Day 37: Return top K frequent integers in an array

Author: thatguyintech

day37/README.md

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+Question of the day: https://leetcode.com/problems/top-k-frequent-elements/#/description
+
+Given a non-empty array of integers, return the k most frequent elements.
+
+For example,  
+Given `[1,1,1,2,2,3]` and `k = 2`, return `[1,2]`.
+
+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.
+
+## Ideas
+
+Can't do a normal sort, since that alone will take `O(nlogn)` runtime.
+The input array isn't sorted, so we need to keep track of a count and organize
+that count somehow as we iterate through the integers in the array. There
+doesn't seem to be any constraints on the types of integers on the array,
+so I'll assume that possible elements in the array range from -maxInt to maxInt
+. 
+
+I think I can actually use radix sort again. Same idea as the challenge from
+[Day 36](../day36).
+
+
+## Code
+[Python](./topKFrequent.py)
+
+## Follow-up
+
+redo this using python `collections` library

day37/topKFrequent.py

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+def topKFrequent(nums, k):
+    if len(nums) == 0:
+        return []
+
+    freqs = dict()
+    for num in nums:
+        if num in freqs:
+            freqs[num] += 1
+        else:
+            freqs[num] = 1
+
+    buckets = [None for i in xrange(max(freqs.keys())+1)]
+
+    for num in freqs:
+        freq = freqs[num]
+        if buckets[freq]:
+            buckets[freq].append(num)
+        else:
+            buckets[freq] = [num]
+
+    ret = []
+    i = len(buckets)-1
+    addedCount = 0
+    offset = 0
+    while addedCount < k:
+        while not buckets[i-offset-addedCount]:
+            offset += 1
+        ret += buckets[i-offset-addedCount]
+        addedCount += 1
+
+    return ret
+
+def testTopKFrequent():
+    assert set(topKFrequent([], 0)) == set([])
+    assert set(topKFrequent([1], 1)) == set([1])
+    assert set(topKFrequent([1,1,1,2,2,3], 2)) == set([1, 2])
+    assert set(topKFrequent([-1,-1,-1,2,2,3], 2)) == set([-1, 2])
+    assert set(topKFrequent([1,1,1,2,2,3], 3)) == set([1, 2, 3])
+    assert set(topKFrequent([1, 2, 3, 4, 5], 1)) == set([1, 2, 3, 4, 5])
+
+def main():
+    testTopKFrequent()
+
+if __name__ == "__main__":
+    main()