806. Number of Lines To Write String

Description

You are given a string s of lowercase English letters and an array widths denoting how many pixels wide each lowercase English letter is. Specifically, widths[0] is the width of 'a', widths[1] is the width of 'b', and so on.

You are trying to write s across several lines, where each line is no longer than 100 pixels. Starting at the beginning of s, write as many letters on the first line such that the total width does not exceed 100 pixels. Then, from where you stopped in s, continue writing as many letters as you can on the second line. Continue this process until you have written all of s.

Return an array result of length 2 where:

• result[0] is the total number of lines.
• result[1] is the width of the last line in pixels.

 

Example 1:

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: You can write s as follows:
abcdefghij  // 100 pixels wide
klmnopqrst  // 100 pixels wide
uvwxyz      // 60 pixels wide
There are a total of 3 lines, and the last line is 60 pixels wide.

Example 2:

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: You can write s as follows:
bbbcccdddaa  // 98 pixels wide
a            // 4 pixels wide
There are a total of 2 lines, and the last line is 4 pixels wide.

 

Constraints:

• widths.length == 26
• 2 <= widths[i] <= 10
• 1 <= s.length <= 1000
• s contains only lowercase English letters.

Solutions

Solution 1: Simulation

We define two variables lines and last, representing the number of lines and the width of the last line, respectively. Initially, lines = 1 and last = 0.

We iterate through the string $s$. For each character $c$, we calculate its width $w$. If $last + w \leq 100$, we add $w$ to last. Otherwise, we increment lines by one and reset last to $w$.

Finally, we return an array consisting of lines and last.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

Python3JavaC++GoTypeScriptRust

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

class Solution {
    public int[] numberOfLines(int[] widths, String s) {
        int lines = 1, last = 0;
        for (int i = 0; i < s.length(); ++i) {
            int w = widths[s.charAt(i) - 'a'];
            if (last + w <= 100) {
                last += w;
            } else {
                ++lines;
                last = w;
            }
        }
        return new int[] {lines, last};
    }
}

class Solution {
public:
    vector<int> numberOfLines(vector<int>& widths, string s) {
        int lines = 1, last = 0;
        for (char c : s) {
            int w = widths[c - 'a'];
            if (last + w <= 100) {
                last += w;
            } else {
                ++lines;
                last = w;
            }
        }
        return {lines, last};
    }
};

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

function numberOfLines(widths: number[], s: string): number[] {
    let [lines, last] = [1, 0];
    for (const c of s) {
        const w = widths[c.charCodeAt(0) - 'a'.charCodeAt(0)];
        if (last + w <= 100) {
            last += w;
        } else {
            ++lines;
            last = w;
        }
    }
    return [lines, last];
}

impl Solution {
    pub fn number_of_lines(widths: Vec<i32>, s: String) -> Vec<i32> {
        let mut lines = 1;
        let mut last = 0;

        for c in s.chars() {
            let idx = ((c as u8) - b'a') as usize;
            let w = widths[idx];
            if last + w <= 100 {
                last += w;
            } else {
                lines += 1;
                last = w;
            }
        }

        vec![lines, last]
    }
}

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