Decode Ways

A message containing letters fromA-Zis 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)

状态转移方程 dp[i] = if (s.substring(i, i + 1) is valid decode) then dp[i - 1] else 0 + if (s.substring(i - 1, i + 1) is valid decode) then dp[i - 2] else 0

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

Ref: https://leetcode.com/problems/decode-ways/discuss/30358/Java-clean-DP-solution-with-explanation

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

和第二种方法类似，但是用charAt(i)来得到对应位置字符，运行效率更高：

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;

并且可以用滚动数组优化空间复杂度为O(1)。

时间复杂度 O(n)。

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++) {
            letters.add(Integer.toString(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];
    }
}