How do you solve Number of Lines To Write String on LeetCode?

Number of Lines To Write String (LeetCode #806) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Each character's width is determined by its position in the alphabet.

What is the time complexity of Number of Lines To Write String?

The optimal solution for Number of Lines To Write String runs in O(n) time and uses O(1) space. The optimal solution runs in O(n) time because we only traverse the string once, making it efficient for longer strings.

What is the brute force solution for Number of Lines To Write String?

The brute force approach for Number of Lines To Write String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through each character of the string and adding its width to the current line's width until we exceed 100 pixels. We then start a new line and continue this process until all characters are processed. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Lines To Write String?

Number of Lines To Write String uses the following concepts: Array, String. The recommended patterns to study are: Array, Greedy.

What are the interview tips for Number of Lines To Write String?

Clarify the constraints and edge cases before coding. Explain your thought process as you code to show understanding. Test your code with different inputs to ensure correctness.

What are common mistakes when solving Number of Lines To Write String?

Not resetting the current width after exceeding 100 pixels. Confusing character indices with their widths.

#806

Number of Lines To Write String

Easy

ArrayStringArrayGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach is similar to the brute-force method but focuses on minimizing unnecessary checks. We keep track of the current line width and only check if we need to start a new line when adding a character, ensuring we only loop through the string once.

⚙️

Algorithm

6 steps

1Step 1: Initialize a counter for lines and a variable for the current line width.
2Step 2: Iterate through each character in the string.
3Step 3: For each character, calculate its width using the widths array.
4Step 4: If adding the character's width exceeds 100, increment the line counter and reset the current width.
5Step 5: Continue adding characters to the current line until all characters are processed.
6Step 6: Return the total number of lines and the width of the last line.

solution.py11 lines

1def numberOfLines(widths, s):
2    lines = 1
3    current_width = 0
4    for char in s:
5        width = widths[ord(char) - ord('a')]
6        if current_width + width > 100:
7            lines += 1
8            current_width = width
9        else:
10            current_width += width
11    return [lines, current_width]

ℹ

Complexity note: The optimal solution runs in O(n) time because we only traverse the string once, making it efficient for longer strings.

1Each character's width is determined by its position in the alphabet.
2The maximum width of a line is capped at 100 pixels.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.