leetcode 1365. 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).

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

 

题目大意：给你一个数组nums，对于其中每个元素nums[i]，请你统计数组中比它小的所有数字的数目.也就是说，对于每个nums[i]你必须计算出有效的j的数量，其中j满足j != i 且 nums[j] < nums[i]. 以数组形式返回答案.

 

思路一：暴力法，对于每个nums[i], 统计数组中所有小于nums[i]的个数，时间复杂度$O(n^2)$.

C++代码：

class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        int len = nums.size();
        vector<int> cnt(len, 0);
        for (int index = 0; index < len; ++index) {
            for (int i = 0; i < len; ++i) {
                if (i != index && nums[i] < nums[index])
                    cnt[index]++;
            }
        }
        return cnt;
    }
};

python3代码：

 

思路二：由于数组中的数属于[0,100], 可以利用计数排序的方式，先将数组排好序。

C++代码：时间复杂度$O(n)$

class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        vector<int> cnt(101, 0);
        for (int i = 0; i < nums.size(); ++i) {//计数， cnt[i]此时统计的是数组中数i的个数
             cnt[nums[i]]++;
        }
        for (int i = 1; i < 101; ++i) {//cnt[i]统计的是数组中小于等于数i的个数
             cnt[i] += cnt[i - 1];
        }
        vector<int> ans(nums.size(), 0);
        for (int i = 0; i < nums.size(); ++i) {
            //如果nums[i] == 0, 说明数组中小于0的个数为0，否则小于nums[i]的个数为cnt[nums[i] - 1];
            ans[i] = (nums[i] == 0 ? 0 : cnt[nums[i] - 1]);
        }
        return ans;
    }
};

python3代码：

时间复杂度O(nlogn)

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        indics = {}
        for index, num in enumerate(sorted(nums)):
            indics.setdefault(num, index)
        return [indics[num] for num in nums]

时间复杂度O(n):

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        count = collections.Counter(nums)

        for i in range(1,101):
            count[i] += count[i-1]
        
        return [count[x-1] for x in nums]