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Sentence Screen Fitting in C++
Suppose we have a rows x cols screen and a sentence represented by a list of non-empty words, so we have to find how many times the given sentence can be fitted on the screen. There are certain properties −
A word will not be split into two lines.
The order of words in the sentence must not be changed.
There will be only one space between two words.
The total number of words in the sentence won't exceed 100.
The length of each word is greater than 0 but less than 10.
1 ≤ rows, cols ≤ 20,000.
So if the input is like rows = 3 and cols = 6, and the sentence is [“a”, “bcd”, “e”], then the output will be 2.
To solve this, we will follow these steps −
Define a map dp, set ret := 0, n := size of sentence array
while row is not 0
start := ret mod n, len := -l and cnt := 0
if start is not present in dp, then
while 1 + len + size of sentence[(start + cnt) mod n] <= cols
len := 1 + len + sentence[(start + cnt) mod n]
increase cnt by 1
dp[start] := cnt
ret := ret + cnt
otherwise ret := ret + dp[start]
row := row – 1
return ret/n
Example (C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int wordsTyping(vector<string>& sentence, int rows, int cols) {
unordered_map <int, int> dp;
int ret = 0;
int n = sentence.size();
while(rows--){
int start = ret % n;
int len = -1;
int cnt = 0;
if(!dp.count(start)){
while(1 + len + (int)sentence[(start + cnt) % n].size() <= cols){
len = 1 + len + sentence[(start + cnt) % n].size();
cnt++;
}
dp[start] = cnt;
ret += cnt;
}
else{
ret += dp[start];
}
}
return ret / n;
}
};
main(){
vector<string> v = {"a","bcd","e"};
Solution ob;
cout << (ob.wordsTyping(v, 3, 6));
}Input
["a","bcd","e"] 3 6
Output
2
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