91. Decode Ways

LeetCode 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 answer is guaranteed to fit 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 = "0"
Output: 0
Explanation: There is no character that is mapped to a number starting with 0.
The only valid mappings with 0 are 'J' -> "10" and 'T' -> "20", neither of which start with 0.
Hence, there are no valid ways to decode this since all digits need to be mapped.

Constraints:

• 1 <= s.length <= 100

• s contains only digits and may contain leading zero(s).

Tags

String, Dynamic Programming

Solution

We initialize an array f with len(s)+1 elements to keep track of Decode Ways of each substring starting from 0. There are 2 cases need to be considered:

• f[i] += f[i-1] when s[i-1] != '0'

• f[i] += f[i-2] when s[i-2] != '0' and s[i-2]*10+s[i-1] <= 26

Also, we need consider the edge case f[0]=1 because the empty string only has 1 decode way.

Since the state f[i] only depends on its previous 2 states, we can use 3 variables to substitute an array to save space overhead.

Complexity

• Time complexity: $O(n)$

• Space complexity: $O(1)$

Code

// Space optimized
func numDecodings(s string) int {
	a, b, c := 0, 1, 0 // idx=i-2,i-1,i
	for i := 1; i <= len(s); i++ {
		c = 0
		if s[i-1] != '0' {
			c += b
		}
		if i > 1 && s[i-2] != '0' && (s[i-2]-'0')*10+(s[i-1]-'0') <= 26 {
			c += a
		}
		a, b = b, c
	}
	return c
}

// DP
func numDecodings(s string) int {
	f := make([]int, len(s)+1)
	f[0] = 1
	for i := 1; i <= len(s); i++ {
		if s[i-1] != '0' {
			f[i] += f[i-1]
		}
		if i > 1 && s[i-2] != '0' && (s[i-2]-'0')*10+(s[i-1]-'0') <= 26 {
			f[i] += f[i-2]
		}
	}
	return f[len(s)]
}

Reference

1. 解码方法