Given a sequence of words, check whether it forms a valid word square.
A sequence of words forms a valid word square if thekthrow and column read the exact same string, where 0 ≤k< max(numRows, numColumns).
Note:
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 alphabeta-z.
Example 1:
Input:
[
"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:
[
"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:
[
"ball",
"area",
"read",
"lady"
]
Output:
false
Explanation:
The third row reads "read" while the third column reads "lead".
Therefore, it is NOT a valid word square.
The Idea: Just verify the each column matches each row across the diagonal.
Complexity: O(n^2) time and O(1) space
string get_col(const vector<string> &words, const int c) {
string vertical = "";
int row_iter = 0;
for (int row_iter = 0; row_iter < words.size(); row_iter++) {
string horizontal = words[row_iter];
if (c >= horizontal.size()) return vertical;
else vertical += horizontal[c];
}
return vertical;
}
bool validWordSquare(vector<string>& words) {
if (words.size() == 0) return true;
const int sq_size = words.size();
for (int i = 0; i < sq_size; i++) {
string horizontal = words.at(i);
string vertical = get_col(words, i);
if (horizontal != vertical) return false;
}
return true;
}