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!