# 1365. How Many Numbers Are Smaller Than the Current Number

## Description

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

## Solutions

Solution 1: Sorting + Binary Search

We can make a copy of the array $nums$, denoted as $arr$, and then sort $arr$ in ascending order.

Next, for each element $x$ in $nums$, we can use binary search to find the index $j$ of the first element that is greater than or equal to $x$. Then $j$ is the number of elements that are smaller than $x$. We can store $j$ in the answer array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

Solution 2: Counting Sort + Prefix Sum

We notice that the range of elements in the array $nums$ is $[0, 100]$. Therefore, we can use the counting sort method to first count the number of each element in the array $nums$. Then we calculate the prefix sum of the counting array. Finally, we traverse the array $nums$. For each element $x$, we directly add the value of the element at index $x$ in the counting array to the answer array.

The time complexity is $O(n + M)$, and the space complexity is $O(M)$. Where $n$ and $M$ are the length and the maximum value of the array $nums$, respectively.

• class Solution {
public int[] smallerNumbersThanCurrent(int[] nums) {
int[] arr = nums.clone();
Arrays.sort(arr);
for (int i = 0; i < nums.length; ++i) {
nums[i] = search(arr, nums[i]);
}
return nums;
}

private int search(int[] nums, int x) {
int l = 0, r = nums.length;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}

• class Solution {
public:
vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
vector<int> arr = nums;
sort(arr.begin(), arr.end());
for (int i = 0; i < nums.size(); ++i) {
nums[i] = lower_bound(arr.begin(), arr.end(), nums[i]) - arr.begin();
}
return nums;
}
};

• class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
arr = sorted(nums)
return [bisect_left(arr, x) for x in nums]


• func smallerNumbersThanCurrent(nums []int) (ans []int) {
arr := make([]int, len(nums))
copy(arr, nums)
sort.Ints(arr)
for i, x := range nums {
nums[i] = sort.SearchInts(arr, x)
}
return nums
}

• function smallerNumbersThanCurrent(nums: number[]): number[] {
const search = (nums: number[], x: number) => {
let l = 0,
r = nums.length;
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
const arr = nums.slice().sort((a, b) => a - b);
for (let i = 0; i < nums.length; ++i) {
nums[i] = search(arr, nums[i]);
}
return nums;
}