# DP solution, O(n) time constant space

#1

Comment body goes here.`class Solution: # @param A : tuple of integers # @return an integer def maxProfit(self, A): n = len(A) if n == 0: return 0 if n==1: return 0 if n ==2: return max(0, A[1]- A[0]) dp = [[[float("-inf") for _ in range(2)] for _ in range(2)] for _ in range(2)] flag = 1 maximum = 0 dp[flag^1][0][0] = -A[0] #buy dp[flag][0][0] = max(dp[flag^1][0][0] , -A[1]) dp[flag][0][1] = dp[flag^1][0][0] + A[1] maximum = max(maximum,dp[flag^1][0][1]) flag = flag^1 #dp.pop(0) #dp.append([[float("-inf") for _ in range(2)] for _ in range(2)]) for i in range(2,n): dp[flag][0][0] = max(dp[flag^1][0][0] , -A[i]) #0_Buy dp[flag][0][1] = max(dp[flag^1][0][1] , dp[flag^1][0][0] + A[i]) #0_sell maximum = max(maximum, dp[flag][0][1]) dp[flag][1][0] = max(dp[flag^1][1][0],dp[flag^1][0][1] - A[i]) #1_buy dp[flag][1][1] = max(dp[flag^1][1][1] , dp[flag^1][1][0] + A[i]) #1_sell = 2 transactions maximum = max(maximum, dp[flag][1][1]) flag = flag^1 return(maximum)`