Decode Ways
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# LeetCode #91: Decode Ways (C/C++).

medium

source: https://leetcode.com/problems/decode-ways/
C/C++ Solution to LeetCode problem 91. Decode Ways.

## 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.

## Examples

### 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).

## Solution

To solve this, we will use dynamic programming.

• Each group can consist of 1 or 2 digits (ranges: [A, Z] -> [1, 26]).
• If the group starts with 0, it can’t be decoded.
• If the group in its integer value is greater than 26 it can’t be decoded.

With this constraints, we will call recursivelly our decode method by moving 1 or 2 positions in the original string.
Once we are at the end of the string, it was successfully decoded, so, we return 1.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 class Solution { private: int decode(string s, int index, vector<int>& dp) { if (index >= s.size()) return 1; if (s[index] == '0') return 0; if (dp[index] != -1) return dp[index]; dp[index] = 0; dp[index] += decode(s, index + 1, dp); if ((index < s.size() - 1) && stoi(s.substr(index, 2)) < 27) dp[index] += decode(s, index + 2, dp); return dp[index]; } public: int numDecodings(string s) { vector<int> dp(s.size(), -1); int waysToDecode = decode(s, 0, dp); return waysToDecode; } };