LeetCode 158: Read N Characters Given Read4 II - Call Multiple Times Solution in Python Explained

Problem Statement

In LeetCode 158, you’re tasked with implementing a function read using a provided API read4:

• int read4(char[] buf4): Reads up to 4 characters from a file into buf4, returning the number read (0 to 4).
• int read(char[] buf, int n): Reads n characters into buf, returning the number read, calling read4 as needed, with state persisting across calls.

The function will be called multiple times on the same file, requiring you to track unread characters. This builds on LeetCode 157: Read N Characters Given Read4, adding persistence, and differs from tree manipulation like LeetCode 156: Binary Tree Upside Down.

Constraints

• 1 <= n <= 10^4
• buf is large enough to hold n characters
• Multiple calls on the same file

Example

Let’s see some cases (assuming a file "abcdefgh"):

Copy codeInput: file = "abcdefgh", calls = ["read","read","read"], n = [1,2,1]
Output: [1,2,1]
Explanation:
<ul>
<li>read(buf, 1): "a" (queue "bcd"), returns 1.</li>
<li>read(buf, 2): "bc" (queue "d"), returns 2.</li>
<li>read(buf, 1): "d" (queue empty), returns 1.</li>
</ul>

Copy codeInput: file = "abc", calls = ["read"], n = [5]
Output: [3]
Explanation: Reads "abc", returns 3 (file ends).

Copy codeInput: file = "ab", calls = ["read","read"], n = [5,2]
Output: [2,0]
Explanation:
<ul>
<li>read(buf, 5): "ab", returns 2 (file ends).</li>
<li>read(buf, 2): "", returns 0.</li>
</ul>

These examples show we’re reading with state persistence.

Understanding the Problem

How do you solve LeetCode 158: Read N Characters Given Read4 II - Call Multiple Times in Python? We need to read n characters using read4, which fetches 4 at a time, persisting leftovers across calls. For "abcdefgh", read(buf, 1) gets "a", queues "bcd"; read(buf, 2) uses "bc", queues "d". This isn’t a stack problem like LeetCode 155: Min Stack; it’s about buffered reading with state. We’ll use: 1. Buffered Read with Queue: O(n) time, O(1) space—our best solution. 2. Buffered Read with Array: O(n) time, O(1) space—an alternative.

Let’s dive into the best solution.

Best Solution: Buffered Read with Queue Approach

Explanation

Buffered Read with Queue uses a persistent queue to store leftover characters from read4 calls:

• Maintain a queue (size ≤ 4) as a class variable.
• Dequeue existing characters into the buffer until n or queue empties.
• Call read4 to refill, enqueue extras, dequeue into buffer as needed.

This is the best solution due to its O(n) time complexity (each character processed once), O(1) space (fixed-size queue), and robust handling of multiple calls, making it efficient and stateful.

For "abcdefgh", read(buf, 1):

• read4 → "abcd", use "a", queue "bcd".
• read(buf, 2) → dequeue "bc", queue "d".

Step-by-Step Example

Example: file = "abcdefgh", calls = ["read","read","read"], n = [1,2,1]

Goal: Return [1,2,1].

• Setup: self.queue = [].
• Step 1: read(buf, 1).
• queue = [], read4 → "abcd", count = 4.
• Copy "a" to buf, queue = ['b','c','d'], return 1.
• buf = "a".
• Step 2: read(buf, 2).
• queue = ['b','c','d'], dequeue "b", "c".
• buf = "bc", queue = ['d'], return 2.
• Step 3: read(buf, 1).
• queue = ['d'], dequeue "d".
• buf = "d", queue = [], return 1.
• Finish: Output [1,2,1].

Example: file = "abc", call = ["read"], n = [5]

Goal: Return [3].

• Step 1: read(buf, 5).
• queue = [], read4 → "abc", count = 3.
• Copy "abc", queue = [], file ends, return 3.
• buf = "abc".
• Finish: Return 3.

Example: file = "ab", calls = ["read","read"], n = [5,2]

Goal: Return [2,0].

• Step 1: read(buf, 5).
• queue = [], read4 → "ab", count = 2.
• Copy "ab", queue = [], return 2.
• Step 2: read(buf, 2).
• queue = [], read4 → "", count = 0, return 0.
• Finish: Return [2,0].

How the Code Works (Buffered Read with Queue) – Detailed Line-by-Line Explanation

Here’s the Python code with a thorough breakdown, assuming read4 is provided:

Copy codeclass Solution:
    def __init__(self):
        # Line 1: Initialize persistent queue
        self.queue = []
        # Stores leftover chars (e.g., [])

    def read(self, buf: list[str], n: int) -> int:
        # Line 2: Initialize variables
        read_count = 0
        # Total chars read (e.g., 0)
        buf4 = [''] * 4
        # Temp buffer for read4 (e.g., ['', '', '', ''])

        # Line 3: Use queued characters
        while self.queue and read_count < n:
            # Process leftovers (e.g., queue=['b','c','d'])
            buf[read_count] = self.queue.pop(0)
            # Copy to buf (e.g., buf[0]='b')
            read_count += 1
            # Increment count (e.g., 1)

        # Line 4: Read from file while needed
        while read_count < n:
            # Continue until n or file ends (e.g., n=2)
            count = read4(buf4)
            # Call read4 (e.g., "abcd", count=4)
            if count == 0:
                # File exhausted (e.g., after "ab")
                break

            # Line 5: Copy needed chars to buf
            chars_to_copy = min(count, n - read_count)
            # e.g., min(4, 2-0)=2
            for i in range(chars_to_copy):
                # Copy chars (e.g., "ab")
                buf[read_count + i] = buf4[i]
            read_count += chars_to_copy
            # Update count (e.g., 2)

            # Line 6: Queue leftover chars
            for i in range(chars_to_copy, count):
                # e.g., if n=2, count=4, queue "cd"
                self.queue.append(buf4[i])

        # Line 7: Return number of chars read
        return read_count
        # e.g., 2 for "bc"

This detailed breakdown clarifies how buffered read with a queue efficiently manages persistent reading.

Alternative: Buffered Read with Array Approach

Explanation

Buffered Read with Array uses a persistent array with a pointer to track leftovers:

• Store up to 4 characters from read4 in a class-level array.
• Use a pointer to track the current position in the buffer.
• Refill when the buffer is exhausted.

It’s a practical alternative, O(n) time with O(1) space (fixed-size array), but requires managing an index, slightly less flexible than a queue.

For "abcdefgh", read(buf, 1):

• read4 → "abcd", use "a", buffer "bcd", ptr=1.
• read(buf, 2) → use "bc", buffer "d", ptr=1.

Step-by-Step Example (Alternative)

For "abcdefgh", [1,2,1]:

• Step 1: read(buf, 1).
• buffer = [], read4 → "abcd", buffer = "abcd", ptr = 0.
• Copy "a", ptr = 1, return 1.
• Step 2: read(buf, 2).
• buffer = "abcd", ptr = 1, copy "bc", ptr = 3, return 2.
• Step 3: read(buf, 1).
• buffer = "abcd", ptr = 3, copy "d", ptr = 4, refill "efgh", copy "e", return 1.

How the Code Works (Array)

Copy codeclass SolutionArray:
    def __init__(self):
        self.buffer = [''] * 4
        self.ptr = 0
        self.count = 0

    def read(self, buf: list[str], n: int) -> int:
        read_count = 0
        while read_count < n:
            if self.ptr >= self.count:
                self.count = read4(self.buffer)
                self.ptr = 0
                if self.count == 0:
                    break
            chars_to_copy = min(self.count - self.ptr, n - read_count)
            for i in range(chars_to_copy):
                buf[read_count + i] = self.buffer[self.ptr + i]
            read_count += chars_to_copy
            self.ptr += chars_to_copy
        return read_count

Complexity

• Buffered Read with Queue:
• Time: O(n) – processes each char once.
• Space: O(1) – fixed-size queue (max 4).
• Buffered Read with Array:
• Time: O(n) – processes each char once.
• Space: O(1) – fixed-size array.

Efficiency Notes

Buffered Read with Queue is the best solution with O(n) time and O(1) space, offering flexibility and simplicity—Buffered Read with Array matches complexity but requires pointer management, making it slightly less intuitive.

Key Insights

• Queue: Natural state management.
• Array: Index-based buffering.
• Persistence: Key to multiple calls.

Additional Example

For file = "abcdef", n = [3,2]:

• Queue: read(3) → "abc", queue="def", read(2) → "de".

Edge Cases

• Short File: "a", n=5 → 1.
• Multiple Calls: "ab", [1,2] → [1,1].
• Empty: "", n=1 → 0.

Both solutions handle these well.

Final Thoughts

LeetCode 158: Read N Characters Given Read4 II in Python is a practical persistence challenge. The Buffered Read with Queue solution excels with its efficiency and clarity, while Buffered Read with Array offers a compact alternative. Want more? Try LeetCode 157: Read N Characters Given Read4 for the basics or LeetCode 94: Binary Tree Inorder Traversal for tree skills. Ready to practice? Solve LeetCode 158 on LeetCode with "abcdefgh" and n=[1,2,1], aiming for [1,2,1]—test your skills now!