Hard
String Compression II — Python
Full explanation · Time O(n^2 * k) · Space O(n * k)
# Time: O(n^2 * k)
# Space: O(n * k)
class Solution(object):
def getLengthOfOptimalCompression(self, s, k):
"""
:type s: str
:type k: int
:rtype: int
"""
def length(cnt):
l = 2 if cnt >= 2 else 1
while cnt >= 10:
l += 1
cnt //= 10
return l
dp = [[len(s)]*(k+1) for _ in xrange(len(s)+1)]
dp[0][0] = 0
for i in xrange(1, len(s)+1):
for j in xrange(k+1):
if i-1 >= 0 and j-1 >= 0:
dp[i][j] = min(dp[i][j], dp[i-1][j-1])
keep = delete = 0
for m in xrange(i, len(s)+1):
if s[i-1] == s[m-1]:
keep += 1
else:
delete += 1
if j+delete <= k:
dp[m][j+delete] = min(dp[m][j+delete], dp[i-1][j]+length(keep))
return dp[len(s)][k]