Adding Efficient Solution for Problem - 91 - Decode Ways

Author: Vanditg

Decode_Ways/EfficientSolution.py

@@ -0,0 +1,40 @@
+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 91
+## Problem Name: Decode Ways   
+##===================================
+#
+#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.
+#
+#Example 1:
+#
+#Input: "12"
+#Output: 2
+#Explanation: It could be decoded as "AB" (1 2) or "L" (12).
+#
+#Example 2:
+#
+#Input: "226"
+#Output: 3
+#Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
+class Solution:
+    def numDecodings(self, s): 
+        if len(s) == 0 or s[0] == '0':	#Condition-check: If str is empty or 1st digit is zero
+            return 0	#We return zero
+        tmpA, tmpB = 1, 0	#Initialize tmpA, tmpB counter 
+        for i in range(len(s)):	#Loop through string 
+            tmp = 0	#Initialize final counter tmp
+            if s[i] != '0':	#Condition-check: If str's 1st digit is not zero
+                tmp = tmpA	#Update tmp by assigining tmpA
+            if i > 0 and (s[i - 1] == '1' or (s[i - 1] == '2' and s[i] <= '6')):	#Condition-check: If i is greater than zero and previous digit is one or previous and current digit is between 2 and 6
+                tmp += tmpB	#Update tmp  by adding tmpB into it 
+            tmpA, tmpB = tmp, tmpA	#Update tmpA, and tmpB by assigining tmp and tmpA
+        return tmpA	#Return tmpA