Count of Smaller Numbers After Self
MediumP3Coding
For every element of nums, count how many elements to its right are strictly smaller. Return the counts as an array counts, where counts[i] is the number of j > i with nums[j] < nums[i].
Example 1
- Input
- nums = [5,2,6,1]
- Output
- [2,1,1,0]
- Why
- 5 has two smaller elements to its right (2 and 1), 2 has one (1), 6 has one (1), 1 has none.
Example 2
- Input
- nums = [-1]
- Output
- [0]
Example 3
- Input
- nums = [-1,-1]
- Output
- [0,0]
Constraints
- 0 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Hints
Give it a real attempt first. Run shows which examples fail.
Saved automatically
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