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Formatted question description: https://leetcode.ca/all/806.html

# 806. Number of Lines To Write String (Easy)

We are to write the letters of a given string S, from left to right into lines. Each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, it is written on the next line. We are given an array widths, an array where widths is the width of 'a', widths is the width of 'b', ..., and widths is the width of 'z'.

Now answer two questions: how many lines have at least one character from S, and what is the width used by the last such line? Return your answer as an integer list of length 2.

Example :
Input:
widths = [10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]
S = "abcdefghijklmnopqrstuvwxyz"
Output: [3, 60]
Explanation:
All letters have the same length of 10. To write all 26 letters,
we need two full lines and one line with 60 units.

Example :
Input:
widths = [4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]
S = "bbbcccdddaaa"
Output: [2, 4]
Explanation:
All letters except 'a' have the same length of 10, and
"bbbcccdddaa" will cover 9 * 10 + 2 * 4 = 98 units.
For the last 'a', it is written on the second line because
there is only 2 units left in the first line.
So the answer is 2 lines, plus 4 units in the second line.


Note:

• The length of S will be in the range [1, 1000].
• S will only contain lowercase letters.
• widths is an array of length 26.
• widths[i] will be in the range of [2, 10].

## Solution 1.

• class Solution {
public int[] numberOfLines(int[] widths, String S) {
final int MAX_WIDTH = 100;
int linesCount = 1, curWidth = 0;
int length = S.length();
for (int i = 0; i < length; i++) {
char c = S.charAt(i);
int width = widths[c - 'a'];
if (curWidth + width <= MAX_WIDTH)
curWidth += width;
else {
linesCount++;
curWidth = width;
}
}
return new int[]{linesCount, curWidth};
}
}

############

class Solution {
private static final int MAX_WIDTH = 100;

public int[] numberOfLines(int[] widths, String s) {
int last = 0, row = 1;
for (char c : s.toCharArray()) {
int w = widths[c - 'a'];
if (last + w <= MAX_WIDTH) {
last += w;
} else {
++row;
last = w;
}
}
return new int[] {row, last};
}
}

• // OJ: https://leetcode.com/problems/number-of-lines-to-write-string/
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> numberOfLines(vector<int>& widths, string S) {
int lineWidth = 0, cnt = 1;
for (char c : S) {
if (lineWidth + widths[c - 'a'] > 100) {
lineWidth = 0;
++cnt;
}
lineWidth += widths[c - 'a'];
}
return { cnt, lineWidth };
}
};

• class Solution:
def numberOfLines(self, widths: List[int], s: str) -> List[int]:
last, row = 0, 1
for c in s:
w = widths[ord(c) - ord('a')]
if last + w <= 100:
last += w
else:
row += 1
last = w
return [row, last]

############

class Solution(object):
def numberOfLines(self, widths, S):
"""
:type widths: List[int]
:type S: str
:rtype: List[int]
"""
lines = 1
last = 0
for s in S:
width = widths[ord(s) - ord('a')]
last += width
if last > 100:
lines += 1
last = width
return [lines, last]

• func numberOfLines(widths []int, s string) []int {
last, row := 0, 1
for _, c := range s {
w := widths[c-'a']
if last+w <= 100 {
last += w
} else {
row++
last = w
}
}
return []int{row, last}
}

• impl Solution {
pub fn number_of_lines(widths: Vec<i32>, s: String) -> Vec<i32> {
let mut count = 1;
let mut sum = 0;
for c in s.as_bytes() {
let width = widths[(c - b'a') as usize];
if sum + width > 100 {
sum = 0;
count += 1;
}
sum += width;
}
vec![count, sum]
}
}