Leetcode #157: Read N Characters Given Read4
In this guide, we solve Leetcode #157 Read N Characters Given Read4 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
Given a file and assume that you can only read the file using a given method read4, implement a method to read n characters. Method read4: The API read4 reads four consecutive characters from file, then writes those characters into the buffer array buf4.
Quick Facts
- Difficulty: Easy
- Premium: Yes
- Tags: Array, Interactive, Simulation
Intuition
The rules are explicit, so simulating the process step by step is safest.
Careful state updates prevent subtle bugs.
Approach
Translate the rules into state updates and apply them in order.
Track the final state or aggregate as required.
Steps:
- Translate rules into state updates.
- Iterate for each step.
- Return the final state.
Example
Parameter: char[] buf4
Returns: int
buf4[] is a destination, not a source. The results from read4 will be copied to buf4[].
Python Solution
"""
The read4 API is already defined for you.
@param buf4, a list of characters
@return an integer
def read4(buf4):
# Below is an example of how the read4 API can be called.
file = File("abcdefghijk") # File is "abcdefghijk", initially file pointer (fp) points to 'a'
buf4 = [' '] * 4 # Create buffer with enough space to store characters
read4(buf4) # read4 returns 4. Now buf = ['a','b','c','d'], fp points to 'e'
read4(buf4) # read4 returns 4. Now buf = ['e','f','g','h'], fp points to 'i'
read4(buf4) # read4 returns 3. Now buf = ['i','j','k',...], fp points to end of file
"""
class Solution:
def read(self, buf, n):
"""
:type buf: Destination buffer (List[str])
:type n: Number of characters to read (int)
:rtype: The number of actual characters read (int)
"""
i = 0
buf4 = [0] * 4
v = 5
while v >= 4:
v = read4(buf4)
for j in range(v):
buf[i] = buf4[j]
i += 1
if i >= n:
return n
return i
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
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.