# 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:

- The problem output is an array.
- 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:

- Initialize the output array called output. It has the same length as nums array.
- Create a for loop to iterate the current number nums[i].
- Initialize an int count to track smaller numbers.
- 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.
- 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!

## No Comments Yet