Strings Problem Solving in Codekata using Python(7 - 15)
Hey guys, Welcome to FORMAL INFINITY Channel, In this video we would discuss the solution for Codekata programing problems using python in Guvi. In this specific video we have discussed Input and Output related problems in Guvi. We would continue this series of video for covering all the topics under codekata.
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Question_1:-
you are given a string made up of parenthesis only.Your task is to check whether parenthesis are balanced or not.If they are balanced print 1 else print 0
Sample Input :
{({})}
Sample Output :
1
*****************
n = input()
r = 0
m = 0
for i in n:
if i in '({[':
r += 1
if i in ')}]':
m += 1
if r == m:
print(1)
else:
print(0)
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Question_2:-
In XYZ country there is rule that car’s engine no. depends upon car’ number plate. Engine no is sum of all the integers present on car’s Number plate.The issuing authority has hired you in order to provide engine no. to the cars.Your task is to develop an algorithm which takes input as in form of string(Number plate) and gives back Engine number.
Sample Input :
HR05-AA-2669
Sample Output :
28
**************
a = input()
r = 0
for i in a:
if i in ('1234567890'):
r = int(i)+r
else:
continue
print(r)
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Question_3:-
Write a program to get a string S, Type of conversion (1 - Convert to Lowercase, 2 - Convert to Uppercase) T, and integer P . Convert the case of the letters in the positions which are multiples of P.(1 based indexing).
Sample Input :
ProFiLe
1
2
Sample Output :
Profile
******************
a = input()
T = int(input())
P = int(input())
arr = []
for i in a:
arr.append(i)
if T == 1:
for i in range(1, int(len(a)/P)+1):
arr[(P*i)-1] = arr[P*i-1].lower()
if T == 2:
for i in range(1, int(len(a)/P)+1):
arr[(P*i)-1] = arr[P*i-1].upper()
for i in arr:
print(i, end='')
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Question 4:-
Jennyfer is fond of strings. She wants to read the character from right to left (reverse the string), so she wants you to design a suitable algorithm which satisfy her desire.
Sample Input :
jennyfer
Sample Output :
Refynnej
*********************
a = input()
print(a[::-1].capitalize())
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Question 5:-
Given a string S, print it without using semicolon in your program.
Sample Testcase :
INPUT
hello world
OUTPUT
hello world
**********************
a = input()
print(a)
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Question 6:-
Given a string S, print 'yes' if it has a vowel in it else print 'no'.
Sample Testcase :
INPUT
codekata
OUTPUT
yes
*********************
a = input()
count = 0
for i in a:
if i in 'AEIOUaeiou':
count += 1
if count 'use grater than sign' 0:
print('yes')
else:
print('no')
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Question 7:-
Given a string S, print it after changing the middle element to * (if the length of the string is even, change the 2 middle elements to *).
Sample Testcase :
INPUT
hello
OUTPUT
he*lo
********************
s = input()
d = len(s)//2
if len(s)%2 == 1:
print(s[0:d]+'*'+s[d+1:])
else:
print(s[0:d-1]+'**'+s[d+1:])
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Question 8:-
You are given string ‘s’. Your task is to modify the string as mentioned below:-
1)The string should not have three consecutive same characters.
2)You can add any number of characters anywhere in the string. Find the minimum number of characters which Ishaan must insert in the string.
Sample Input :
aabbbcc
Sample Output :
1
******************
a = input()
arr = []
condition = 0
for i in a:
arr.append(i)
for i in range(len(a)-2):
if arr[i] == arr[i+1] and arr[i] == arr[i+2]:
condition += 1
print(condition)
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Question 9:-
You are given a string ‘s’.Your task is to print the string in the order they are present and then sum of digits.
Sample Input :
AC30BD40
Sample Output :
ACBD7
*******************
a = input()
r = 0
arr = []
for i in a:
if i in '1234567890':
r += int(i)
else:
print(i, end='')
print(r)
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