Valid word square¶
Time: O(MxN); Space: O(1); easy
Given a sequence of words, check whether it forms a valid word square.
A sequence of words forms a valid word square if the k^th row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns).
The number of words given is at least 1 and does not exceed 500. Word length will be at least 1 and does not exceed 500. Each word contains only lowercase English alphabet a-z. Have you met this question in a real interview?
Example 1
Input: words =
[
"abcd",
"bnrt",
"crmy",
"dtye"
]
Output: True
Explanation:
The first row and first column both read “abcd”,
The second row and second column both read “bnrt”,
The third row and third column both read “crmy”,
The fourth row and fourth column both read “dtye”,
Therefore, it is a valid word square.
Example 2:
Input: words =
[
"abcd",
"bnrt",
"crm",
"dt"
]
Output: True
Explanation:
The first row and first column both read “abcd”,
The second row and second column both read “bnrt”,
The third row and third column both read “crm”,
The fourth row and fourth column both read “dt”,
Therefore, it is a valid word square.
Example 3:
Input: words =
[
"ball",
"area",
"read",
"lady"
]
Output: False
Explanation:
The third row reads “read”,
Therefore, it is NOT a valid word square.
[1]:
class Solution1(object):
def validWordSquare(self, words):
"""
:type words: List[str]
:rtype: bool
"""
for i in range(len(words)):
for j in range(len(words[i])):
if j >= len(words) or i >= len(words[j]) or \
words[j][i] != words[i][j]:
return False
return True
[2]:
s = Solution1()
words = [
"abcd",
"bnrt",
"crmy",
"dtye"
]
assert s.validWordSquare(words) == True
words = [
"abcd",
"bnrt",
"crm",
"dt"
]
assert s.validWordSquare(words) == True
words = [
"ball",
"area",
"read",
"lady"
]
assert s.validWordSquare(words) == False