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

Input:[ 1 1 3 5 2 2 2 3 ]Output:NOExplanation: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:NOExplanation:The given elements of the cannot be converted to Contiguous Distinct Set ArrayIt is of the form [9, 9] [4, 4, 4] [1] [6, 6] So the elements are present contiguousBut the no. of 9’s=2 which is also equal to the no. of 6’s=2Hence the no.s of different elements of the array are not Distincthence 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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