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.
public class Solution {
public bool IsOneBitCharacter(int[] bits) {
return Helper(bits,0);
}
private bool Helper(int[] bits, int start){
if(start == bits.Length-1){
return bits[start]==0;
}
if(start > bits.Length-1){
return false;
}
var valid1 = false;
if(bits[start]==1){
valid1 = Helper(bits,start+2);
}
var valid2 = false;
if(bits[start]==0){
valid2 = Helper(bits,start+1);
}
return valid1||valid2;
}
}
Time Complexity: O(2^n) since it is a recursive call ( we can optimize it using Dynamic Programming)
Space Complexity: O(n) for stack trace


