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Python – Check whether the given List forms Contiguous Distinct Sub-Array or Not
• Last Updated : 12 Jun, 2020

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]  [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]  [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 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)`

Output:

`YES`

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