0091. 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.

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

Example 1:

Input: s = "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).

Example 2:

Input: s = "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

Example 3:

Input: s = "0"
Output: 0
Explanation: There is no character that is mapped to a number starting with '0'. We cannot ignore a zero when we face it while decoding. So, each '0' should be part of "10" --> 'J' or "20" --> 'T'.

Example 4:

Input: s = "1"
Output: 1

Constraints:

• 1 <= s.length <= 100
• s contains only digits and may contain leading zero(s).

java

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; // # of ways to decode for empty s
        dp[1] = s.charAt(0) != '0' ? 1 : 0; // 0 cannot be leading element for a number
        for(int i = 2; i <= n; i++) {
            int first = Integer.valueOf(s.substring(i-1, i)); // s[i-1:i]
            int second = Integer.valueOf(s.substring(i-2, i));// s[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];
    }
}

dp[i]: number of ways to decode

1. dp[i] += dp[i-1] if s[i-1:i]is valid (single digit: range from 0-9)
2. dp[i] += dp[i-2] if s[i-2:i] is valid (two digits: range from 10-26)

code

int numDecodings(string s) {
    int n = s.size();
    vector<int> dp(n+1);
    dp[n] = 1;
    for(int i = n - 1; ~i; i--) { // from back to front
        if(s[i]=='0') dp[i]=0;
        else {
            dp[i] = dp[i+1];
            if(i<n-1 && (s[i]=='1'||s[i]=='2'&&s[i+1]<'7')) dp[i]+=dp[i+2];
        }
    }
    return s.empty()? 0 : dp[0];
}

dp[i] += dp[i+2] if 1. s[i] == 1, so that any s[i+2] is valid, s[i:i+1] is in range 10-19 2. s[i] == 2 and s[i+1] < 7, so that s[i:i+1] is in range 20-26 if s[i] == 0, then dp[i] = 0, because 0 cannot be leading

optimize with constant space

class Solution {
public:
    int numDecodings(string s) {
        int n = s.size();
        int pre = 1, prepre = 1, curr = 1;
        for(int i=n-1;i>=0;i--) { // from back to front
            if(s[i]=='0') curr=0;
            else {
                curr=pre;
                if(i<n-1 && (s[i]=='1'||s[i]=='2'&&s[i+1]<'7')) 
                    curr+=prepre;
            }
            prepre = pre;
            pre = curr;
            curr = 1;
        }
        return s.empty()? 0 : pre;
    }
};

only needs two more states from the current state, so 1. pre: s[i:i+1] 2. prepre: s[i:i+2]

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Last update: April 1, 2022