1365. How many number less than the current number

To give you an array nums, all figures for the number of each element nums [i], you count the array smaller than it therein.

In other words, for each nums [i] You must calculate the effective number of j, where j satisfies j! = I and nums [j] <nums [i].

The answer is returned in an array.

prompt:

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

  class Solution {
      public int[] smallerNumbersThanCurrent(int[] nums) {
          int [] a = 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])
                      {
                          a[i] = a[i] + 1;
                      }
                  }
              }
          }
          return a;
      }
  }