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 $nums$.
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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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