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Author: hogan-tech

1482-how-many-numbers-are-smaller-than-the-current-number/1482-how-many-numbers-are-smaller-than-the-current-number.py

@@ -0,0 +1,39 @@
+# time complexity: O(nlogn)
+# space complexity: O(n)
+from collections import defaultdict
+from typing import List
+
+
+class Solution:
+    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
+        freq = defaultdict(int)
+        for num in nums:
+            freq[num] += 1
+        freqList = [(key, count) for key, count in freq.items()]
+        freqList.sort()
+        currCount = 0
+
+        prefixFreq = defaultdict(int)
+        for key, count in freqList:
+            prefixFreq[key] = currCount
+            currCount += count
+
+        for i, num in enumerate(nums):
+            nums[i] = prefixFreq[num]
+
+        return nums
+
+
+'''
+
+[(1, 0), (2, 1), (3, 3), (8, 4)]
+[(1, 1), (2, 2), (3, 1), (8, 1)]
+
+'''
+
+nums = [8, 1, 2, 2, 3]
+print(Solution().smallerNumbersThanCurrent(nums))
+nums = [6, 5, 4, 8]
+print(Solution().smallerNumbersThanCurrent(nums))
+nums = [7, 7, 7, 7]
+print(Solution().smallerNumbersThanCurrent(nums))