Python – 检查给定的列表是否形成连续的不同子数组

给定一个由 A1、A2、A3…….An 形式的元素组成的数组。任务是找出数组是否可以形成为连续的不同子数组。您需要确定数组是否可以转换为由相似元素组成的连续子数组，并且每个元素都有不同的数量。

一旦遇到的元素不应出现在数组的后面，因为它不会是连续的

例子：

Input:[ 1 1 3 6 6 6 ] 
Output: YES 
Explanation: 
The given elements of the can be converted to Contiguous Distinct Set Array in the form of [1, 1] [3] [6, 6, 6]and also 
no. of 1’s = 2 
no. of 3’s = 1 
no. of 6’s = 3 
which are distinct 

Input:[ 1 1 3 5 2 2 2 3 ] 
Output: NO 
Explanation: 
The given elements of the cannot be converted to Contiguous Distinct Set Array as sub array [3 5 2 2 2 3] 
violates the condition(elements need to be contiguous) 3 again appears after 5 and 2.

Input:[9 9 4 4 4 1 6 6] 
Output: NO 
Explanation: 
The given elements of the cannot be converted to Contiguous Distinct Set Array 
It is of the form [9, 9] [4, 4, 4] [1] [6, 6] So the elements are present contiguous 
But the no. of 9’s=2 which is also equal to the no. of 6’s=2 
Hence the no.s of different elements of the array are not Distinct 
hence the Answer is NO
 

解决方案：

Python3

from collections import Counter
 
# function to check contiguous
# distinct set array
def contig_distinct_setarr(l, n):
     
    c = Counter(l)
    a = list(set(l))
      
    b =[]
    flag = True
     
    for j in c.values():
         
        # iterating and moving it to another
        # array if already present we print NO
        # when it finds no. of different elements
        # to be equal if no. of x's = no. of y's
        # So we break of the loop
        if j not in b:
            b.append(j)
        else:
            print("NO")
            flag = False
            break
             
    # If already there are similar elements
    # in c.values()- the count of elements
    # flag = False and we dont need to check
    # the below condition If not flag = False
    # then the iterate through the array loop
    if (flag != False):
        i, k = 0, 0
          
        # for each elements in set a
        # cou stores the count of a[i]
        while k

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输出：

YES