Medium
Count Number of Texts — Python
Full explanation · Time O(n) · Space O(1)
# Time: O(n)
# Space: O(1)
# dp
class Solution(object):
def countTexts(self, pressedKeys):
"""
:type pressedKeys: str
:rtype: int
"""
MOD = 10**9+7
dp = [1]*5
for i in xrange(1, len(pressedKeys)+1):
dp[i%5] = 0
for j in reversed(xrange(max(i-(4 if pressedKeys[i-1] in "79" else 3), 0), i)):
if pressedKeys[j] != pressedKeys[i-1]:
break
dp[i%5] = (dp[i%5]+dp[j%5])%MOD
return dp[len(pressedKeys)%5]