From e56bbeddd3aae9e0f179b744d4196296a1364a89 Mon Sep 17 00:00:00 2001
From: Tanishq sharma <166993496+tvnisxq@users.noreply.github.com>
Date: Sun, 10 Aug 2025 00:36:19 +0530
Subject: [PATCH] Create jewelsAndStones.py
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Added another Leetcode solution to a Problem from topic 𝗦𝘁𝗿𝗶𝗻𝗴𝘀
---
 .../jewelsAndStones.py                        | 34 +++++++++++++++++++
 1 file changed, 34 insertions(+)
 create mode 100644 Strings/Jewels & Stones- LC 771 /jewelsAndStones.py

diff --git a/Strings/Jewels & Stones- LC 771 /jewelsAndStones.py b/Strings/Jewels & Stones- LC 771 /jewelsAndStones.py
new file mode 100644
index 0000000..2aa3a43
--- /dev/null
+++ b/Strings/Jewels & Stones- LC 771 /jewelsAndStones.py	
@@ -0,0 +1,34 @@
+from collections import defaultdict
+class Solution:
+    def numJewelsInStones(self, jewels: str, stones: str) -> int:
+        '''
+            Brute Force Approach:
+                If n = len(stones) & m = len(jewels)
+                ▷ Time: O(n.m) -> Iterating through stones and jewels once. String lookup is linear 
+                ▷ Space: O(1) -> Only a variable count is used.
+        '''
+        count = 0
+        # Iterating through string literal stones character by character
+        for stone in stones:
+            # if that very character is also present in the string literal jewels. 
+            if stone in jewels:
+                count += 1 # Increment the count varaible by 1.
+        return count
+
+
+
+      '''
+        Optimal Approach: O(1) constant time lookup to check is something is in the set.
+                ▷ Time: O(n+m):
+                    - n for set creation + m for iterating through stones string literal & doing O(1).
+                ▷ Space: O(n) :
+                    - because of n jewels which store n characters if all are unique.
+      '''
+      setJewels = set(jewels) # {'a', 'A',...}
+        # count variable to keep track of stones which are jewels as well.
+        count = 0
+
+        for stone in stones:
+            if stone in setJewels:
+                count += 1
+        return count