Create 05-00347-top-k-frequent-elements.py

Author: leetlab11

01-Arrays and Hashing/05-00347-top-k-frequent-elements.py

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+# Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.
+
+# Example 1:
+# Input: nums = [1,1,1,2,2,3], k = 2
+# Output: [1,2]
+
+# Example 2:
+# Input: nums = [1], k = 1
+# Output: [1]
+ 
+# Constraints:
+# 1 <= nums.length <= 105
+# -104 <= nums[i] <= 104
+# k is in the range [1, the number of unique elements in the array].
+# It is guaranteed that the answer is unique.
+ 
+# Follow up: Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
+
+# -------------------------------------------------------------------------------------------------------------------------------------------------
+
+# HASHMAP
+
+# create a hashmap where num is key and it's freq is the value
+# create a bucket list where index is freq and value at each index is a list of nums that has that freq
+# we are creating a bucket of frequencies because at max, all the elments in the nums list will be same, and will be in the last bucket
+# eg if there are nine 7's in the nums list and length of nums list = 9, so freq bucket list will from 0th index to 9th index = length 10
+# bucket list will be length of nums + 1, because there's a 0th index
+# create an o/p list res that is empty
+# iterate a loop on the bucket list in reverse order till 1st index(because 0th index will be empty)
+# because we want top frequent elements, so the later in list the higher the freq highest
+# iterate loop in the inner list(each bucket)
+# append in the res list
+# when length of res == k, return res
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+
+        numToFreq = {}
+        freqBuckets = [[] for i in range(len(nums) + 1)]
+
+        for n in nums:
+            numToFreq[n] = 1 + numToFreq.get(n, 0)
+        
+        for num, count in numToFreq.items():
+            freqBuckets[count].append(num)
+
+        
+        res = []
+        for i in range(len(freqBuckets) - 1, 0, -1):
+            for j in freqBuckets[i]:
+                res.append(j)
+                if len(res) == k:
+                    return res
+                  
+
+Time complexity = O(n)
+Space complexity = O(n)