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 <= 100scontains 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;
}
Input
Output
#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];
}
};
Input
Output
#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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Output
#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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Output
#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
Input
#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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Output