Fill The Matrix Problem Statement
Given a fixed-size matrix and a sentence represented by a list of strings, return the number of times the given sentence can be filled on the matrix.
- A single string should not be split into multiple rows.
- The order of the strings in the sentence remains unchanged.
- Each cell in the matrix should contain either a single character or a space.
- A single space must separate two consecutive strings in a line.
Example One
{
"rows": 8,
"cols": 5,
"sentence": ["abcd", "ef", "gh", "ijk", "l", "m"]
}
Output:
2
Given sentence can be filled twice on the screen of size 8 x 5:
abcd.
ef.gh
ijk.l
m....
abcd.
ef.gh
ijk.l
m....
Assume dot(.) represents an empty space.
Example Two
{
"rows": 5,
"cols": 5,
"sentence": ["this", "is", "the", "topic", "string"]
}
Output:
0
Notes
Constraints:
- 1 <= number of words in the sentence <= 102
- 1 <= length of each word in the sentence <= 10
- 1 <= number of rows, number of columns in the matrix <= 104
- Each string in the sentence will only contain lowercase English characters.
Code For Fill The Matrix Solution: Dynamic Programming
/*
Asymptotic complexity in terms of \`n\` = length of sentence array,
\`lengthAverage\` is the average length of the words in the input array:
* Time: O(n * (cols / lengthAverage) + rows).
* Auxiliary space: O(n).
* Total space: O(n).
*/
/*
The idea behind this implementation is summarized below:
If we know how many words will be covered in a row starting
with sentence[i], then we can iterate through the rows, and by starting at the very
beginning word, try to count the number of times we can fit the complete sentence in the screen.
*/
int number_of_times(int rows, int cols, vector<string>& sentence) {
int n = sentence.size();
// \`dp[i]\` denotes the index of the word from which the next row will begin if the current row begins at sentence[i].
vector<int> dp(n, 0);
int total_fit = 0;
for (int i = 0; i < n; i++) {
int length = 0;
// \`count\` shows how many words are covered, but since we need to use it as an index in sentence, we need to decrement it.
int count = 0;
int start = i;
while (length <= cols) {
length += sentence[start].size();
start = (start + 1) % n;
count++;
if (length < cols) {
length++;
}
}
dp[i] = count - 1;
}
int k = 0;
for (int i = 0; i < rows; i++) {
// \`total_fit\` keep track of the number of covered words.
total_fit += dp[k];
k = (k + dp[k]) % n;
}
return (total_fit / n);
}
We hope that these solutions to the fill the matrix problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.
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