SmallerNumber.java

import java.util.Arrays;

/*
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).
 */
public class SmallerNumber {
    public static void main(String[] args) {
        int[] nums={8,1,2,2,3};
        int[] result=smallerNumbersThanCurrent(nums);
        System.out.println(Arrays.toString(result));
    }
    public static int[] smallerNumbersThanCurrent(int[] 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[i]>nums[j]){
                    count++;
                }
            }
            ans[i]=count;
        }
        return ans;

    }
}