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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 <= 5000 <= 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.
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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; }