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  1. Prepare
  2. Algorithms
  3. Constructive Algorithms
  4. Gaming Array
  5. Discussions

Gaming Array

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137 Discussions

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  • oscarrleo585
     + 0 comments

    I found this game logic really interesting, especially how it uses XOR to determine the winner. It reminds me of the decision-making patterns in some custom Brawl-style mods I’ve explored recently. If anyone is interested in alternate game mechanics and strategies, you can check out Finx and Lumi – two unique brawler characters with special abilities designed to challenge your thinking.

    Would love to hear thoughts on how their abilities could relate to XOR-style game outcomes!

  • macohip968
     + 0 comments

    Just like in Minecraft Premium Apk where every move counts, Andy and Bob’s game is all about strategy! Bob starts strong, but Andy turns the tables in the first game, winning with a clever play. In the second round, Bob takes the crown. It’s a fun twist, much like crafting the perfect pickaxe in Minecraft Premium Apk—timing is everything!

  • wifivop476
     + 0 comments

    xQc’s setup is known for its top-tier performance and flashy design, built to handle intense streaming and gaming sessions with ease. The gaming PC xQc uses features high-end components like an NVIDIA RTX 4090 and a powerful CPU, making it one of the most capable rigs out there.

  • 7urkm3n
     + 0 comments
    func findMax(arr *[]int32) int {
    max := (*arr)[0]
    index := 0
    
    for i:=1; i< len(*arr); i++ {
        if max < (*arr)[i]{
            max = (*arr)[i]
            index = i
        }   
    }
    return index
}


func gamingArray(arr []int32) string {
	i := 0

	for 0 < len(arr) {
		index := findMax(&arr)
		arr = arr[:index]
		i++
	}

	if i%2 == 0 {
		return "ANDY"
	}
	return "BOB"
}

  • leoisaac_lameda
     + 1 comment

    Python3

    def gamingArray(arr):
    # Write your code here
    helper = sorted(arr, reverse=True)
    d = dict()
    for i in range(len(arr)):
        d[arr[i]] = i
    count = 0
    n = len(arr) - 1
    for maxVal in helper:
        if d[maxVal] <= n:
            n = d[maxVal] - 1
            count += 1
        if n < 0:
            break
    return 'BOB' if count % 2 == 1 else 'ANDY'