How Many Numbers Are Smaller Than the Current Number

Problem 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

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]

Constraints

• 2 <= nums.length <= 500

Solution Approach

Intuition:

To efficiently determine the Numbers Are Smaller Than the Current Number

Solution Implementation

Code In Different Languages:

JavaScript

Written by @Ishitamukherjee2004

 
class Solution {
 smallerNumbersThanCurrent(nums) {
     let n = nums.length;
     let ans = new Array(n);
     for (let i = 0; i < n; i++) {
         let cnt = 0;
         for (let j = 0; j < n; j++) {
             if (nums[i] > nums[j]) {
                 cnt++;
             }
         }
         ans[i] = cnt;
     }
     return ans;
 }
}

TypeScript

Written by @Ishitamukherjee2004

 class Solution {
 smallerNumbersThanCurrent(nums: number[]): number[] {
     let n = nums.length;
     let ans: number[] = new Array(n);
     for (let i = 0; i < n; i++) {
         let cnt = 0;
         for (let j = 0; j < n; j++) {
             if (nums[i] > nums[j]) {
                 cnt++;
             }
         }
         ans[i] = cnt;
     }
     return ans;
 }
}

Python

Written by @Ishitamukherjee2004

 class Solution:
 def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
     n = len(nums)
     ans = [0] * n
     for i in range(n):
         cnt = 0
         for j in range(n):
             if nums[i] > nums[j]:
                 cnt += 1
         ans[i] = cnt
     return ans

Java

Written by @Ishitamukherjee2004

 public class Solution {
 public int[] smallerNumbersThanCurrent(int[] nums) {
     int n = nums.length;
     int[] ans = new int[n];
     for (int i = 0; i < n; i++) {
         int cnt = 0;
         for (int j = 0; j < n; j++) {
             if (nums[i] > nums[j]) {
                 cnt++;
             }
         }
         ans[i] = cnt;
     }
     return ans;
 }
}

C++

Written by @Ishitamukherjee2004

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

Complexity Analysis

• Time Complexity: $O(n^2)$
• Space Complexity: $O(n)$
• The time complexity is $O(log(n))$ where n is the length of the input array nums. This is because the algorithm uses a nested loop structure, where each element in the array is compared to every other element.
• The space complexity is $O(n)$ where n is the length of the input array nums. This is because the algorithm creates a new array ans of the same length as the input array to store the results.