Richie Rich – Hackerrank Challenge – Python Solution
This is the python solution for the Hackerrank problem – Richie Rich – Hackerrank Challenge – Python Solution.
Source – Ryan Fehr’s repository.
#!/bin/python3
'''
Problem: https://www.hackerrank.com/challenges/richie-rich
Python 3
Thoughts: Not a problem with elegant solution.
Lots of edge cases and if else's required.
First pass turning the string into palindrome by
converting mismatch digit to larger of the two mismatch digits.
marking with '-' to keep track of flipped digits.
Taking second pass to maximise the number, turning digits to '9'
and keeping track of number of flips based on already flipped count.
Time Complexity: O(N)
Space Complexity: O(N)
'''
import sys
n,k = input().strip().split(' ')
n,k = [int(n),int(k)]
num = input().strip()
def make_palin(n, k, number):
number_str = str(number)
number = [number_str[i] for i in range(len(number_str))]
if n == 1 and k > 0:
print('9')
return
for i in range(int(n/2)):
if number[i] != number[n-i-1]:
max_num = max(int(number[i]),int(number[n-i-1]))
number[i] = str(-1 * max_num)
number[n-i-1] = str(max_num)
k = k-1
if k<0 :
print('-1')
return
flips_available = k
for i in range(int(n/2)):
dig_1 = int(number[i])
dig_2 = int(number[n-i-1])
if dig_1 < 0 and flips_available > 0:
decrement = 1
if dig_2 == 9:
decrement = 0
number[i] = str(9)
number[n-i-1] = str(9)
flips_available = flips_available - decrement
elif flips_available > 0:
decrement = 2
if dig_1 == 9:
decrement = decrement - 1
if dig_2 == 9:
decrement = decrement - 1
if flips_available >= decrement:
number[i] = str(9)
number[n-i-1] = str(9)
flips_available = flips_available - decrement
elif dig_1 < 0 and flips_available <= 0:
number[i] = str(-1 * dig_1)
if n%2 != 0 and flips_available>0:
number[int(n/2)] = str(9)
print(''.join(number))
make_palin(n , k, num)
