91. Decode Ways

Description

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).

Solutions

Solution 1: Dynamic Programming

We define $f[i]$ to represent the number of decoding methods for the first $i$ characters of the string. Initially, $f[0]=1$, and the rest $f[i]=0$.

Consider how $f[i]$ transitions.

If the $i$th character (i.e., $s[i-1]$) forms a code on its own, it corresponds to one decoding method, i.e., $f[i]=f[i-1]$. The premise is $s[i-1] \neq 0$.
If the string formed by the $i-1$th character and the $i$th character is within the range $[1,26]$, then they can be treated as a whole, corresponding to one decoding method, i.e., $f[i] = f[i] + f[i-2]$. The premise is $s[i-2] \neq 0$, and $s[i-2]s[i-1]$ is within the range $[1,26]$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.

Python Code

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class Solution:
    def numDecodings(self, s: str) -> int:
        n = len(s)
        f = [1] + [0] * n
        for i, c in enumerate(s, 1):
            if c != "0":
                f[i] = f[i - 1]
            if i > 1 and s[i - 2] != "0" and int(s[i - 2 : i]) <= 26:
                f[i] += f[i - 2]
        return f[n]

Java Code

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class Solution {
    public int numDecodings(String s) {
        int n = s.length();
        int[] f = new int[n + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            if (s.charAt(i - 1) != '0') {
                f[i] = f[i - 1];
            }
            if (i > 1 && s.charAt(i - 2) != '0' && Integer.valueOf(s.substring(i - 2, i)) <= 26) {
                f[i] += f[i - 2];
            }
        }
        return f[n];
    }
}

C++ Code

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class Solution {
public:
    int numDecodings(string s) {
        int n = s.size();
        int f[n + 1];
        memset(f, 0, sizeof(f));
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            if (s[i - 1] != '0') {
                f[i] = f[i - 1];
            }
            if (i > 1 && (s[i - 2] == '1' || s[i - 2] == '2' && s[i - 1] <= '6')) {
                f[i] += f[i - 2];
            }
        }
        return f[n];
    }
};

Go Code

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func numDecodings(s string) int {
	n := len(s)
	f := make([]int, n+1)
	f[0] = 1
	for i := 1; i <= n; i++ {
		if s[i-1] != '0' {
			f[i] = f[i-1]
		}
		if i > 1 && (s[i-2] == '1' || (s[i-2] == '2' && s[i-1] <= '6')) {
			f[i] += f[i-2]
		}
	}
	return f[n]
}

TypeScript Code

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function numDecodings(s: string): number {
    const n = s.length;
    const f: number[] = new Array(n + 1).fill(0);
    f[0] = 1;
    for (let i = 1; i <= n; ++i) {
        if (s[i - 1] !== '0') {
            f[i] = f[i - 1];
        }
        if (i > 1 && (s[i - 2] === '1' || (s[i - 2] === '2' && s[i - 1] <= '6'))) {
            f[i] += f[i - 2];
        }
    }
    return f[n];
}

C# Code

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public class Solution {
    public int NumDecodings(string s) {
        int n = s.Length;
        int[] f = new int[n + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            if (s[i - 1] != '0') {
                f[i] = f[i - 1];
            }
            if (i > 1 && (s[i - 2] == '1' || (s[i - 2] == '2' && s[i - 1] <= '6'))) {
                f[i] += f[i - 2];
            }
        }
        return f[n];
    }
}

We notice that the state $f[i]$ is only related to the states $f[i-1]$ and $f[i-2]$, and is irrelevant to other states. Therefore, we can use two variables to replace these two states, reducing the original space complexity from $O(n)$ to $O(1)$.