N-Repeated_Element_in_Size_2N_Array.py

# You are given an integer array nums with the following properties:

# nums.length == 2 * n.
# nums contains n + 1 unique elements.
# Exactly one element of nums is repeated n times.
# Return the element that is repeated n times.

 

# Example 1:

# Input: nums = [1,2,3,3]
# Output: 3
# Example 2:

# Input: nums = [2,1,2,5,3,2]
# Output: 2
# Example 3:

# Input: nums = [5,1,5,2,5,3,5,4]
# Output: 5
 

# Constraints:

# 2 <= n <= 5000
# nums.length == 2 * n
# 0 <= nums[i] <= 104
# nums contains n + 1 unique elements and one of them is repeated exactly n times.

class Solution:
    def repeatedNTimes(self, nums: List[int]) -> int:
        # collect the frequency of each number in a dictionary
        freq_map = {}
        for num in nums:
            if num not in freq_map:
                freq_map[num] = 1
            else:
                freq_map[num] += 1
        
        # check which number has frequncy equal to n
        for num in nums:
            if freq_map[num] == len(nums)//2:
                return num