# Leetcode - how-many-numbers-are-smaller-than-the-current-number

violet posted @ Mar 10, 2020 07:37:16 AM in 算法 with tags Algorithm array ruby count sort , 735 阅读

https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/

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).
```

Constraints:

• `2 <= nums.length <= 500`
• `0 <= nums[i] <= 100`

This is a count sort problem actually. It already provides the number range is from 0 to 100. So just creating two arrays, `pre_count` and `count`, with 101 length to store all the numbers, index is number, value is count. And then modify `count` from 1 to 101, sum count from the previous one for each point. And then iterate the numbers and store `count[val] - pre_count[val]`

```# @param {Integer[]} nums
# @return {Integer[]}
def smaller_numbers_than_current(nums)
count = Array.new(101, 0)
pre_count = Array.new(101, 0)
nums.each do |val|
count[val] += 1
pre_count[val] += 1
end
i = 1
while i < 101
count[i] = count[i] + count[i-1]
i += 1
end

result = []
nums.each do |val|
result << count[val] - pre_count[val]
end

return result
end
```

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