LeetCode 1365. How Many Numbers Are Smaller Than the Current Number

Description

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).

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

Explanation

Two loops comparison.

Python Solution

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        results = [0 for i in range(len(nums))]
        
        i = 0
        
        while i < len(nums):
            for j, num in enumerate(nums):
                if j != i and num < nums[i]:
                    results[i] += 1
            i += 1
                        
        return results
  • Time Complexity: O(N^2)
  • Space Complexity: O(N)

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