Dynamic Programming - Medium Level - Question 1

Dynamic Programming - Medium Level - Question 1

Leetcode 91 Decode ways

A message containing letters from A-Z can be encoded into numbers using the following mapping:

'A' -> "1"

'B' -> "2"

...

'Z' -> "26"

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into:

"AAJF" with the grouping (1 1 10 6)

"KJF" with the grouping (11 10 6)

Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".

Given a string s containing only digits, return the number of ways to decode it.

The answer is guaranteed to fit in a 32-bit integer.

Constraints:

1 <= s.length <= 100

s contains only digits and may contain leading zero(s).

Analysis:

We can define f(i) as the number of ways to decode the string to the ith position.

Similar to Climbing Stairs, f(i) can be derived from f(i-1) and f(i-2). (or for each char, we at most just need to consider the previous char).

case 1. If s[i] itself is a valid choice, such as '1', '2', ... '9'; then one way to decode the string to the ith position is decoding the s[0, i-1] and the ith, and the number of ways to decode in this way is f(i-1). 

case 2. If s[i] can be connected with the previous char, then we can decode the string to the ith position as s[0, i-2] + s[i-1, i], and the number of ways to decode in this way is f(i-2).

For each char, 

if only 1 valid, then f(i) = f(i-1); 

else if only 2 valid, then f(i) = f(i-2);

else if both valid, then f(i) = f(i-1) + f(i-2)

else return 0. 

So now the question becomes how to categorize the question based on the value of s[i].  

At each position i, we have 10 possibilities, s[i] could be 0, 1, 2, ..., 9, which gives the range of discussion.

If the first char is '0', then there is no way to decode, game over;

In general, if s[i] == '0', then the s[i-1] has to be '1' or '2' (otherwise, there is no way to decode), and it has to connect with the previous char (since itself cannot be decoded), which is case 2; 

If s[i-1] == '0' or s[i-1] > '2', then s[i-1] cannot be connected with s[i], which is case 1;

If s[i-1] == '2' or s[i] > '6', then s[i-1] cannot be connected with s[i] neither, which is case 1;

For all the rest case,  both 1 and 2 ways are valid, so f[i] = f[i-1] + f[i-2].

See the code below:

class Solution {
public:
    int numDecodings(string s) {
        if(!s.size() || s[0] == '0') return 0;
        int n = s.size();
        vector<int> dp(n+1, 0);
        dp[0] = dp[1] = 1;
        for(int i=2; i<=n; ++i) {
            int cur = s[i-1] - '0', pre = s[i-2] - '0';
            // case 2
            if(cur == 0) dp[i] = (pre == 0 || pre > 2) ? 0 : dp[i-2];
            // case 1
            else if(pre > 2 || pre == 0 || cur > 6 && pre == 2) dp[i] = dp[i-1];
            // case 1 + 2
            else dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }
};

Since we just need the last two elements of the dp array, the space can be reduced to constant.

See the code below,

class Solution {
public:
    int numDecodings(string s) {
        if(!s.size() || s[0] == '0') return 0;
        int n = s.size();
        int a = 1, b = 1, c = 1;
        for(int i=1; i<n; ++i) {
            int cur = s[i] - '0', pre = s[i-1] - '0';
            if(cur == 0) c = (pre == 0 || pre > 2) ? 0 : a;
            else if(pre > 2 || pre == 0 || cur > 6 && pre == 2) c = b;
            else c = a + b;
            a = b;
            b = c;
        }
        return c;
    }
};

Upper Layer