Decode Ways Leetcode Solution

In this post, we are going to solve the Decode Ways Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

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).

Now, let’s see the leetcode solution of Decode Ways Leetcode Solution.

Decode Ways Leetcode Solution 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 Solution 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 Solution 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 Chase2learn This tutorial is only for Educational and Learning purposes.

NEXT: Reverse Linked List II Leetcode Solution