SmallerThan.java

//1365. How Many Numbers Are Smaller Than the Current Number

//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

import java.util.Arrays;

public class SmallerThan {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int[] arr = new int[nums.length];
        for(int i = 0; i < nums.length; i++){
            for(int j = 0; j < nums.length; j++){
                if(i != j) {
                    if (nums[i] > nums[j]) {
                        arr[i]++;
                    }
                }
            }
        }
        return arr;
    }

    public static void main(String[] args) {
        SmallerThan st = new SmallerThan();
        int[] nums = {8, 1, 2, 2, 3};
        System.out.println(Arrays.toString(st.smallerNumbersThanCurrent(nums)));
    }
}