diff --git a/greedy_methods/rearranging_fruits.py b/greedy_methods/rearranging_fruits.py
index ec87d94c348f..4bf6eba49561 100644
--- a/greedy_methods/rearranging_fruits.py
+++ b/greedy_methods/rearranging_fruits.py
@@ -1,22 +1,37 @@
+# it's a leetcode question no. 2561 where You have two fruit baskets containing n fruits each. You are given two 0-indexed integer arrays basket1 and basket2 representing the cost of fruit in each basket. 
+# You want to make both baskets equal. To do so, you can use the following operation as many times as you want:
+
+# Chose two indices i and j, and swap the ith fruit of basket1 with the jth fruit of basket2.
+# The cost of the swap is min(basket1[i],basket2[j]).
+# Two baskets are considered equal if sorting them according to the fruit cost makes them exactly the same baskets.
+
+# Return the minimum cost to make both the baskets equal or -1 if impossible.
+
+from typing import List
+from collections import defaultdict
+
 class Solution:
     def minCost(self, basket1: List[int], basket2: List[int]) -> int:
         n = len(basket1)
         freq = defaultdict(int)
-        mn = float("inf")
+        mn = float('inf')
         for i in range(n):
             freq[basket1[i]] += 1
             freq[basket2[i]] -= 1
             mn = min(mn, basket1[i], basket2[i])
-
+        
         to_swap = []
-        for j, k in freq.items():
+        for j,k in freq.items():
             if k % 2 != 0:
                 return -1
             to_swap += [j] * (abs(k) // 2)
-
+        
         to_swap.sort()
         res = 0
         for i in range(len(to_swap) // 2):
             res += min(to_swap[i], 2 * mn)
-
+        
         return res
+
+s = Solution()
+print(s.minCost([4, 2, 2, 2], [1, 4, 1, 2]))  # Output: 1