717. 1-bit and 2-bit Characters #

Problem: #

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).

Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.

Example 1:

Input: 
bits = [1, 0, 0]
Output: True
Explanation: 
The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.

Example 2:

Input: 
bits = [1, 1, 1, 0]
Output: False
Explanation: 
The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.

Note:

• 1 <= len(bits) <= 1000.
• bits[i] is always 0 or 1.

Problem Summary #

There are 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).

Now given a string consisting of several bits. Determine whether the last character must be a one-bit character. The given string always ends with 0.

Note:

• 1 <= len(bits) <= 1000.
• bits[i] is always 0 or 1.

Solution Approach #

• Given an array whose elements are only 0 and 1, and the last element of the array must be 0. There are 2 special characters: the first type of character is “0”, and the second type of character is “11” and “10”. Determine whether the last element of this array must belong to the first type of character.
• According to the problem statement, 0 can come from 2 places: it can be the first type of character, or it can be part of the second type of character. 1 can come from only 1 place: it must be from the second type of character. There are 2 cases where the last 0 is the first type of character. In the first case, there is a 0 before it, but the 0 and 1 are paired to form the second type of character. In the second case, no 0 appears before it. The common point of these two cases is that, excluding the last element, all previous 1s in the array are “paired up”. Therefore, using the feature of the second type of character, “1X”, traverse the entire array. If “1” is encountered, jump 2 steps, because whatever number appears after 1 (0 or 1) does not need to be considered. If i can stop at len(bits) - 1 (the last element of the array), then this corresponds to case one or case two: the previous 0s have all been matched with 1s, and the last 0 must be the first type of character. If i stops at len(bit) (beyond the array index), it means bits[len(bits) - 1] == 1; at this point, the last 0 must belong to the second type of character.

Code #

package leetcode

func isOneBitCharacter(bits []int) bool {
	var i int
	for i = 0; i < len(bits)-1; i++ {
		if bits[i] == 1 {
			i++
		}
	}
	return i == len(bits)-1
}