# Leetcode – Number of Lines To Write String Solution in Python

The “Number of Lines To Write String” is a Leetcode problem that challenges programmers to understand basic string manipulations while considering space constraints. It’s a great problem for anyone looking to delve deeper into the realm of character positioning and layout calculations. This article will provide an in-depth look into the problem, solution methodologies, and the implementation in Python.

1. Problem Statement
2. Solution Strategies
3. Python Implementation
4. Testing and Analysis
6. Conclusion

### 1. Problem Statement

Given a string S and an integer array widths that indicates the number of pixels required for each character (a to z), the goal is to determine how many lines the string will occupy and the width of the last line.

The constraints are:

1. The characters are written horizontally.
2. Only the lowercase English alphabet is used.
3. Each word is separated by a single space.
4. A line can only contain 100 units.

You need to return two integers: the number of lines and the width of the last line.

### 2. Solution Strategies

1. Traversal and Accumulation: Traverse through each character of the string, accumulate the widths, and consider line breaks when the width surpasses 100.
2. Tracking Progress: Maintain counters for the current line width and the total number of lines required. Update them based on the accumulated width.

### 3. Python Implementation

def numberOfLines(widths, S):
lines, width_of_last_line = 1, 0

for char in S:
# Find the width of the current character
char_width = widths[ord(char) - ord('a')]

# Check if adding the current character exceeds line width
if width_of_last_line + char_width > 100:
lines += 1
width_of_last_line = 0

width_of_last_line += char_width

return [lines, width_of_last_line]

#### How the Code Works:

1. Initialize counters: lines to 1 (since we’ll at least need one line) and width_of_last_line to 0.
2. Loop through each character in string S.
3. Use the ord function to determine the character’s index in the widths array and fetch its width.
4. Check if adding this width to the current width exceeds 100. If so, increment the line count and reset the width for the new line.
5. Continue until all characters in S are processed.

### 4. Testing and Analysis

Testing the function with an example:

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"
print(numberOfLines(widths, S))  # Expected output: [3, 60]

Time Complexity: O(n), where n is the length of string S. We traverse the string once and perform constant-time operations during each iteration.

Space Complexity: O(1), as we use a fixed amount of space regardless of the input size.

1. Boundary Checks: Always ensure that the widths array has a length of 26 (for each lowercase English letter).
2. Edge Cases: What if the string S is empty? Our current solution would return [1, 0], which is accurate since we’d need one line, but its width is zero.