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Top K Frequent Elements in Python
Suppose we have a non-empty array of integer numbers. we have to return the kth most frequent elements. So if the elements are [1,1,1,1,2,2,3,3,3] and k = 2, then the result will be
Formally the function should −
- Return true if there exists i, j, k
- such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
To solve this, we will follow these steps −
- num_freq = an empty map, freq_list := an empty map
- for each element i in nums
- if i is not in num_freq, then num_freq[i] := 1, otherwise increase num_freq[i] by 1
- for each key-value pair in num_freq map
- if value is not present in freq_list, then freq_list[value] := a list with [key], otherwise insert key into freq_list[value] array
- res := empty list
- for i := length of numbers down to 0
- if i in freq_list, then add elements of freq_list[i] into res
- if length of res >= k, then break
- return result
Let us see the following implementation to get better understanding −
Example
class Solution(object):
def topKFrequent(self, nums, k):
number_frequency = {}
frequency_list ={}
for i in nums:
if i not in number_frequency:
number_frequency[i] = 1
else:
number_frequency[i] += 1
for key,value in number_frequency.items():
if value not in frequency_list:
frequency_list[value] = [key]
else:
frequency_list[value].append(key)
result = []
for i in range(len(nums),0,-1):
if i in frequency_list:
result.extend(frequency_list[i])
if len(result) >=k:
break
return result
ob1 = Solution()
print(ob1.topKFrequent([1,1,1,1,2,2,3,3,3], 2))Input
[1,1,1,1,2,2,3,3,3] 2
Output
[1, 3]
Updated on: 04-May-2020
2K+ Views
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