1365. How Many Numbers Are Smaller Than the Current Number(Solution || Leetcode easy || Java)

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]

Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0]

Constraints:

• 2 <= nums.length <= 500
• 0 <= nums[i] <= 100

SOLUTION:

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        //create a new array to store numbers less than the element
        // at that index in nums
        int[] ans = 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++;
                }
            }
            ans[i] = count;
        }
        
        return ans;
    }
}

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