Question

IN PYTHON

Rondo Form is a type of musical structure, in which there is a recurring theme/refrain. Your job is to determine whether a given input is a valid Rondo Form or not. Here are the rules for valid Rondo forms:

• Rondo forms always start and end with an A section.
• In between the A sections, there should be contrasting sections (of arbitrary length) notated as B (or BB, BBB, BBBBB, etc), then C, then D, etc... No letter should be skipped.
• There shouldn't be any repeats in the sequence (such as ABBACCA).

Create a program which validates whether a given string is a valid Rondo Form.

• Inputs will be given as all uppercase.
• For the purpose of this problem, accept ABA (ACA, ADA, etc) as valid Rondo Forms. In other words, we allow a single section to be repeated in this problem; however, double, triple, etc. sections can NOT be repeated (per 3rd bullet above).

Examples

ABACADAEAFAGAHAIAJA ➞ True ABA ➞ True ABBACCA ➞ False ACAC ➞ False A ➞ False ABBACCCADAEA ➞ True

Sample Input 1

ABBACADA

Sample Output 1

True

Answer 1

Approach :

1. If lenght of string is 1 then false as it can't have 2 A's

2. If first or last character is not A then False

3. Remove all A's from string and check if new string is sorted or not.

4. if there exist a index i such that new[i] > new [i +1] then string is not sorted else string is sorted

5. store frequency of each character in frequency array eg BBCCC should be [ 2,3]

6. Check if there exist a duplicate in frequency other than if it exist then ans is false else true

Below is implementation of above approach :

# input string
s=input() 
# check length of string
n = len(s)
# if first or last character is not A then print false
# or lenght of string =1 
if s[0] != 'A' or s[n-1] !='A' or n==1:
        print("False")
else :
        # new is a string containing 0 A
        new="" 
        for character in s:
                # if character is A then skip
                if character=='A':
                        continue
                new+=character
        #check if new is sorted or not
        is_sorted = 1
        for idx in range(len(new)-1):
                if new[idx] > new [idx+1]:
                        is_sorted = 0
                        break
        if is_sorted == 0:
                print("False")
        else :
                frequency_array = []
                cur_lenght=1
                #Build frequency array
                for idx in range(1,len(new)):
                        if new[idx]==new[idx-1]:
                                cur_lenght+=1
                        else :
                                frequency_array.append(cur_lenght)
                                cur_lenght=1
                if cur_lenght > 0:
                        frequency_array.append(cur_lenght)
                # Use count array to check for duplicate
                count = []
                duplicate = 0
                n=len(frequency_array)
                for i in range(n+1):
                        count.append(0)
                for element in frequency_array:
                        count[element]+=1
                        #check for duplicate 
                        if count[element] > 1 and element!=1:
                                duplicate = 1;
                                break
                if duplicate==0 :
                        print("True")
                else :
                        print("False")

Below is screenshot of code along with the output