Decode Ways

, , , , ,

Question

A message containing letters from A-Z is being encoded to numbers using the following mapping:

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'A' -> 1
'B' -> 2
...
'Z' -> 26

Example

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Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).

Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

Solution

In this question, we have to try different length of string, such as one or two because the number belongs to 1 to 26.

Code

Brute force (DFS)

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// Time:O(2^n), Space:O(n) 
public int numDecodings(String s) {
        if (s == null || s.length() == 0) return 0;
        return dfs(s, 0);
    }
    
    public int dfs(String s, int start) {
        if (start > s.length()) return 0;
        if (start == s.length()) return 1;
        // move one char
        int res = 0;
        
        if (start + 1 <= s.length()) {
            int oneNum = Integer.parseInt(s.substring(start, start + 1));
            if (oneNum >= 0 && oneNum <= 9) {
                res += dfs(s, start + 1);
            }
        }
        
        // move two char
        if (start + 2 <= s.length()) {
            int twoNum = Integer.parseInt(s.substring(start, start + 2));
            if (twoNum >= 10 && twoNum <= 26) {
                res += dfs(s, start + 2);
            }
        }
        return res;
    }

DFS + Memo

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// time:O(n) space:O(n)
public int numDecodings(String s) {
        int n = s.length();
        if (s == null || n == 0) return 0;
        
        int[] memo = new int[n + 1];
        Arrays.fill(memo, -1);
        return dfs(s, 0, memo);
    }
    
    public int dfs(String s, int start, int[] memo) {
        if (memo[start] != -1) return memo[start];
        if (start > s.length()) return 0;
        if (start == s.length()) return 1;
        // move one char
        int res = 0;
        
        if (start + 1 <= s.length()) {
            int oneNum = Integer.parseInt(s.substring(start, start + 1));
            if (oneNum >= 1 && oneNum <= 9) {
                res += dfs(s, start + 1, memo);
            }
        }
        
        // move two char
        if (start + 2 <= s.length()) {
            int twoNum = Integer.parseInt(s.substring(start, start + 2));
            if (twoNum >= 10 && twoNum <= 26) {
                res += dfs(s, start + 2, memo);
            }
        }
        memo[start] = res;
        return memo[start];
    }

Bottom up DP

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// time:O(n) space:O(n) 
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' ? 0 : 1; // 对应上s的第0位。
        for (int i = 2; i <= n; i++) {
            int oneDigit = s.charAt(i - 1) - '0';
            int twoDigits = (s.charAt(i - 2) - '0') * 10 +
                    (s.charAt(i - 1) - '0');
            if (oneDigit > 0 && oneDigit <= 9) {
                dp[i] += dp[i - 1];
            }
            if (twoDigits >= 10 && twoDigits <= 26) {
                dp[i] += dp[i - 2];
            }
        }
        return dp[n];
    }