LeetCode 157: Read N Characters Given Read4 Solution in Python Explained

Problem Statement

In LeetCode 157, you’re tasked with implementing a function read using a provided API read4:

• int read4(char[] buf4): Reads 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.

The function may be called multiple times, and you must handle file exhaustion. This differs from tree transformation like LeetCode 156: Binary Tree Upside Down, focusing on file reading rather than tree manipulation.

Constraints

• 1 <= n <= 10^4
• buf is large enough to hold n characters
• Multiple calls possible, file may exhaust

Example

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

Copy codeInput: file = "abcdefghijk", calls = ["read","read"], n = [4,3]
Output: [4,3]
Explanation:
<ul>
<li>read(buf, 4): Reads "abcd" into buf, returns 4.</li>
<li>read(buf, 3): Reads "efg" into buf, returns 3.</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 = [1,2]
Output: [1,1]
Explanation:
<ul>
<li>read(buf, 1): "a", returns 1.</li>
<li>read(buf, 2): "b", returns 1 (file ends).</li>
</ul>

These examples show we’re reading n characters with read4.

Understanding the Problem

How do you solve LeetCode 157: Read N Characters Given Read4 in Python? We need to read n characters using read4, which fetches 4 at a time, handling cases where n isn’t a multiple of 4 or the file ends. For "abcdefghijk", read(buf, 4) gets "abcd", then read(buf, 3) gets "efg". Multiple calls require tracking leftover characters. This isn’t a stack problem like LeetCode 155: Min Stack; it’s about buffered file reading. We’ll use: 1. Buffered Read with Queue: O(n) time, O(1) space—our best solution. 2. Direct 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 small queue (size ≤ 4) to store leftover characters from read4 calls:

• Read 4 characters into a temporary buffer via read4.
• Enqueue characters until n are read or the file ends.
• Dequeue into the target buffer, reusing leftovers across calls.

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

For "abcdefghijk", read(buf, 4):

• read4 → "abcd", all used.
• read(buf, 3) → read4 "efgh", use "efg", queue "h".

Step-by-Step Example

Example: file = "abcdefghijk", calls = ["read","read"], n = [4,3]

Goal: Return [4,3].

• Setup: Assume read4 reads "abcd", "efgh", etc.
• Step 1: read(buf, 4).
• queue = [], read4 → "abcd", count = 4.
• Copy "abcd" to buf, queue empty, return 4.
• buf = "abcd".
• Step 2: read(buf, 3).
• queue = [], read4 → "efgh", count = 4.
• Need 3, copy "efg" to buf, enqueue "h".
• queue = ['h'], buf = "efg", return 3.
• Finish: Output [4,3].

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

Goal: Return [3].

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

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

Goal: Return [1,1].

• Step 1: read(buf, 1).
• queue = [], read4 → "ab", count = 2.
• Copy "a" to buf, queue = ['b'], return 1.
• Step 2: read(buf, 2).
• Dequeue "b", need 1 more, read4 → "", count = 0.
• Copy "b" to buf, file ends, return 1.
• Finish: Return [1,1].

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 queue for leftovers
        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: Process queued characters
        while self.queue and read_count < n:
            # Use leftovers (e.g., queue=['h'])
            buf[read_count] = self.queue.pop(0)
            # Copy to buf (e.g., buf[0]='h')
            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=4)
            count = read4(buf4)
            # Call read4 (e.g., "abcd", count=4)
            if count == 0:
                # File exhausted (e.g., after "abc")
                break

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

            # Line 6: Queue leftover chars
            for i in range(chars_to_copy, count):
                # e.g., if n=3, count=4, queue "h"
                self.queue.append(buf4[i])

        # Line 7: Return number of chars read
        return read_count
        # e.g., 4 for "abcd"

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

Alternative: Direct Read with Array Approach

Explanation

Direct Read with Array uses a temporary array to store read4 results, copying directly to the buffer without a persistent queue:

• Read 4 characters at a time, copy up to n, discard extras.
• No state between calls, simpler but less flexible.

It’s a practical alternative, O(n) time with O(1) space (fixed-size array), but doesn’t handle multiple calls with leftovers as elegantly.

For "abcdefghijk", read(buf, 4):

• read4 → "abcd", copy all, discard none.

Step-by-Step Example (Alternative)

For "abc", read(buf, 5):

• Step 1: read4 → "abc", count = 3.
• Step 2: Copy "abc" to buf, return 3.
• Finish: Return 3.

How the Code Works (Direct Read)

Copy codeclass SolutionDirect:
    def read(self, buf: list[str], n: int) -> int:
        read_count = 0
        buf4 = [''] * 4
        while read_count < n:
            count = read4(buf4)
            if count == 0:
                break
            chars_to_copy = min(count, n - read_count)
            for i in range(chars_to_copy):
                buf[read_count + i] = buf4[i]
            read_count += 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).
• Direct 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 for multiple calls with leftovers—Direct Read with Array matches complexity but lacks state persistence, making it simpler but less robust.

Key Insights

• Queue: Manages leftovers.
• Direct: Discards extras.
• O(1): Space critical.

Additional Example

For file = "abcdef", n = 2, then n = 3:

• Queue: read(2) → "ab", queue="cd", read(3) → "cde".
• Direct: read(2) → "ab", read(3) → "abcd" (file reset issue).

Edge Cases

• Short File: "a", n=5 → 1.
• Exact Multiple: "abcd", n=4 → 4.
• Multiple Calls: Handled by queue.

Both solutions handle these, but queue excels with state.

Final Thoughts

LeetCode 157: Read N Characters Given Read4 in Python is a practical file-reading challenge. The Buffered Read with Queue solution excels with its efficiency and state management, while Direct Read with Array offers a simpler approach. Want more? Try LeetCode 158: Read N Characters Given Read4 II for call persistence or LeetCode 94: Binary Tree Inorder Traversal for tree skills. Ready to practice? Solve LeetCode 157 on LeetCode with "abcdefghijk" and n=[4,3], aiming for [4,3]—test your skills now!