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 1 Problem 1 : Group Sold Products by the Date	(https://leetcode.com/problems/group-sold-products-by-the-date/)
 
+----------------------------------------------
+
+rite a solution to find for each date the number of different products sold and their names.
+
+The sold products names for each date should be sorted lexicographically.
+
+Return the result table ordered by sell_date.
+
+The result format is in the following example.
+
+------------------------------------------------
+import pandas as pd
+
+def group_sold_products(sold_products: pd.DataFrame) -> pd.DataFrame:
+    result = sold_products.groupby('sold_date')['product'].apply(lambda x: ','.join(sorted(x))).reset_index()
+    return result
+
+
+------------------------------------------------
+
 
 2 Problem 2 : Daily Leads and Partners	(	https://leetcode.com/problems/daily-leads-and-partners/ )
 
+----------------------------------------------
+For each date_id and make_name, find the number of distinct lead_id's and distinct partner_id's.
+
+Return the result table in any order.
+
+The result format is in the following example.
+
+
+------------------------------------------------
+import pandas as pd
+
+def daily_leads_partners(leads: pd.DataFrame) -> pd.DataFrame:
+    grouped = leads.groupby(['lead_date', 'partner_id']).size().reset_index(name='leads')
+    return grouped.sort_values(['lead_date', 'partner_id'])
+
+
+------------------------------------------------
 
 3 Problem 3 : Actors and Directors who Cooperated At Least Three Times	(https://leetcode.com/problems/actors-and-directors-who-cooperated-at-least-three-times/)
+----------------------------------------------
+
+rite a solution to find all the pairs (actor_id, director_id) where the actor has cooperated with the director at least three times.
+
+Return the result table in any order.
+
+The result format is in the following example.
+
+------------------------------------------------
+import pandas as pd
+
+def actors_directors(actor_director: pd.DataFrame) -> pd.DataFrame:
+    counts = actor_director.groupby(['actor_id', 'director_id']).size().reset_index(name='num_movies')
+    result = counts[counts['num_movies'] >= 3][['actor_id', 'director_id']]
+    return result.sort_values(['actor_id', 'director_id'])
+
+
+------------------------------------------------