查找发生超过⌊N/3⌋次的所有数组元素

给定一个由N个整数组成的数组arr [] ，任务是查找给定数组中所有频率大于⌊N/3⌋的数组元素。

例子：

Input: arr[] = {5, 3, 5}
Output: 5
Explanation:
The frequency of 5 is 2, which is more than N/3( = 3/3 = 1).

Input: arr[] = {7, 7, 7, 3, 4, 4, 4, 5}
Output: 4 7
Explanation:
The frequency of 7 and 4 in the array is 3, which is more than N/3( = 8/3 = 2).

方法：解决问题的方法是使用分而治之技术。请按照以下步骤解决问题：

1. 初始化函数majorityElement()，将自左向右的任何索引返回大多数元件的计数在数组中。
2. 将给定的数组arr []分成两半，然后将其重复传递给函数mostElementElement() 。
3. 将低和高分别初始化为0和(N – 1) 。
4. 使用以下步骤计算多数元素：
   • 如果低=高：返回arr [low]作为多数元素。
   • 找到中间的索引，例如mid (= (low + high)/ 2 )。
   • 递归地将左右两个子数组分别作为多数元素(arr，low，mid)和多数元素(arr，mid + 1，high) 。
   • 完成上述步骤后，合并两个子数组并返回多数元素。
5. 每当找到所需的多数元素时，请将其附加到结果列表中。
6. 打印列表中存储的所有多数元素。

下面是上述方法的实现：

Python3

# Python program for the above approach
class Solution:
 
    # Function to find all elements that
    # occurs >= N/3 times in the array
    def majorityElement(self, a):
 
        # If array is empty return
        # empty list
        if not a:
            return []
 
        # Function to find the majority
        # element by Divide and Conquer
        def divideAndConquer(lo, hi):
            if lo == hi:
                return [a[lo]]
 
            # Find mid
            mid = lo + (hi - lo)//2
 
            # Call to the left half
            left = divideAndConquer(lo, mid)
 
            # Call to the right half
            right = divideAndConquer(mid + 1, hi)
 
            # Stores the result
            result = []
            for numbers in left:
                if numbers not in right:
                    result.append(numbers)
 
            result.extend(right)
 
            # Stores all majority elements
            ans = []
 
            for number in result:
                count = 0
 
                # Count of elements that
                # occurs most
                for index in range(lo, hi + 1):
                    if a[index] == number:
                        count += 1
 
                # If the number is a
                # majority element
                if count > (hi - lo + 1)//3:
                    ans.append(number)
 
            # Return the list of element
            return ans
 
        # Function Call
        print(divideAndConquer(0, len(a) - 1))
 
 
# Driver Code
if __name__ == "__main__":
 
    # Given array a[]
    a = [7, 7, 7, 3, 4, 4, 4, 6]
    object = Solution()
 
    # Function Call
    object.majorityElement(a)

输出：

[7, 4]

时间复杂度： O(N * log N)
辅助空间： O(log N)

另一种方法：使用内置的Python函数：

• 使用Counter()函数计算每个元素的频率。
• 遍历频率阵列并打印所有出现在n / 3次以上的元素。

下面是实现：

Python3

# Python3 program for the above approach
from collections import Counter
 
# Function to find the number of array
# elements with frequency more than n/3 times
def printElements(arr, n):
 
    # Calculating n/3
    x = n//3
 
    # Counting frequency of every element using Counter
    mp = Counter(arr)
     
    # Traverse the map and print all
    # the elements with occurrence atleast n/3 times
    for it in mp:
        if mp[it] > x:
            print(it, end=" ")
 
 
# Driver code
arr = [7, 7, 7, 3, 4, 4, 4, 6]
 
# Size of array
n = len(arr)
 
# Function Call
printElements(arr, n)
 
# This code is contributed by vikkycirus

输出：

7 4

时间复杂度： O(N)

辅助空间： O(N)