Update decode_ways.cpp

cpp/neetcode_150/13_1-d_dynamic_programming/decode_ways.cpp

@@ -1,43 +1,3 @@
-/*
- Given a string w/ only digits, return # ways to decode it (letter -> digit)
- Ex. s = "12" -> 2 (AB 1 2 or L 12), s = "226" -> 3 (2 26 or 22 6 or 2 2 6)
-
- DP: At each digit, check validity of ones & tens, if valid add to # ways
- Recurrence relation: dp[i] += dp[i-1] (if valid) + dp[i-2] (if valid)
-
- Time: O(n)
- Space: O(n)
-*/
-/*
-class Solution {
-public:
- int numDecodings(string s) {
- if (s[0] == '0') {
- return 0;
- }
- 
- int n = s.size();
- 
- vector<int> dp(n + 1);
- dp[0] = 1;
- dp[1] = 1;
- 
- for (int i = 2; i <= n; i++) {
- int ones = stoi(s.substr(i - 1, 1));
- if (ones >= 1 && ones <= 9) {
- dp[i] += dp[i - 1];
- }
- int tens = stoi(s.substr(i - 2, 2));
- if (tens >= 10 && tens <= 26) {
- dp[i] += dp[i - 2];
- }
- }
- 
- return dp[n];
- }
-};
-*/
-
 /*
  Given a string w/ only digits, return # ways to decode it (letter -> digit)
  Ex. s = "12" -> 2 (AB 1 2 or L 12), s = "226" -> 3 (2 26 or 22 6 or 2 2 6)
@@ -74,3 +34,41 @@ class Solution {
  return curr;
  }
 };
+
+/*
+ Given a string w/ only digits, return # ways to decode it (letter -> digit)
+ Ex. s = "12" -> 2 (AB 1 2 or L 12), s = "226" -> 3 (2 26 or 22 6 or 2 2 6)
+
+ DP: At each digit, check validity of ones & tens, if valid add to # ways
+ Recurrence relation: dp[i] += dp[i-1] (if valid) + dp[i-2] (if valid)
+
+ Time: O(n)
+ Space: O(n)
+*/
+class Solution2 {
+public:
+ int numDecodings(string s) {
+ if (s[0] == '0') {
+ return 0;
+ }
+
+ int n = s.size();
+
+ vector<int> dp(n + 1);
+ dp[0] = 1;
+ dp[1] = 1;
+
+ for (int i = 2; i <= n; i++) {
+ int ones = stoi(s.substr(i - 1, 1));
+ if (ones >= 1 && ones <= 9) {
+ dp[i] += dp[i - 1];
+ }
+ int tens = stoi(s.substr(i - 2, 2));
+ if (tens >= 10 && tens <= 26) {
+ dp[i] += dp[i - 2];
+ }
+ }
+
+ return dp[n];
+ }
+};