Decode Ways LeetCode Solution | Easy Approach

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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).

 Decode Ways Solutions

Time: O(n)
Space: O(n)

C++

 Will be updated Soonclass Solution {
 public:
  int numDecodings(string s) {
    const int n = s.length();
    // dp[i] := # of ways to decode s[i..n)
    vector<int> dp(n + 1);
    dp[n] = 1;  // ""
    dp[n - 1] = isValid(s[n - 1]);

    for (int i = n - 2; i >= 0; --i) {
      if (isValid(s[i]))
        dp[i] += dp[i + 1];
      if (isValid(s[i], s[i + 1]))
        dp[i] += dp[i + 2];
    }

    return dp[0];
  }

 private:
  bool isValid(char c) {
    return c != '0';
  }

  bool isValid(char c1, char c2) {
    return c1 == '1' || c1 == '2' && c2 < '7';
  }
};
 Will be updated Soon

Java

 

class Solution {
  public int numDecodings(String s) {
    final int n = s.length();
    // dp[i] := # of ways to decode s[i..n)
    int[] dp = new int[n + 1];
    dp[n] = 1; // ""
    dp[n - 1] = isValid(s.charAt(n - 1)) ? 1 : 0;

    for (int i = n - 2; i >= 0; --i) {
      if (isValid(s.charAt(i)))
        dp[i] += dp[i + 1];
      if (isValid(s.charAt(i), s.charAt(i + 1)))
        dp[i] += dp[i + 2];
    }

    return dp[0];
  }

  private boolean isValid(char c) {
    return c != '0';
  }

  private boolean isValid(char c1, char c2) {
    return c1 == '1' || c1 == '2' && c2 < '7';
  }
}

Python

  Will be updated Soon
class Solution {
  public int numDecodings(String s) {
    final int n = s.length();
    // dp[i] := # of ways to decode s[i..n)
    int[] dp = new int[n + 1];
    dp[n] = 1; // ""
    dp[n - 1] = isValid(s.charAt(n - 1)) ? 1 : 0;

    for (int i = n - 2; i >= 0; --i) {
      if (isValid(s.charAt(i)))
        dp[i] += dp[i + 1];
      if (isValid(s.charAt(i), s.charAt(i + 1)))
        dp[i] += dp[i + 2];
    }

    return dp[0];
  }

  private boolean isValid(char c) {
    return c != '0';
  }

  private boolean isValid(char c1, char c2) {
    return c1 == '1' || c1 == '2' && c2 < '7';
  }
}

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