How Many Numbers Are Smaller Than the Current Number

Problem Description:


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


Solution:


class Solution {


    public int[] smallerNumbersThanCurrent(int[] nums) {


        int count = 0;


        int[] ary= new int[nums.length];


        for(int i=0; i<nums.length; i++){


            count = 0;


            for(int j=0;j<nums.length;j++){


                if(j!=i && nums[i]>nums[j]){


                    count++;


                }


            }


            ary[i] = count;


        }


        return ary;

    }


}


Reference👉How Many Numbers Are Smaller Than the Current Number




                                                              Happy Coding💻:-)


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