Decode Ways II

This page explains Java solution to problem `Decode Ways II` using `Dynamic Programming`.

Problem Statement

A message containing letters from A-Z is being encoded to numbers using the following mapping way:

'A' -> 1
'B' -> 2
...
'Z' -> 26

Beyond that, now the encoded string can also contain the character `*`, which can be treated as one of the numbers from `1` to `9`.

Given the encoded message containing digits and the character `*`, return the total number of ways to decode it.

Also, since the answer may be very large, you should return the output mod `109 + 7`.

Example 1:

Input: "*"
Output: 9
Explanation: The encoded message can be decoded to the string: "A", "B", "C", "D", "E", "F", "G", "H", "I".

Example 2:

Input: 1*"
Output: 9 + 9 = 18

Solution

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``````
package com.vc.hard;

class DecodeWaysIi {
public int numDecodings(String s) {
if(s == null || s.length() == 0) return 0;

int MOD = (int)1e9 + 7;
int n = s.length();
long[] dp = new long[n + 1];

dp[0] = 1; // how many different ways empty string can be decoded
dp[1] = helper(s.charAt(0) + "");

//dp[i] means how many ways we can decode string ending at position i
for(int i = 2; i <= n; i++) {
long oneDigit = helper(s.substring(i - 1, i));
long twoDigit = helper(s.substring(i - 2, i));

if(oneDigit != 0) dp[i] += (dp[i - 1] * oneDigit) % MOD;
if(twoDigit != 0) dp[i] += (dp[i - 2] * twoDigit) % MOD;

dp[i] %= MOD;
}
return (int)dp[n];
}

private long helper(String str) {
if(str.length() == 1) {
if(str.charAt(0) == '*') return 9;
else if(str.charAt(0) == '0') return 0; //Zero can't be decoded as anything on it's own
else return 1;
}
else if(str.equals("**")) {
return 15; //11 to 26 except 20 as * can't be Zero
}
else if(str.charAt(0) == '*') {
/**
Consider str.charAt(1) = 0
10
20

Consider str.charAt(1) = 1
11
21

Consider str.charAt(1) = 2
12
22

Consider str.charAt(1) = 3
13
23

Consider str.charAt(1) = 4
14
24

Consider str.charAt(1) = 5
15
25

Consider str.charAt(1) = 6
16
26

After 6 we can only make 1 number
Consider str.charAt(1) = 7
17

Consider str.charAt(1) = 8
18

Consider str.charAt(1) = 9
19
*/
if(str.charAt(1) >= '0' && str.charAt(1) <= '6') return 2;
else return 1;
}
else if(str.charAt(1) == '*') {
if(str.charAt(0) == '1') return 9;
else if(str.charAt(0) == '2') return 6;
else return 0;
}
else {
int num = Integer.parseInt(str);
return (num >= 10 && num <= 26) ? 1 : 0;
}
}
}
``````

Time Complexity

O(N) Where
N length of input string S

Space Complexity

O(N) Where
N length of input string S