Class Solution


  • public class Solution
    extends Object
    1365 - How Many Numbers Are Smaller Than the Current Number.

    Easy

    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
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • smallerNumbersThanCurrent

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