TCS DIGITAL| Advanced Coding

  Advanced Coding

 Question 1 (Even-Odd series): Given a string and it contains the digits as well as non-digits. We have to find the count of non-digits. If it is odd then remove all the non-digits and print the string as in even-odd order.If it is even then print the string as in odd-even order.

E.g. The given string is */24#5%7&9*3@. We have to count the non-digit. It’s 7, odd. Then remove all the digits from the string and output will become (in a string) 254739. In the problem, we have only 2 even and 4 odd numbers then after the even number of completion print the remaining odd numbers.

Solution: This is a basic question and the time complexity is O(n)

·         Python

string = input()  characters = [] even = [] odd = [] for i in string:     if i.isdigit():         if int(i)%2==0:             even.append(i)         else:             odd.append(i)     else:         characters.append(i)     charlen = len(characters) minlen = min(len(odd),len(even)) result = [] if charlen%2 == 0:     for i in range(minlen):         result.append(odd.pop(0))         result.append(even.pop(0))     result.extend(even)     result.extend(odd) else:     for i in range(minlen):         result.append(even.pop(0))         result.append(odd.pop(0))     result.extend(even)     result.extend(odd)        print("".join(result))