[백준] 9251. LCS / 골드5Dynamic Programming시간복잡도 : O(N X M) *N : length s1, M : length s2 def find_lis(): s1 = "0" + str(input()) s2 = "0" + str(input()) dp = [[0]*(len(s2)) for _ in range(len(s1))] for i in range(1, len(s1)): for j in range(1, len(s2)): if s1[i] == s2[j]: dp[i][j] = dp[i-1][j-1] + 1 else: dp[i][j] = ma..