Meta — Read N Characters Given read4 (Buffered I/O Design)

Problem Statement#

The API read4(buf4) reads up to 4 characters from a file into buf4 and returns the count of characters actually read. Using only read4, implement read(buf, n) which reads exactly n characters.

Constraints:

• 1 <= file size <= 500
• 1 <= n <= 500
• Version 1: read() is called only once
• Version 2: read() may be called multiple times

Input:  file = "abcde", n = 5
Output: 5, buf = ['a','b','c','d','e']

Input:  file = "abc", n = 4
Output: 3, buf = ['a','b','c']

Why This Problem Matters#

Read N Characters Given read4 (LeetCode 157 and 158) is a Meta-exclusive interview question that rarely appears elsewhere. Meta uses it specifically because it models real-world file I/O buffering — the pattern behind C's fread, Python's io.BufferedReader, and Java's BufferedInputStream. Engineers at Meta work extensively with large-scale data pipelines where buffer management is critical to performance.

Version 1 (call once) is relatively straightforward. Version 2 (multiple calls) is where candidates typically struggle. The key challenge is that read4 may have leftover characters from the previous call that must be consumed before calling read4 again. This requires storing internal state between calls — a design challenge that tests object-oriented thinking.

This problem is also a proxy for understanding API contracts: you must respect that read4 advances a file pointer you do not directly control, making state management non-trivial.

The Core Insight#

For Version 1: repeatedly call read4 into a temporary 4-char buffer. Copy min(count, remaining) characters to the output. Stop when read4 returns less than 4 (EOF) or you have read enough.

For Version 2: maintain an internal buffer (buf4) and two pointers (buf4_idx and buf4_count) as instance variables. On each read() call, first drain leftover characters from the internal buffer, then call read4 only when more data is needed. This preserves cross-call state correctly.

Visual Dry Run#

File: "abcdefgh", Version 2 calls: read(buf, 5) then read(buf, 3)

Solution (Optimal)#

# Version 1 — called once
def read(self, buf, n):
    total = 0
    tmp = [''] * 4
    while total < n:
        count = read4(tmp)
        count = min(count, n - total)
        buf[total:total + count] = tmp[:count]
        total += count
        if count < 4:
            break
    return total
 
 
# Version 2 — may be called multiple times
class Solution:
    def __init__(self):
        self.buf4 = [''] * 4
        self.buf4_idx = 0
        self.buf4_count = 0
 
    def read(self, buf, n):
        total = 0
        while total < n:
            if self.buf4_idx == self.buf4_count:
                self.buf4_count = read4(self.buf4)
                self.buf4_idx = 0
                if self.buf4_count == 0:
                    break
            buf[total] = self.buf4[self.buf4_idx]
            self.buf4_idx += 1
            total += 1
        return total

// Version 1
var read = function(buf, n) {
    let total = 0;
    const tmp = new Array(4);
    while (total < n) {
        const count = Math.min(read4(tmp), n - total);
        for (let i = 0; i < count; i++) buf[total++] = tmp[i];
        if (count < 4) break;
    }
    return total;
};
 
// Version 2
class Solution {
    constructor() {
        this.buf4 = new Array(4);
        this.buf4Idx = 0;
        this.buf4Count = 0;
    }
 
    read(buf, n) {
        let total = 0;
        while (total < n) {
            if (this.buf4Idx === this.buf4Count) {
                this.buf4Count = read4(this.buf4);
                this.buf4Idx = 0;
                if (this.buf4Count === 0) break;
            }
            buf[total++] = this.buf4[this.buf4Idx++];
        }
        return total;
    }
}

Time: O(N) — proportional to characters read Space: O(1) — only a 4-element internal buffer, no additional storage

Common Mistakes#

• In Version 2, calling read4 every time instead of draining the internal buffer first — corrupts file pointer
• Forgetting to reset buf4_idx = 0 after each fresh read4 call
• Not handling the case where read4 returns 0 (EOF reached) — causes infinite loop
• Copying more characters than n - total remaining — buffer overflow
• Not returning the actual count (returning n even when file ends early)

Interview Tips#

• Immediately clarify Version 1 vs Version 2 — they require fundamentally different designs
• For Version 2, draw the internal buffer state diagram before coding
• Mention that the class-level state (buf4_idx, buf4_count) is the entire crux of Version 2
• Discuss thread safety as a follow-up: two threads sharing the same Solution instance would corrupt state
• Frame this as a "producer-consumer with 4-element chunks" to show systems thinking

Follow-up Questions#

• How would you handle multi-threaded reads? — Each thread needs its own buffer state or a mutex
• What if read4 could return up to 8 characters? — Replace 4 with the new chunk size throughout
• How would you implement write4? — Similar buffer, but flush when full instead of fill
• What is the space-time tradeoff of increasing buffer size? — Larger buffer reduces read4 calls but uses more memory
• How does this relate to Java BufferedInputStream? — Same pattern: internal byte buffer drained before refilling

Key Takeaways#

• Version 1 is a simple loop; Version 2 requires instance-level state to preserve leftover characters between calls
• The internal buffer index and count are the two pointers that define "how much leftover is usable"
• When buf4_idx == buf4_count the internal buffer is empty — only then call read4 again
• This problem models real buffered I/O used in every file system, HTTP chunked transfer, and stream API
• Meta tests this to verify understanding of API contracts and state machine design
• Time is O(N) and space is O(1) — the fixed 4-element buffer does not grow with input size
• The pattern generalizes: replace read4 with any chunked-read primitive and apply the same state machine