Given the array
nums, for each
nums[i] find out how many numbers in the array are smaller than it. That is, for each
nums[i] you have to count the number of valid
j's such that
j != i and
nums[j] < nums[i].
Return the answer in an array.
Input: nums = [8,1,2,2,3] Output: [4,0,1,1,3] Explanation: For nums=8 there exist four smaller numbers than it (1, 2, 2 and 3). For nums=1 does not exist any smaller number than it. For nums=2 there exist one smaller number than it (1). For nums=2 there exist one smaller number than it (1). For nums=3 there exist three smaller numbers than it (1, 2 and 2).
Input: nums = [6,5,4,8] Output: [2,1,0,3]
Input: nums = [7,7,7,7] Output: [0,0,0,0]
2 <= nums.length <= 500
0 <= nums[i] <= 100
Two loops comparison.
class Solution: def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]: results = [0 for i in range(len(nums))] i = 0 while i < len(nums): for j, num in enumerate(nums): if j != i and num < nums[i]: results[i] += 1 i += 1 return results
- Time Complexity: O(N^2)
- Space Complexity: O(N)