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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

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 $nums$.

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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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:
    def smallerNumbersThanCurrent(self,