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61 lines
1.5 KiB
Python
61 lines
1.5 KiB
Python
"""
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1365. How Many Numbers Are Smaller Than the Current Number
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Difficulty: Easy
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https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
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──────────────────────────────────────────────────
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Given the array nums, for each nums[i] find out how many numbers in
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the array are smaller than it. That is, for each nums[i] you have to
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count the number of valid j's such that j != i and nums[j] < nums[i].
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Return the answer in an array.
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Example 1:
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Input: nums = [8,1,2,2,3]
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Output: [4,0,1,1,3]
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Explanation:
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For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and
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3).
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For nums[1]=1 does not exist any smaller number than it.
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For nums[2]=2 there exist one smaller number than it (1).
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For nums[3]=2 there exist one smaller number than it (1).
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For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
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Example 2:
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Input: nums = [6,5,4,8]
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Output: [2,1,0,3]
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Example 3:
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Input: nums = [7,7,7,7]
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Output: [0,0,0,0]
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Constraints:
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• 2 <= nums.length <= 500
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• 0 <= nums[i] <= 100
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"""
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from collections import Counter
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class Solution:
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def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
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freq = Counter(nums)
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sorted_nums = sorted(freq.keys())
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smaller = {}
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count = 0
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for num in sorted_nums:
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smaller[num] = count
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count += freq[num]
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return [smaller[num] for num in nums]
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