refactor 1365

fishercoder1534 · fishercoder1534 · commit ebc7f880da29 · 2020-10-19T06:50:42.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_1365.java b/src/main/java/com/fishercoder/solutions/_1365.java
@@ -4,35 +4,6 @@
 import java.util.Arrays;
 import java.util.List;
 
-/**
- * 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
- * */
 public class _1365 {
     public static class Solution1 {
         public int[] smallerNumbersThanCurrent(int[] nums) {
