Skip to content
Related Articles

Related Articles

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

View Discussion
Improve Article
Save Article
  • Last Updated : 21 Sep, 2021
View Discussion
Improve Article
Save Article

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 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
 

Solution:

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<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)

Output:

YES

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!