A message containing letters from A-Z is being encoded to numbers using the following mapping:
('A' -> 1), ('B' -> 2), ..... ('Z' - > 26)
Given a non-empty string containing only digits, determine the total number of ways to decode it.
Constraints
Approach
Links
YouTube
Examples
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).
Solutions
/**
* Time complexity : O(N), where N is length of the string. We iterate the
* length of dp array which is N+1.
* Space complexity : O(N). The length of the DP array.
*/
class Solution {
public int numDecodings(String s) {
if(s.charAt(0) == '0') return 0;
int n = s.length();
int[] dp = new int[n+1];
dp[0] = 1;
dp[1] = (s.charAt(0) == '0')? 0: 1;
for(int i = 2; i < n+1; i++) {
if(s.charAt(i-1) != '0')
dp[i] += dp[i-1];
int twoDigit = Integer.valueOf(s.substring(i-2, i));
if(twoDigit >= 10 && twoDigit <= 26) {
dp[i] += dp[i-2];
}
}
return dp[n];
}
}
/**
* Time complexity : O(N), where N is length of the string. We iterate the
* length of dp array which is N+1.
* Space complexity : O(1).
*/
class Solution {
public int numDecodings(String s) {
if(s == null || s.charAt(0) == '0') return 0;
int[] dp = new int[2];
dp[0] = 1;
dp[1] = (s.charAt(0) == '0')? 0: 1;
for(int i = 2; i < s.length()+1; i++) {
char curr = s.charAt(i-1);
char prev = s.charAt(i-2);
int noOfWays = 0;
// Check if successful single digit decode is possible.
if(curr != '0') {
noOfWays += dp[1];
}
// Check if successful two digit decode is possible.
if(prev == '1' || (prev == '2' && curr <= '6')) {
noOfWays += dp[0];
}
dp[0] = dp[1];
dp[1] = noOfWays;
}
return dp[1];
}
}