Leetcode #1618: Maximum Font to Fit a Sentence in a Screen
In this guide, we solve Leetcode #1618 Maximum Font to Fit a Sentence in a Screen in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a string text. We want to display text on a screen of width w and height h.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, String, Binary Search, Interactive
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
interface FontInfo {
// Returns the width of character ch on the screen using font size fontSize.
// O(1) per call
public int getWidth(int fontSize, char ch);
// Returns the height of any character on the screen using font size fontSize.
// O(1) per call
public int getHeight(int fontSize);
}
Python Solution
# """
# This is FontInfo's API interface.
# You should not implement it, or speculate about its implementation
# """
# class FontInfo(object):
# Return the width of char ch when fontSize is used.
# def getWidth(self, fontSize, ch):
# """
# :type fontSize: int
# :type ch: char
# :rtype int
# """
#
# def getHeight(self, fontSize):
# """
# :type fontSize: int
# :rtype int
# """
class Solution:
def maxFont(
self, text: str, w: int, h: int, fonts: List[int], fontInfo: 'FontInfo'
) -> int:
def check(size):
if fontInfo.getHeight(size) > h:
return False
return sum(fontInfo.getWidth(size, c) for c in text) <= w
left, right = 0, len(fonts) - 1
ans = -1
while left < right:
mid = (left + right + 1) >> 1
if check(fonts[mid]):
left = mid
else:
right = mid - 1
return fonts[left] if check(fonts[left]) else -1
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.