From 2ac1dbb0eaff365d9257ecb29e286aaaab85a3b0 Mon Sep 17 00:00:00 2001
From: Rohit Jadhav <34981708+Rohit177@users.noreply.github.com>
Date: Thu, 7 Sep 2023 14:31:48 +0530
Subject: [PATCH] Create am_i_N_repeated_in_2N_?_solution.py

---
 am_i_N_repeated_in_2N_?_solution.py | 38 +++++++++++++++++++++++++++++
 1 file changed, 38 insertions(+)
 create mode 100644 am_i_N_repeated_in_2N_?_solution.py

diff --git a/am_i_N_repeated_in_2N_?_solution.py b/am_i_N_repeated_in_2N_?_solution.py
new file mode 100644
index 0000000..24621d4
--- /dev/null
+++ b/am_i_N_repeated_in_2N_?_solution.py
@@ -0,0 +1,38 @@
+# Define a function named 'check' that takes a list 'b', [b is our string] as input
+def check(b):
+    # Create an empty dictionary 'temp' to store counts of each element in 'b'
+    temp = {}
+
+    # Loop through each element 'i' in the list 'b'
+    for i in b:
+        # Check if 'i' is already in the 'temp' dictionary
+        if i in temp:
+            # If it is, increment its count by 1
+            temp[i] += 1
+        else:
+            # If it's not, add it to the dictionary with a count of 1
+            temp[i] = 1
+
+    # Initialize variables 'count' and 'rep_num' to keep track of the most common element and its count
+    rep_num = 0
+    count = 0
+
+    # Loop through the items in the 'temp' dictionary
+    for n, c in temp.items():
+        # Check if the count 'c' is greater than the current maximum count 'count'
+        if c > count:
+            # If it is, update 'count' and 'rep_num' with the new maximum values
+            count = c
+            rep_num = n
+
+    # Return the most common element 'rep_num' as the result
+    return rep_num
+
+# Input: Read an integer 'a'
+a = int(input())
+
+# Input: Read a list of integers 'b'
+b = list(map(int, input().split()))
+
+# Call the 'check' function with the list 'b' and print the result
+print(check(b))