Update decode_ways.cpp

Author: Ahmad-A0

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];
+    }
+};