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.
Example 1:
Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation:
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3).
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1).
For nums[3]=2 there exist one smaller number than it (1).
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
Example 2:
Input: nums = [6,5,4,8]
Output: [2,1,0,3]
Example 3:
Input: nums = [7,7,7,7]
Output: [0,0,0,0]
Constraints:
-
2 <= nums.length <= 500
-
0 <= nums[i] <= 100
SOLUTION:
import bisect
class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
n = len(nums)
ans = [0] * n
order = sorted([(num, i) for i, num in enumerate(nums)])
for i in range(n):
pos = bisect.bisect_left(order, (order[i][0], float('-inf')))
ans[order[i][1]] = i - max(i - pos, 0)
return ans
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