leetcode 1657 added

Author: Kondareddy1209

Leetcode/leetcode_1657.py

@@ -0,0 +1,54 @@
+"""
+Problem: Close Strings
+
+Two strings are considered "close" if:
+1. They have the same set of characters
+2. Their character frequencies can be rearranged to match
+
+This implementation uses dictionaries (no Counter).
+"""
+
+
+class Solution:
+    def closeStrings(self, word1: str, word2: str) -> bool:
+        # Step 1: If lengths are different → cannot be close
+        if len(word1) != len(word2):
+            return False
+        
+        # Step 2: Create frequency dictionaries
+        dici1 = {}
+        dici2 = {}
+        
+        # Count frequency of characters in word1
+        for i in word1:
+            if i in dici1:
+                dici1[i] += 1
+            else:
+                dici1[i] = 1
+                
+        # Count frequency of characters in word2
+        for i in word2:
+            if i in dici2:
+                dici2[i] += 1
+            else:
+                dici2[i] = 1
+        
+        # Step 3: Check if both words have same unique characters
+        # If not → cannot transform one into another
+        if set(dici1.keys()) != set(dici2.keys()):
+            return False
+        
+        # Step 4: Compare sorted frequency values
+        # Order doesn't matter, only distribution matters
+        return sorted(dici1.values()) == sorted(dici2.values())
+
+
+# -----------------------------------
+# Example Test Cases
+# -----------------------------------
+if __name__ == "__main__":
+    sol = Solution()
+    
+    print(sol.closeStrings("abc", "bca"))        # True
+    print(sol.closeStrings("aabb", "bbcc"))      # False
+    print(sol.closeStrings("cabbba", "abbccc"))  # True

OOPS/Product_of_array.py

@@ -13,4 +13,5 @@
 for i in range(n-1,-1,-1):
     post[i]*=postfix
     postfix*=nums[i]
-print(post)
+print(post)
+