Python – Check whether the given List forms Contiguous Distinct Sub-Array or Not

You are given an array consisting of elements in the form A1, A2, A3.......An. The task is to find whether  the array can be formed as a Contiguous Distinct Sub Array or Not. You need to find whether the array can be converted to contiguous sub-arrays that consist of similar elements and there are a distinct number of each element.

The elements once encountered should not appear later in the array as it would not be contiguous

Example:

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 arrray [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

Solution:

Python3

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from collections import Counter
  
# function to check contiguous 
# distinct set arrray
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<len(a):
            cou = c[a[i]]
            x = l.index(a[i])
  
            # here we extract thecontiguous
            # sub array of length equal to 
            # the count of the element 
            temp =(l[x:x + cou])
  
            # if the number of elements of
            # subsequences is not equal to
            # the value of element in the 
            # dictionary print NO
            if len(temp) != c[a[i]]:
                print("NO")
                flag = False
                break
                  
            k+= 1
            i+= 1
              
        # if we have iterated all over the array 
        # and the condition is satisfied we print
        # YES
        if flag == True:
            print("YES")
# initialize the array and its length
n = 6
l =[1, 1, 3, 6, 6, 6]
contig_distinct_setarr(l, n)

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Output:

YES



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