Answer to this problem is simply the longest common sub-sequence between A and sorted(set(A)), where set(A) contains only one copy of every distinct element in A. Eg.: A = [4,1,1,2,3,3], set(A) = [4,1,2,3].
NOTE: We can further reduce the space complexity to O(n).
class Solution: # @param A : tuple of integers # @return an integer def LCS(self,A,B): n = len(A) m = len(B) dp = [[0 for j in range(n+1)] for i in range(m+1)] for i in range(1,m+1): for j in range(1,n+1): if A[j-1]==B[i-1]: dp[i][j] = 1 + dp[i-1][j-1] else: dp[i][j] = max(dp[i][j-1], dp[i-1][j]) return(dp[m][n]) def lis(self, A): B = sorted(list(set(A))) #since we want a strictly increasing sequence #We get rid of the duplicates return(self.LCS(A,B))