Question

1365. How Many Numbers Are Smaller Than the Current Number

中文文档

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 $nums$.

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

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;
  }
};

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;
}

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.

class Solution:
  def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
    cnt = [0] * 102
    for x in nums:
      cnt[x + 1] += 1
    s = list(accumulate(cnt))
    return [s[x] for x in nums]

class Solution {
  public int[] smallerNumbersThanCurrent(int[] nums) {
    int[] cnt = new int[102];
    for (int x : nums) {
      ++cnt[x + 1];
    }
    for (int i = 1; i < cnt.length; ++i) {
      cnt[i] += cnt[i - 1];
    }
    int n = nums.length;
    int[] ans = new int[n];
    for (int i = 0; i < n; ++i) {
      ans[i] = cnt[nums[i]];
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
    int cnt[102]{};
    for (int& x : nums) {
      ++cnt[x + 1];
    }
    for (int i = 1; i < 102; ++i) {
      cnt[i] += cnt[i - 1];
    }
    vector<int> ans;
    for (int& x : nums) {
      ans.push_back(cnt[x]);
    }
    return ans;
  }
};

func smallerNumbersThanCurrent(nums []int) (ans []int) {
  cnt := [102]int{}
  for _, x := range nums {
    cnt[x+1]++
  }
  for i := 1; i < len(cnt); i++ {
    cnt[i] += cnt[i-1]
  }
  for _, x := range nums {
    ans = append(ans, cnt[x])
  }
  return
}

function smallerNumbersThanCurrent(nums: number[]): number[] {
  const cnt: number[] = new Array(102).fill(0);
  for (const x of nums) {
    ++cnt[x + 1];
  }
  for (let i = 1; i < cnt.length; ++i) {
    cnt[i] += cnt[i - 1];
  }
  const n = nums.length;
  const ans: number[] = new Array(n);
  for (let i = 0; i < n; ++i) {
    ans[i] = cnt[nums[i]];
  }
  return ans;
}