Decode Ways II LeetCode Solution

Here, We see Decode Ways II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Decode Ways II LeetCode Solution

Decode Ways II LeetCode Solution

Problem Statement

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".

In addition to the mapping above, an encoded message may contain the '*' character, which can represent any digit from '1' to '9' ('0' is excluded). For example, the encoded message "1*" may represent any of the encoded messages "11""12""13""14""15""16""17""18", or "19". Decoding "1*" is equivalent to decoding any of the encoded messages it can represent.

Given a string s consisting of digits and '*' characters, return the number of ways to decode it.

Since the answer may be very large, return it modulo 109 + 7.

Example 1:
Input: s = “*”
Output: 9
Explanation: The encoded message can represent any of the encoded messages “1”, “2”, “3”, “4”, “5”, “6”, “7”, “8”, or “9”. Each of these can be decoded to the strings “A”, “B”, “C”, “D”, “E”, “F”, “G”, “H”, and “I” respectively.
Hence, there are a total of 9 ways to decode “*”.

Example 2:
Input: s = “1*”
Output: 18
Explanation: The encoded message can represent any of the encoded messages “11”, “12”, “13”, “14”, “15”, “16”, “17”, “18”, or “19”. Each of these encoded messages have 2 ways to be decoded (e.g. “11” can be decoded to “AA” or “K”).
Hence, there are a total of 9 * 2 = 18 ways to decode “1*”.

Example 3:
Input: s = “2*”
Output: 15
Explanation: The encoded message can represent any of the encoded messages “21”, “22”, “23”, “24”, “25”, “26”, “27”, “28”, or “29”. “21”, “22”, “23”, “24”, “25”, and “26” have 2 ways of being decoded, but “27”, “28”, and “29” only have 1 way.
Hence, there are a total of (6 * 2) + (3 * 1) = 12 + 3 = 15 ways to decode “2*”.

Decode Ways II LeetCode Solution C++

class Solution {
public:
    int MOD = 1e9+7;
    int dp[100001];
    int helper(int ind, string &s){
        if(ind == s.size()){
            return 1;
        }
        if(s[ind] == '0'){
            return 0;
        }
        if(dp[ind] != -1){
            return dp[ind];
        }
        long long first = helper(ind+1, s);
        if(s[ind] == '*'){
            first = (first*9)%MOD;
        }
        long long second = 0;
        long long third = 0;
        if(ind+1 < s.size()){
            if(s[ind] == '1' || s[ind] == '*'){
                if(s[ind+1] == '*'){
                    second = helper(ind+2, s);
                    second = (second*9)%MOD;
                }
                else if(s[ind+1] != '*'){
                    second = helper(ind+2, s);
                }
            }
            if(s[ind] == '2' || s[ind] == '*'){
                if(s[ind+1] == '*') {
                    third = helper(ind+2, s);
                    third = (third*6)%MOD;
                }
                else if(s[ind+1]-'0' <= 6){
                    third = helper(ind+2, s);
                }
            }
        }
        return dp[ind] = ((first+second)%MOD+third)%MOD;
    }
    int numDecodings(string s) {
        memset(dp, -1, sizeof(dp));
        return helper(0, s);
    }
};Code language: PHP (php)

Decode Ways II LeetCode Solution Java

class Solution {
    public int numDecodings(String s) {
        if (s == null) {
            throw new IllegalArgumentException("Input string is null");
        }
        if (s.length() == 0 || s.charAt(0) == '0') {
            return 0;
        }
        long pre = 1; // dp[i-2]
        long cur = s.charAt(0) == '*' ? 9 : 1;
        for (int i = 1; i < s.length(); i++) {
            long sum = 0; // dp[i]
            char curChar = s.charAt(i);
            char preChar = s.charAt(i - 1);
            if (curChar != '0') {
                sum = cur * (curChar == '*' ? 9 : 1);
            }
            if (preChar == '*') {
                if (curChar == '*') {
                    sum += pre * 15;
                } else if (curChar <= '6') {
                    sum += pre * 2;
                } else {
                    sum += pre;
                }
            } else {
                if (curChar == '*') {
                    if (preChar == '1') {
                        sum += pre * 9;
                    } else if (preChar == '2') {
                        sum += pre * 6;
                    }
                } else {
                    int num = Integer.parseInt(s.substring(i - 1, i + 1));
                    if (num >= 10 && num <= 26) {
                        sum += pre;
                    }
                }
            }
            pre = cur;
            cur = sum % 1000000007;
        }
        return (int) cur;
    }
}Code language: JavaScript (javascript)

Decode Ways II LeetCode Solution JavaScript

var numDecodings = function(s) {
    let mod = 1e9 + 7;
    let [e0, e1, e2, f0] = [1, 0, 0, 0];
    for (const c of s) {
        if (c == '*') {
            f0 = 9 * e0 + 9 * e1 + 6 * e2;
            e1 = e0;
            e2 = e0;
        } else {
            f0 = (c > '0') * e0 + e1 + (c <= '6') * e2;
            e1 = (c == '1') * e0;
            e2 = (c == '2') * e0;
        }
        e0 = f0 % mod;
    }
    return e0;  
};Code language: JavaScript (javascript)

Decode Ways II LeetCode Solution Python

class Solution(object):
    def numDecodings(self, S):
        answer = 1
        prev_answer = 1
        one, two = False, False
        for i in S:
            if i == '*':
                new = answer*9
                if one:
                    new += 9*prev_answer
                if two:
                    new += 6*prev_answer
                one, two = True, True
            else:
                new = answer if (i > '0') else 0
                if one:
                    new += prev_answer
                if (two and i <= '6'):
                    new += prev_answer
                one = (i == '1')
                two = (i == '2')
            prev_answer = answer
            answer = new % (10**9 + 7)
        return answerCode language: HTML, XML (xml)
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