Solution Explanation: Number of Lines to Write String

This problem involves calculating the number of lines needed to write a given string, s, considering the width of each character specified in the widths array. Each line has a maximum width of 100 pixels.

Approach:

The most efficient way to solve this problem is through a single pass simulation. We iterate through the input string, character by character, keeping track of the current line's width. If adding the current character's width exceeds the 100-pixel limit, we increment the line count and start a new line.

Algorithm:

1. Initialization:

   • lines: Initialize a counter for the number of lines to 1 (we start with one line).
   • last: Initialize a variable to store the width of the last line to 0.

2. Iteration:

   • Iterate through each character c in the input string s.
   • Get the width w of the character c from the widths array using the character's ASCII value (e.g., widths[c - 'a'] in many languages).
   • Check if adding w to the current last line's width exceeds 100 pixels (last + w > 100).
     • If it exceeds 100, increment lines (start a new line), and reset last to w (the new line starts with the current character's width).
     • Otherwise, add w to last (add the character to the current line).

3. Result:

   • After iterating through all characters, return an array [lines, last], containing the total number of lines and the width of the last line.

Time and Space Complexity:

• Time Complexity: O(n), where n is the length of the input string s. We iterate through the string once.
• Space Complexity: O(1). We use a constant amount of extra space to store variables (lines, last).

Code Implementation (Python):

def numberOfLines(widths, s):
    lines = 1
    last_line_width = 0
    for char in s:
        char_width = widths[ord(char) - ord('a')]
        if last_line_width + char_width <= 100:
            last_line_width += char_width
        else:
            lines += 1
            last_line_width = char_width  # Start a new line
    return [lines, last_line_width]
 

Code Implementation (Java):

class Solution {
    public int[] numberOfLines(int[] widths, String s) {
        int lines = 1;
        int lastLineWidth = 0;
        for (char c : s.toCharArray()) {
            int charWidth = widths[c - 'a'];
            if (lastLineWidth + charWidth <= 100) {
                lastLineWidth += charWidth;
            } else {
                lines++;
                lastLineWidth = charWidth;
            }
        }
        return new int[]{lines, lastLineWidth};
    }
}

Code Implementation (C++):

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

The other languages (JavaScript, Go, TypeScript, Rust) would follow a very similar structure, adapting the syntax to the specific language but maintaining the core algorithm. The key is the efficient single-pass iteration and the logic for handling line breaks.