"""Leetcode 347. Top K Frequent Elements

URL: https://leetcode.com/problems/top-k-frequent-elements/

Medium

Given a non-empty array of integers, return the k most frequent elements.

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

Example 2:
Input: nums = [1], k = 1
Output: [1]

Note:
You may assume k is always valid, 1 <= k <= number of unique elements.
Your algorithm's time complexity must be better than O(n log n),
where n is the array's size.
"""

class Solution(object):
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]

        Time complexity: O(n*logn), where n is the number of nums.
        Space complexity: O(n).
        """
        from collections import defaultdict

        # Make num->freq dict.
        num_freq_d = defaultdict(int)

        for n in nums:
            num_freq_d[n] += 1

        # Sort num->freq dict by freq. 
        sorted_num_freqs = sorted(num_freq_d.items(),
                                  key=lambda x: x[1],
                                  reverse=True)

        # Take the top k num.
        topk_nums = [num for (num, freq) in sorted_num_freqs[:k]]
        return topk_nums


def main():
    nums = [1, 1, 1, 2, 2, 3]
    k = 2
    # Should be: [1,2]
    print Solution().topKFrequent(nums, k)

    nums = [1]
    k = 1
    # Should be: [1]
    print Solution().topKFrequent(nums, k)

    nums = [3,2,3,1,2,4,5,5,6,7,7,8,2,3,1,1,1,10,11,5,6,2,4,7,8,5,6]
    k = 10
    # Should be: [1,2,5,3,7,6,4,8,10,11]
    print Solution().topKFrequent(nums, k)


if __name__ == '__main__':
    main()

"""Leetcode 347. Top K Frequent Elements

URL: https://leetcode.com/problems/top-k-frequent-elements/

Medium

Given a non-empty array of integers, return the k most frequent elements.

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

Example 2:
Input: nums = [1], k = 1
Output: [1]

Note:
You may assume k is always valid, 1 <= k <= number of unique elements.
Your algorithm's time complexity must be better than O(n log n),
where n is the array's size.
"""

class Solution(object):
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]

        Time complexity: O(n*logn), where n is the number of nums.
        Space complexity: O(n).
        """
        from collections import defaultdict

        # Make num->freq dict.
        num_freq_d = defaultdict(int)

        for n in nums:
            num_freq_d[n] += 1

        # Sort num->freq dict by freq. 
        sorted_num_freqs = sorted(num_freq_d.items(),
                                  key=lambda x: x[1],
                                  reverse=True)

        # Take the top k num.
        topk_nums = [num for (num, freq) in sorted_num_freqs[:k]]
        return topk_nums


def main():
    nums = [1, 1, 1, 2, 2, 3]
    k = 2
    # Should be: [1,2]
    print Solution().topKFrequent(nums, k)

    nums = [1]
    k = 1
    # Should be: [1]
    print Solution().topKFrequent(nums, k)

    nums = [3,2,3,1,2,4,5,5,6,7,7,8,2,3,1,1,1,10,11,5,6,2,4,7,8,5,6]
    k = 10
    # Should be: [1,2,5,3,7,6,4,8,10,11]
    print Solution().topKFrequent(nums, k)


if __name__ == '__main__':
    main()

Leetcode 347. Top K Frequent Elements  URL: https://leetcode.com/problems/top-k-frequent-elements/  Medium  Given a non-empty array of integers, return the k most frequent elements.  Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2]  Example 2: Input: nums = [1], k = 1 Output: [1]  Note: You may assume k is always valid, 1 <= k <= number of unique elements. Your algorithm's time complexity must be better than O(n log n), where n is the array's size. :type nums: List[int]         :type k: int         :rtype: List[int]          Time complexity: O(n*logn), where n is the number of nums.         Space complexity: O(n). # Make num->freq dict. # Sort num->freq dict by freq. # Take the top k num. # Should be: [1,2] # Should be: [1] # Should be: [1,2,5,3,7,6,4,8,10,11]