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

List of all LeetCode Solution

Decode Ways 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".

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").

Decode Ways Leetcode Solution C++

class Solution {
public:
    int numDecodings(string s) {
        if (s[0] == '0') return 0;
        if (s.size() == 1) return 1;
        int len = s.size(), dp[len];
        dp[0] = 1;
        dp[1] = (s[0] == '1' || s[0] == '2' && s[1] < '7' ? 1 : 0) + (s[1] != '0');
        for (int i = 2; i < len; i++) {
            if (s[i] == '0' && (s[i - 1] > '2' || s[i - 1] == '0')) return 0;
            dp[i] = s[i] != '0' ? dp[i - 1] : 0;
            if (s[i - 1] == '1' || s[i - 1] == '2' && s[i] < '7') dp[i] += dp[i - 2];
        }
        return dp[len - 1];  
    }
};Code language: PHP (php)

Decode Ways Leetcode Solution Java

class Solution {
    public int numDecodings(String s) {
	Integer[] memo = new Integer[s.length() + 1];
	return numDecodings(s, 0, memo);
}

private int numDecodings(String s, int index, Integer[] memo) {
	if (index == s.length()) {
		return 1;
	}
	if (s.charAt(index) == '0') {
		return 0;
	}
	if (memo[index] != null) {
		return memo[index];
	}
	int way1 = numDecodings(s, index + 1, memo);
	int way2 = 0;
	if (index < s.length() - 1 && Integer.parseInt(s.substring(index, index + 2)) <= 26) {
		way2 = numDecodings(s, index + 2, memo);
	}
	memo[index] = way1 + way2;
	return memo[index];
}
}Code language: JavaScript (javascript)

Decode Ways Leetcode Solution JavaScript

var numDecodings = function(s) {
  if (s == null || s.length === 0) return 0;
  if (s[0] === '0') return 0;

  const dp = new Array(s.length + 1).fill(0);

  dp[0] = 1;
  dp[1] = 1;

  for (let i = 2; i <= s.length; i++) {
    const a = Number(s.slice(i - 1, i));  // last one digit
    if (a >= 1 && a <= 9) {
      dp[i] += dp[i - 1];
    }

    const b = Number(s.slice(i - 2, i));  // last two digits
    if (b >= 10 && b <= 26) {
      dp[i] += dp[i - 2];
    }
  }
  return dp[s.length];   
};Code language: JavaScript (javascript)

Decode Ways Leetcode Solution Python

class Solution(object):
    def numDecodings(self, s):
    	if not s:
		return 0

	dp = [0 for x in range(len(s) + 1)] 
	
	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)]