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.