Algorithm


Problem Nmae: 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 test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

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

Example 2:

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

Example 3:

Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").

Constraints:

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

 

Code Examples

#1 Code Example with C Programming

Code - C Programming


int numDecodings(char* s) {
    int a, b, c;
    char p, t;
    
    /* a b c
           2267312
       0 1 
    */
    a = 0;
    b = 1;
    c = 0;
    p = 'x';    // anything other than '1' or '2'
    
    while (t = *s ++) {
        if (t == '0') {
            if (p != '1' && p != '2') {
                return 0;
            }
            c = a;
        } else {
            c = b;
            if (p == '1' || (p == '2' && t  < = '6')) {
                c += a;
            }
        }
        a = b;
        b = c;
        p = t;
    }

    return c;
}
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Input

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s = "12"

Output

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2

#2 Code Example with C++ Programming

Code - C++ Programming


class Solution {
public:
    int numDecodings(string s) {
        /** dp[i] = 
         * value                            Example
         * 0                                00, 30, 80 - invalid ending
         * dp[i-2]                          10, 20     - valid ending with 0
         * dp[i-2]                          08, 09     - s[i - 1] == '0'      
         * dp[i-1] + dp[i-2]                11, 16     - valid ending
         * dp[i-1]                          32, 56     - large ending, decrease i by 1
         */
        if(s.size() == 0 || s[0] == '0') return 0;
        vector<int>dp(s.size());
        dp[0] = 1;
        for(int i = 1; i < s.size(); i++){
            if(s[i] == '0'>{
                if(s[i - 1] == '0' || s[i - 1] - '0' > 2) return 0;
                dp[i] = (i==1) ? dp[0] : dp[i - 2];
            }
            else if(s[i - 1] == '0') dp[i] = dp[i - 2];
            else if(stoi(s.substr(i - 1, 2)) <= 26) dp[i] = (i==1) ? dp[0] + 1 : dp[i - 1] + dp[i - 2];
            else dp[i] = dp[i - 1];
        }
        return dp[s.size(>-1];
    }
};
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Input

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s = "12"

Output

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2

#3 Code Example with Java Programming

Code - Java Programming


class Solution {
  
  public int numDecodings(String s) {
    Map map = new HashMap<>();
    return helper(s, 0, map);
  } 
  
  private int helper(String s, int idx, Map < Integer, Integer> map) {
    if (map.containsKey(idx)) {
      return map.get(idx);
    }
    if (idx == s.length()) {
      return 1;
    }
    if (s.charAt(idx) == '0') {
      return 0;
    }
    if (idx == s.length() - 1) {
      return 1;
    }
    int result = helper(s, idx + 1, map);
    if (Integer.parseInt(s.substring(idx, idx + 2))  < = 26) {
      result += helper(s, idx + 2, map);
    }
    map.put(idx, result);
    return result;
  }
}
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Input

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s = "226"

Output

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3

#4 Code Example with Javascript Programming

Code - Javascript Programming


const numDecodings = function(s) {
  if(s == null || s.length === 0) return 1
  if(s[0] === '0') return 0
  const set = new Set()
  const n = s.length
  for(let i = 1; i <= 26; i++) {
    set.add(`${i}`)
  }
  const dp = Array(n + 1).fill(0)
  dp[0] = dp[1] = 1
  for(let i = 2; i  < = n; i++) {
    if(set.has(s[i - 2] + s[i - 1])) dp[i] += dp[i - 2]
    if(set.has(s[i - 1])> dp[i] += dp[i - 1]
  }
  return dp[n]
};
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Input

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s = "06"

Output

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s = "06"

#5 Code Example with Python Programming

Code - Python Programming


class Solution:
    def numDecodings(self, s):
        if s[0] == "0": return 0
        dp1 = dp2 = 1
        for i in range(1, len(s)):
            if s[i] == "0" and (s[i - 1] == "0" or s[i - 1] >= "3"): return 0
            dp1, dp2 = [dp2, dp1] if s[i] == "0" else [dp2, dp2 + dp1] if "10" <= s[i -1: i + 1] <= "26" else [dp2, dp2]
        return dp2  
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Input

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s = "06"

#6 Code Example with C# Programming

Code - C# Programming


namespace LeetCode
{
    public class _091_DecodeWays
    {
        public int NumDecodings(string s)
        {
            if (s.Length == 0) return 0;

            int[] memo = new int[s.Length + 1];
            memo[s.Length] = 1;
            memo[s.Length - 1] = s[s.Length - 1] == '0' ? 0 : 1;

            for (var i = s.Length - 2; i >= 0; i--)
            {
                if (s[i] == '0') memo[i] = 0;
                else
                    memo[i] = int.Parse(s.Substring(i, 2))  < = 26 ? memo[i + 1] + memo[i + 2] : memo[i + 1];
            }
            return memo[0];
        }
    }
}
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Input

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s = "12"

Output

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2
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