Create 1365_How_Many_Numbers_Are_Smaller_Than_the_Current_Number.md

Author: koji

1365_How_Many_Numbers_Are_Smaller_Than_the_Current_Number.md

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+## 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
+
+
+```python
+def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
+ n = len(nums)
+ result = [0] * 101
+ 
+ for i in range(n):
+ result[nums[i]] +=1
+ 
+ for i in range(1, 101):
+ result[i] += result[i-1]
+ 
+ for i in range(n):
+ pos = nums[i]
+ if pos == 0:
+ nums[i] = 0
+ else:
+ nums[i] = result[pos-1]
+
+ return nums
+```
+
+```
+Runtime: 44 ms, faster than 98.58% of Python3 online submissions for How Many Numbers Are Smaller Than the Current Number.
+Memory Usage: 14.1 MB, less than 100.00% of Python3 online submissions for How Many Numbers Are Smaller Than the Current Number.
+```