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