# LeetCode Challenge #4: How Many Numbers Are Smaller Than the Current Number

### Problem statement: 
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]
```
### Step-by-Step Solution
**Identify Patterns**
After reading the problem and examples, we can observe 2 things:
1. The problem output is an array.
2. The brute force way to solve is a nested for loop.
> Although this problem can be solved multiple ways like using a HashMap, I want to keep this tutorial simple. So no data structures.

**Write the Pseudocode** So let's try writing a solution for this problem without code:
1. Initialize the output array called output. It has the same length as nums array.
2. Create a for loop to iterate the current number nums[i].
3. Initialize an int count to track smaller numbers.
4. Create the nested for loop to iterate through nums array to check if nums[j] < nums[i]. If so, increase count by 1 and let output[i] = count.
5. Finally, at the end of the outer for loop, return the output array.

**Let's get coding!** Now, we can write the Java code as follows:
```
class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        
        int output[] = new int[nums.length];
        
        for(int i = 0; i < nums.length; i++){
            int count = 0;
            for(int j = 0; j < nums.length; j++){
                if(nums[j] < nums[i]){
                    count++;
                    output[i] = count;
                }
            }
        }
        return output;
    }
}
```
### Conclusion
This solution is not optimized both in terms of time and space. The brute force solution is the most beginner-friendly solution I could think of because I want to avoid using data structures and complicated algorithms for this tutorial series. But of course, I encourage everyone to try solving this problem optimally as a challenge. Good luck!