# Decode Ways Leetcode Solution

#### ByBrokenprogrammers

Dec 16, 2022

In this post, you will know how to solve the Decode Ways Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## Problem

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).

### Decode WaysLeetcode Solutions in Python

```def numDecodings(s):
if not s:
return 0
dp = [0 for x in range(len(s) + 1)]

# base case initialization
dp[0] = 1
dp[1] = 0 if s[0] == "0" else 1   #(1)
for i in range(2, len(s) + 1):
# One step jump
if 0 < int(s[i-1:i]) <= 9:    #(2)
dp[i] += dp[i - 1]
# Two step jump
if 10 <= int(s[i-2:i]) <= 26: #(3)
dp[i] += dp[i - 2]
return dp[len(s)]
```

### Decode Ways Leetcode Solutions in CPP

```int numDecodings(string s) {
return s.empty() ? 0: numDecodings(0,s);
}
int numDecodings(int p, string& s) {
int n = s.size();
if(p == n) return 1;
if(s[p] == '0') return 0; // sub string starting with 0 is not a valid encoding
int res = numDecodings(p+1,s);
if( p < n-1 && (s[p]=='1'|| (s[p]=='2'&& s[p+1]<'7'))) res += numDecodings(p+2,s);
return res;
}
```

### Decode Ways Leetcode Solutions in 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;
dp[1] = s.charAt(0) != '0' ? 1 : 0;
for (int i = 2; i <= n; i++) {
int first = Integer.valueOf(s.substring(i - 1, i));
int second = Integer.valueOf(s.substring(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];
}
}```

Note: This problem Decode Ways is generated by Leetcode but the solution is provided by  BrokenProgrammers. This tutorial is only for Educational and Learning purposes.