Adding solution for Problem - 771 - Jewels and Stones

Author: Vanditg

Jewels_Stones/EfficientSolution.py

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+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 771
+## Problem Name: Jewels and Stones
+##===================================
+#
+#You're given strings J representing the types of stones that are jewels, and S representing the stones you have.  
+#
+#Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels.
+#
+#The letters in J are guaranteed distinct, and all characters in J and S are letters. 
+#
+#Letters are case sensitive, so "a" is considered a different type of stone from "A".
+#
+#Example 1:
+#
+#Input: J = "aA", S = "aAAbbbb"
+#Output: 3
+#
+#Example 2:
+#
+#Input: J = "z", S = "ZZ"
+#Output: 0
+
+class Solution:
+    def numJewelsInStone(self, Jewels, Stones):
+        JewelsList = []    #Initialize our empty JewelsList.
+        Number = 0    #Initialize our counter number.
+		
+        for i in range(len(Jewels)):    #Loop through Jewels
+            JewelsList.append(Jewels[i])    #We'll append Jewels[i] in our JewelsList
+        for j in range(len(Stones)):    #Loop through Stones
+            if Stones[j] in JewelsList:    #Condition-check - if Stone is in JewelsList
+                Number += 1    #We'll update our number counter. [Condition enter]
+        return Number    #Finally, we'll return our counter Number.
+
+#Example 1, Jewels = "z", Stones = "ZZ"
+#JewelsList = []
+#Number = 0
+#i = 0; 
+#JewelsList = ["z"] Exit the loop as Jewels has one string character. 
+#j = 0;
+#Condition-check Stones[0] = "Z" is not in JewelsList, we fail the condition-check
+#j = 1; 
+#Condition-check Stones[1] = "Z" is not in JewelsList, we fail the condition-check, exit the loop. 
+#Return the counter Number which is Zero. 
+#Number = 0