README_EN.md

1365. How Many Numbers Are Smaller Than the Current Number

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Description

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

Solutions

Python3

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        cnt = [0] * 101
        for num in nums:
            cnt[num] += 1
        for i in range(1, 101):
            cnt[i] += cnt[i - 1]
        res = []
        for num in nums:
            res.append(0 if num == 0 else cnt[num - 1])
        return res

Java

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int[] cnt = new int[101];
        for (int e : nums) {
            ++cnt[e];
        }
        for (int i = 1; i < 101; ++i) {
            cnt[i] += cnt[i - 1];
        }
        int[] res = new int[nums.length];
        for (int i = 0; i < nums.length; ++i) {
            res[i] = nums[i] == 0 ? 0 : cnt[nums[i] - 1];
        }
        return res;
    }
}

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