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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 <= 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.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
  }
  
  

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