## Challenge#

Given a String S and and an array representing the cost of each deletion. You goal is to minimize the cost needed to remove all consecutive repetive characters.

Example 1:

``````"aba"
[2,3,4]
=> 0
``````

Example 2:

``````"abaa"
[2,3,4,1]
=> 1
``````

## Solution 1 - Brute Force - TLE#

``````class Solution:
def no_consecutive_letters(self, s):

t = ""
for e in s:
if e !="#":
t += e
for i in range(1,len(t)):
if t[i-1] == t[i]:
return False
return True

def helper(self, s, current_cost, cost, index):

if self.no_consecutive_letters(s):
self.min_cost = min(current_cost, self.min_cost)
return 0
if index > self.s_len:
return 0

self.helper(s[:index] + "#" +s[index+1:] , current_cost + cost[index], cost,index+1)
self.helper(s, current_cost, cost,index+1)

return 0

def minCost(self, s: str, cost: List[int]) -> int:
self.s_len = len(s)-1
self.min_cost = float("inf")
self.helper(s, 0, cost, 0)
return self.min_cost
``````

## Solution 2 - Greedy#

``````class Solution:

def remove_n_max(self, n, cost):
c = sorted(cost)
res = []
for i in range(0, n-1):
res.append(c[i])
return sum(res)

def minCost(self, s: str, cost: List[int]) -> int:
total = 0
i = 0
while i < len(s):
arr = []
current = 0
in_loop = False
tmp = []
while i < len(s)-1 and s[i] == s[i+1]:
arr.append(cost[i])
in_loop = True
current +=1
tmp.append(s[i])
i += 1
if in_loop is True:
arr.append(cost[i])
total += self.remove_n_max(current+1, arr)
i += 1