LeetCode [1365] How Many Numbers Are Smaller Than the Current Number

 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

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class Solution { public int[] smallerNumbersThanCurrent(int[] nums) { int[] count = new int[101]; int[] res = new int[nums.length]; for(int i:nums) { count[i] ++; } for(int i=1; i<=100; ++i) { count[i] += count[i-1]; } for(int i=0; i<nums.length; ++i) { if(nums[i]==0) { res[i] = 0; } else { res[i] = count[nums[i]-1]; } } return res; } }