Python O(N^2) Solution
\n\n\n\n   \n\n\nclass Solution(object):\n    def findTheString(self, lcp):\n        """\n        :type lcp: List[List[int]]\n        :rtype: str\n        """\n
![image](https://assets.leetcode.com/users/images/51e8be83-3f98-438f-8488-38818711e0d9_1690658956.1728718.jpeg)\n![image](https://assets.leetcode.com/users/images/edd4e98a-0a42-4dfc-8f89-6b6bda4c3342_1690658956.2422037.jpeg)\n![image](https://assets.leetcode.com/users/images/09db80e9-d087-48e4-b26d-d681013fbf54_1690658956.2269988.jpeg)\n![image](https://assets.leetcode.com/users/images/6e12ddda-59fc-4751-a4dc-15bcb949516d_1690658956.2199922.jpeg)\n   ![image](https://assets.leetcode.com/users/images/ccb56869-56ab-4dc6-b1ee-10198a228572_1690658955.6743765.jpeg)\n\n```\nclass Solution(object):\n    def findTheString(self, lcp):\n        """\n        :type lcp: List[List[int]]\n        :rtype: str\n        """\n        n = len(lcp)\n        \n        """\n        airline pre-check\n        rule 1: lcp[0][0]=dimesnion of matrix;\n\t\t\t# rule 1 is more specific of rule 2 (redundant)\n        rule 2: i == j, lcp[i][j] == n - i\n        rule 3: lcp[i][j]=lcp[j][i] \n        rule 4: lcp[i+1][j+1]=lcp[i][j]-1 if lcp[i][j]>0\n        rule 5: lcp[i][j]<=n- max(i,j)\n        """\n\t\t\n\t\t# O(N^2)\n        # if lcp[0][0] != n:\n           #  print("case 0 ")\n            # return ""\n        for i, row in enumerate(lcp):\n            for j, item in enumerate(row):\n\t\t\t\t# rule 2 \n                if i == j and lcp[i][j] != n - i:\n                    print("where messed up 1")\n                    return ""\n\t\t\t\t\n\t\t\t\t# rule 3 \n                if lcp[i][j] != lcp[j][i]:\n                    print("where messed up 2")\n                    return ""\n\t\t\t\t\n\t\t\t\t# rule 5\n                if lcp[i][j] > n - max(i, j):\n                    print("where messed up 4")\n                    \n                    return ""\n\t\t\t\t\n                if i != len(row) - 1 and j != len(row) - 1:\n                    if lcp[i][j] > 0:\n                        if lcp[i + 1][j + 1] != lcp[i][j] - 1:\n                            print("case 4.5")\n                            return ""\n                    \n        \n        min_char = \'a\'\n        return_list = [\'#\' for i in range(n)]\n        \n\t\t# O(N^2)\n        for i, curr_char in enumerate(return_list):\n            if curr_char != \'#\':\n                continue\n            if min_char > \'z\':\n                print("where messed up 5")\n                return ""\n            \n            for j in range(i, n):\n                if lcp[i][j] > 0:\n                    return_list[j] = min_char\n            min_char = chr(ord(min_char) + 1)\n        \n        """\n        Then, check if the LCP is aligned with actual resulting string.\n\t\tO(N^2)\n        """\n        for i, row in enumerate(return_list):\n            # Checks for Rule 2 again\n\t\t\t# if i != len(return_list) - 1:\n               #  if lcp[i][i] != len(return_list) - i:\n                  #   print(\'case 6\')\n                   #  return ""\n\t\t\t\t   \n\t\t\t# Rule 4 \n            for j, curr_char in enumerate(row):\n                lcp_amount = lcp[i][j]\n                if lcp_amount > 0:\n                    if return_list[i: i + lcp_amount] != return_list[j: j + lcp_amount]:\n                    \n                        print("case 7")\n                        return ""\n                    \n            \n        return "".join(return_list)\n               \n\n                \n                \n        \n\n   ```\n   \n
['Python']