# Decode Ways

A message containing letters from`A-Z`is being encoded to numbers using the following mapping:
'A' -> 1
'B' -> 2
...
'Z' -> 26
Given a non-empty string containing only digits, determine the total number of ways to decode it.
Example 1:
Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).
Example 2:
Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

## Analysis

### Dynamic Programming - 1

`dp[i]`代表string s从0到i能够decode的数目, number of ways to decode string from 0 to i (included)

### Dynamic Programming - 2 (Preferred, more succinct, clear, no need of hashset)

Use a dp array of size `n + 1` to save subproblem solutions.
`dp[0]` - means an empty string will have one way to decode,
`dp[1]`- means the way to decode a string of size 1. I then check one digit and two digit combination and save the results along the way. In the end,
`dp[n]` - will be the end result.

### Dynamic Programming - 3

`dp[i]` - 从`0``i`个字符对应的decode ways。因此`dp[]`的size是`n`，而不是`n+1`

char ch = s.charAt(i);
int num = ch - '0';
int oneVal = 0;
if (num >= 1 && num <= 9) {
oneVal = (i - 1 >= 0 ? dp[i - 1] : 1);
}
int twoVal = 0;
if (i > 0) {
num += (s.charAt(i - 1) - '0') * 10;
if (num >= 10 && num <= 26) {
twoVal = (i - 2 >= 0 ? dp[i - 2] : 1);
}
}
dp[i] = oneVal + twoVal;

## Solution

Dynamic Programming - 1 - O(n) space O(n) time
class Solution {
public int numDecodings(String s) {
if (s == null) return 0;
Set<String> letters = new HashSet<String>();
for (int i = 1; i <= 26; i++) {
}
int[] nums = new int[s.length()];
nums[0] = letters.contains(s.substring(0, 1)) ? 1 : 0;
if (s.length() == 1) return nums[0];
nums[1] = (letters.contains(s.substring(1, 2)) ? nums[0] : 0) +
(letters.contains(s.substring(0, 2)) ? 1 : 0);
for (int i = 2; i < s.length(); i++) {
nums[i] = (letters.contains(s.substring(i, i + 1)) ? nums[i - 1] : 0) +
(letters.contains(s.substring(i - 1, i + 1)) ? nums[i - 2] : 0);
}
return nums[s.length() - 1];
}
}
Dynamic Programming - 2 - O(n) space, O(n) time
substring 是 O(k) 时间复杂度，其中k是substring长度。
public class Solution {
public int numDecodings(String s) {
if(s == null || s.length() == 0) {
return 0;
}
int n = s.length();
int[] dp = new int[n+1];
dp[0] = 1;
dp[1] = s.charAt(0) != '0' ? 1 : 0;
for(int i = 2; i <= n; i++) {
int first = Integer.valueOf(s.substring(i-1, i));
int second = Integer.valueOf(s.substring(i-2, i));
if(first >= 1 && first <= 9) {
dp[i] += dp[i-1];
}
if(second >= 10 && second <= 26) {
dp[i] += dp[i-2];
}
}
return dp[n];
}
}

### *Dynamic Programming - 3 - Reference - O(n) space, O(n) time

0 ms, faster than 100.00%
/*
v
226
num: 26
oneVal: 2
twoVal: 1
dp: 1 2 3
dp[i]=
dp[i-1] 1~9
+
dp[i-2] 10~26
dp[0]=1 1~9 1
dp[1]=dp[0]+(01~26) 2
dp[2]=dp[0]+dp[1] 3
*/
class Solution { // O(N) | O(N)
public int numDecodings(String s) {
int n = s.length();
if (n == 0) {
return 0;
}
int[] dp = new int[n];
for (int i = 0; i < n; ++i) {
char ch = s.charAt(i);
int num = ch - '0';
int oneVal = 0;
if (num >= 1 && num <= 9) {
oneVal = (i - 1 >= 0 ? dp[i - 1] : 1);
}
int twoVal = 0;
if (i > 0) {
num += (s.charAt(i - 1) - '0') * 10;
if (num >= 10 && num <= 26) {
twoVal = (i - 2 >= 0 ? dp[i - 2] : 1);
}
}
dp[i] = oneVal + twoVal;
}
return dp[n - 1];
}
}

### Dynamic Programming - 3 - space optimized - O(1) space, O(n) time

/*
v
226
num: 26
oneVal: 2
twoVal: 1
dp: 1 2 3
dp[i]=
dp[i-1] 1~9
+
dp[i-2] 10~26
dp[0]=1 1~9 1
dp[1]=dp[0]+(01~26) 2
dp[2]=dp[0]+dp[1] 3
*/
-- 根据上一种方法，发现只需要dp[0], dp[1], dp[2] 三个元素存储中间状态即可，因此可以用滚动数组优化空间。
Space: O(1)
Time: O(n)
Memory Usage: 33.5 MB, less than 89.69%
Runtime: 0 ms, faster than 100.00%
class Solution { // O(1) space, O(n) time
public int numDecodings(String s) {
if (s == null || s.isEmpty()) {
return 0;
}
int n = s.length();
int[] dp = new int[3];
for (int i = 0; i < n; i++) {
char ch = s.charAt(i);
int num = ch - '0';
int oneDigit = 0;
if (num >= 1 && num <= 9) {
oneDigit = (i > 0) ? dp[(i - 1) % 3] : 1;
}
int twoDigits = 0;
if (i > 0) {
num += (s.charAt(i - 1) - '0') * 10;
if (num >= 10 && num <= 26) {
twoDigits = (i > 1) ? dp[(i - 2) % 3] : 1;
}
}
dp[i % 3] = oneDigit + twoDigits;
}
return dp[(n - 1) % 3];
}
}