Class Solution
Hard
We define the lcp
matrix of any 0-indexed string word
of n
lowercase English letters as an n x n
grid such that:
lcp[i][j]
is equal to the length of the longest common prefix between the substringsword[i,n-1]
andword[j,n-1]
.
Given an n x n
matrix lcp
, return the alphabetically smallest string word
that corresponds to lcp
. If there is no such string, return an empty string.
A string a
is lexicographically smaller than a string b
(of the same length) if in the first position where a
and b
differ, string a
has a letter that appears earlier in the alphabet than the corresponding letter in b
. For example, "aabd"
is lexicographically smaller than "aaca"
because the first position they differ is at the third letter, and 'b'
comes before 'c'
.
Example 1:
Input: lcp = [[4,0,2,0],[0,3,0,1],[2,0,2,0],[0,1,0,1]]
Output: “abab”
Explanation: lcp corresponds to any 4 letter string with two alternating letters. The lexicographically smallest of them is “abab”.
Example 2:
Input: lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]
Output: “aaaa”
Explanation: lcp corresponds to any 4 letter string with a single distinct letter. The lexicographically smallest of them is “aaaa”.
Example 3:
Input: lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,3]]
Output: ""
Explanation: lcp[3][3] cannot be equal to 3 since word[3,…,3] consists of only a single letter; Thus, no answer exists.
Constraints:
1 <= n == ``lcp.length ==
lcp[i].length
<= 1000
0 <= lcp[i][j] <= n
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Constructor Summary
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Method Summary
Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
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Constructor Details
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Solution
public Solution()
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Method Details
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findTheString
public java.lang.String findTheString(int[][] lcp)
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