i] == '0'):
            block = False
            white += 1
        if (b[i] == '1'):
            if block and white % 2 != 0:
                return 'NO'
            block = True
        elif (b[i] == '0'):
            block = False
            white += 1          
    return 'YES'

  • hnam5
     + 0 comments

    python3, 

    def hexagonalGrid(a, b):
    def explore(name, i):
        if name == 'a':
            if i < len(a) and a[i] == 0:
                a[i] = 1
                return 1 + explore('b', i) + explore('a', i+1)
        if name == 'b':
            if i < len(b) and b[i] == 0:
                b[i] = 1
                return 1 + explore('a', i+1) + explore('b', i+1)
        return 0

    a = [int(i) for i in a]
    b = [int(i) for i in b]

    for i in range(len(a)):
        cnt = 0
        if a[i] == 0:
            cnt = explore('a', i)
        if b[i] == 0:
            cnt = explore('b', i)
        if cnt % 2 != 0:
            return "NO"

    return "YES"