[Python3] Greedy O(n^2) solution with O(n) space cost with some exposition.
Intuition and Approach\n Describe your first thoughts on how to solve this problem. \n## Greedy\nWe can construct a valid string, not necessarily the lexicograp
# Intuition and Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\n## Greedy\nWe can construct a valid string, not necessarily the lexicographically smallest one, by starting to look at the `lcp[i][j]` values for large `i` and `j` and moving toward smaller values. For example, suppose the string we want to construct is `s` and `len(lcp) == n` then we can put an arbitrary letter at `s[n-1]`, say `s[n-1] = \'z\'`. `lcp[n-2][n-1]` is either `0` or `1`. If it is `1` that means `s[n-2] == s[n-1]` and we assign `s[n-2] = \'z\'`. If it is `0` then `s[n-2] != s[n-1]` and we assign a different letter to `s[n-2]`, say `s[n-2] = \'y\'`. Once the last two letters have been decided we can focus on `s[n-3]` and fix it by first comparing it with `s[n-1]` (i.e. by looking at `lcp[n-3][n-1]`), and then comparing it with `s[n-2]` (i.e. by looking at `lcp[n-3][n-2]`). This process can be continued until we have fixed the first letter `s[0]`.\n\nSicne we have to iterate over the 2d array `lcp`, the time complexity is going to be $$O(n^2)$$.\n\n## Lexicographically minimum\nThe `lcp` matrix does not change if we change distinct letters in the string by distinct letters. So, once a valid string has been found it can be turned into the lexicographically smallest one by iterating once over the string. Before we start iterating, we consider all the letters in the alphabet to be avaialble. Then, wehenever we encounter a new letter in the string, we replace it with the smallest available letter, and mark that letter as unavailable. This takes $$O(n)$$ time, which is subleading to the $$O(n^2)$$ cost of iterating over `lcp`.\n\n## Some observations about a valid lcp matrix\n1. It is symmetric, obviously, the lcp of `s[i:]` and `s[j:]` is the same as that of `s[j:]` and `s[i:]`.\n2. `lcp[i][i] == n-i` because the lcp of `s[i]` and `s[i]` is `s[i]` itself. We use these to check if we have a valid `lcp` matrix as we iterate over it.\n3. If `lcp[i][j] != 0`, then `lcp[i][j] == lcp[i+1][j+1]+1` because the lcp of `s[i]` and `s[j]` is simply the lcp of `s[i+1]` and `s[j+1]` preceded by s[i]\n\n# Complexity\n- Time complexity: $$O(n^2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def findTheString(self, lcp: list[list[int]]) -> str:\n        n = len(lcp)\n        \n        # start with the possibility that any letter can be placed anywhere\n        s = [\'abcdefghijklmnopqrstuvwxyz\' for _ in range(n)] \n        # storing this array takes O(26*n) = O(n) space.\n        \n        # arbitrarily assing the final letter in the string\n        s[-1] = \'z\'\n        \n        for row in range(n-1,-1,-1):\n            for col in range(n-1, row-1, -1):\n                # lcp has to be symmetric\n                if lcp[row][col] != lcp[col][row]: return(\'\')\n\n                if row == col:\n                    # lcp[i][i] has to equal n-i\n                    if lcp[row][col] != n-row: return(\'\')\n\n                    # once we have compared s[row] with s[row+1:] and eliminated all letters from the possible choices for s[row], we can freely assign one out of the remaining pool. If nothing remains in the pool, a valid string does not exist.\n                    if len(s[row]): s[row] = s[row][-1]\n                    else: return(\'\')\n                    continue\n                        \n                c = s[col]\n                \n                # if lcp[row][col] is zero, we remove s[col] from the list of possible letters for s[row]\n                if lcp[row][col] == 0:\n                    if s[row][-1] == c: s[row] = s[row][:-1]\n                    if not s[row]: return(\'\')\n                    continue\n                \n                # if lcp[i][j] is nonzero then it must equal lcp[i+1][j+1] + 1 \n                if lcp[row][col] > min(n-row, n-col): return(\'\')\n                if col < n-1 and lcp[row+1][col+1] != lcp[row][col]-1: return(\'\')\n                \n                # once all these checks have been passed and lcp[row][col] is nonzero, we can safely assign s[row] to be equal to s[col]\n                if c not in s[row]: return(\'\')\n                else: s[row] = c\n                    \n        # from the existing string s (a list of characters at the momen), generate the lexicographically smallest one\n        sub = 26*[None]\n        newLetter = \'a\'\n        for i, c in enumerate(s):\n            if sub[ord(c)-ord(\'a\')] is None:\n                sub[ord(c)-ord(\'a\')] = newLetter\n                s[i] = newLetter\n                newLetter = chr(ord(newLetter)+1)\n            else:\n                s[i] = sub[ord(s[i])-ord(\'a\')]\n        \n        return(\'\'.join(s))\n```
['Greedy', 'Python3']