1-bit and 2-bit Characters

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: true

💡 Note: The array decodes as one two-bit character "11" followed by one one-bit character "0". So the last character is a one-bit character.

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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Asked in

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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

✓ Count Two-Bit Characters

⏱️ Time: O(n) Space: O(1)

Instead of tracking position, we can count from the end. We look at all the 1's that appear before the last 0 - each represents a two-bit character. If there are an even number of consecutive 1's before the last 0, then the last 0 is a one-bit character.

Sequential Parsing

⏱️ Time: O(n) Space: O(1)

Start from the beginning and parse each character according to the rules. If we encounter 0, it's a one-bit character. If we encounter 1, the next bit forms a two-bit character (10 or 11).

Count Two-Bit Characters — Algorithm Steps

Start from second-to-last position
Count consecutive 1's going backwards
If count is even, last 0 is one-bit character
If count is odd, last 0 is part of two-bit character

Visualization

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Step-by-Step Walkthrough

Start from End

Begin from second-to-last position

Count 1's

Count consecutive 1's going backward

Check Parity

Even count = one-bit, odd count = two-bit

Code -

solution.c — C

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

bool solution(int* bits, int size) {
    int i = size - 2;  // Start from second-to-last position
    
    // Count consecutive 1's going backwards
    int onesCount = 0;
    while (i >= 0 && bits[i] == 1) {
        onesCount++;
        i--;
    }
    
    // If even number of 1's, last 0 is one-bit character
    // If odd number of 1's, last 0 is part of a two-bit character
    return onesCount % 2 == 0;
}

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)

In worst case, we scan backwards through consecutive 1's

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Count Two-Bit Characters — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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