花花酱 LeetCode 1365. How Many Numbers Are Smaller Than the Current Number

zxi — Mon, 02 Mar 2020 08:52:29 +0000

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

Solution 1: Brute Force

Time complexity: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua
class Solution {
public:
  vector smallerNumbersThanCurrent(vector& nums) {
    const int n = nums.size();
    vector ans(n);
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if (i != j && nums[j] < nums[i]) ++ans[i];          
    return ans;
  }
};

Solution 2: Sort + Binary Search

Time complexity: O(nlogn)
Space complexity: O(n)

C++

// Author: Huahua
class Solution {
public:
  vector smallerNumbersThanCurrent(vector& nums) {
    const int n = nums.size();
    vector s(nums);
    sort(begin(s), end(s));    
    vector ans(n);
    for (int i = 0; i < n; ++i)
      ans[i] = distance(begin(s), lower_bound(begin(s), end(s), nums[i]));
    return ans;
  }
};

Solution 3: Cumulative frequency

Time complexity: O(n)
Space complexity: O(101)

C++

// Author: Huahua
class Solution {
public:
  vector smallerNumbersThanCurrent(vector& nums) {
    vector f(101);
    for (int x : nums) ++f[x];
    partial_sum(begin(f), end(f), begin(f));    
    for (int& x : nums) 
      x = x == 0 ? 0 : f[x - 1];
    return nums;
  }
};

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花花酱 LeetCode 1365. How Many Numbers Are Smaller Than the Current Number

Solution 1: Brute Force

C++

Solution 2: Sort + Binary Search

C++

Solution 3: Cumulative frequency

C++