1-bit and 2-bit Characters - Practice Coding Problems

1-bit and 2-bit Characters - Problem

Array Easy

We have two special characters:

The first character can be represented by one bit 0
The second character can be represented by two bits (10 or 11)

Given a binary array bits that ends with 0, return true if the last character must be a one-bit character.

Note: The array is guaranteed to end with 0.

Input & Output

Example 1 — Two-bit then One-bit

$ Input: bits = [1,0,0]

› Output: true

💡 Note: The first character is a two-bit character "10", and the second character is a one-bit character "0". So the last character is a one-bit character.

Example 2 — Two consecutive 1's

$ Input: bits = [1,1,0]

› Output: false

💡 Note: The only way to decode this is as one two-bit character "11" followed by "0". However, this means the last 0 cannot be reached as a separate one-bit character through proper parsing.

Example 3 — Single character

$ Input: bits = [0]

› Output: true

💡 Note: The array contains only one bit "0", which is a one-bit character.

Constraints

1 ≤ bits.length ≤ 1000
bits[i] is either 0 or 1
bits[bits.length - 1] == 0

Visualization

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Understanding the Visualization

Input

Binary array ending with 0

Parse

Follow encoding: 0→1 step, 1→2 steps

Check

Land exactly on last 0?

Key Takeaway

🎯 Key Insight: Parse from start following encoding rules - if we land exactly on the last 0, it's a one-bit character

Asked in

G Google 15 f Facebook 12

The key insight is that we need to parse characters from the beginning following encoding rules (0 = one-bit, 1x = two-bit) and check if we land exactly on the last 0. Best approach is sequential parsing with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Sequential Parsing	O(n)	O(1)	Parse characters from beginning, tracking current position
Count Two-Bit Characters	O(n)	O(1)	Count how many two-bit characters appear before the last 0

Sequential Parsing — Algorithm Steps

Start at index 0
If current bit is 0, move 1 position (one-bit character)
If current bit is 1, move 2 positions (two-bit character)
Check if we land exactly on the last 0

Visualization

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Step-by-Step Walkthrough

Start Parsing

Begin at index 0, follow encoding rules

Skip Characters

0 = move 1 step, 1 = move 2 steps

Check Landing

If we land on last 0, it's one-bit character

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(int* bits, int size) {
    int i = 0;
    // Parse until we reach the second-to-last position
    while (i < size - 1) {
        if (bits[i] == 0) {
            i += 1;  // One-bit character
        } else {
            i += 2;  // Two-bit character (10 or 11)
        }
    }
    
    // If we land exactly on the last position, it's a one-bit character
    return i == size - 1;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    int bits[100];
    int size = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        bits[size++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    bool result = solution(bits, size);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array, visiting each position at most once

✓ Linear Growth

Space Complexity

O(1)

Only using a single pointer variable to track position

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Sequential Parsing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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