Question

2552. Count Increasing Quadruplets

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Description

Given a 0-indexed integer array nums of size n containing all numbers from 1 to n, return _the number of increasing quadruplets_.

A quadruplet (i, j, k, l) is increasing if:

• 0 <= i < j < k < l < n, and
• nums[i] < nums[k] < nums[j] < nums[l].

 

Example 1:

Input: nums = [1,3,2,4,5]
Output: 2
Explanation: 
- When i = 0, j = 1, k = 2, and l = 3, nums[i] < nums[k] < nums[j] < nums[l].
- When i = 0, j = 1, k = 2, and l = 4, nums[i] < nums[k] < nums[j] < nums[l]. 
There are no other quadruplets, so we return 2.

Example 2:

Input: nums = [1,2,3,4]
Output: 0
Explanation: There exists only one quadruplet with i = 0, j = 1, k = 2, l = 3, but since nums[j] < nums[k], we return 0.

 

Constraints:

• 4 <= nums.length <= 4000
• 1 <= nums[i] <= nums.length
• All the integers of nums are unique. nums is a permutation.

Solutions

Solution 1: Enumeration + Preprocessing

We can enumerate $j$ and $k$ in the quadruplet, then the problem is transformed into, for the current $j$ and $k$:

• Count how many $l$ satisfy $l > k$ and $nums[l] > nums[j]$;
• Count how many $i$ satisfy $i < j$ and $nums[i] < nums[k]$.

We can use two two-dimensional arrays $f$ and $g$ to record these two pieces of information. Where $f[j][k]$ represents how many $l$ satisfy $l > k$ and $nums[l] > nums[j]$, and $g[j][k]$ represents how many $i$ satisfy $i < j$ and $nums[i] < nums[k]$.

Therefore, the answer is the sum of all $f[j][k] \times g[j][k]$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Where $n$ is the length of the array.

class Solution:
  def countQuadruplets(self, nums: List[int]) -> int:
    n = len(nums)
    f = [[0] * n for _ in range(n)]
    g = [[0] * n for _ in range(n)]
    for j in range(1, n - 2):
      cnt = sum(nums[l] > nums[j] for l in range(j + 1, n))
      for k in range(j + 1, n - 1):
        if nums[j] > nums[k]:
          f[j][k] = cnt
        else:
          cnt -= 1
    for k in range(2, n - 1):
      cnt = sum(nums[i] < nums[k] for i in range(k))
      for j in range(k - 1, 0, -1):
        if nums[j] > nums[k]:
          g[j][k] = cnt
        else:
          cnt -= 1
    return sum(
      f[j][k] * g[j][k] for j in range(1, n - 2) for k in range(j + 1, n - 1)
    )

class Solution {
  public long countQuadruplets(int[] nums) {
    int n = nums.length;
    int[][] f = new int[n][n];
    int[][] g = new int[n][n];
    for (int j = 1; j < n - 2; ++j) {
      int cnt = 0;
      for (int l = j + 1; l < n; ++l) {
        if (nums[l] > nums[j]) {
          ++cnt;
        }
      }
      for (int k = j + 1; k < n - 1; ++k) {
        if (nums[j] > nums[k]) {
          f[j][k] = cnt;
        } else {
          --cnt;
        }
      }
    }
    long ans = 0;
    for (int k = 2; k < n - 1; ++k) {
      int cnt = 0;
      for (int i = 0; i < k; ++i) {
        if (nums[i] < nums[k]) {
          ++cnt;
        }
      }
      for (int j = k - 1; j > 0; --j) {
        if (nums[j] > nums[k]) {
          g[j][k] = cnt;
          ans += (long) f[j][k] * g[j][k];
        } else {
          --cnt;
        }
      }
    }
    return ans;
  }
}

const int N = 4001;
int f[N][N];
int g[N][N];

class Solution {
public:
  long long countQuadruplets(vector<int>& nums) {
    int n = nums.size();
    memset(f, 0, sizeof f);
    memset(g, 0, sizeof g);
    for (int j = 1; j < n - 2; ++j) {
      int cnt = 0;
      for (int l = j + 1; l < n; ++l) {
        if (nums[l] > nums[j]) {
          ++cnt;
        }
      }
      for (int k = j + 1; k < n - 1; ++k) {
        if (nums[j] > nums[k]) {
          f[j][k] = cnt;
        } else {
          --cnt;
        }
      }
    }
    long long ans = 0;
    for (int k = 2; k < n - 1; ++k) {
      int cnt = 0;
      for (int i = 0; i < k; ++i) {
        if (nums[i] < nums[k]) {
          ++cnt;
        }
      }
      for (int j = k - 1; j > 0; --j) {
        if (nums[j] > nums[k]) {
          g[j][k] = cnt;
          ans += 1ll * f[j][k] * g[j][k];
        } else {
          --cnt;
        }
      }
    }
    return ans;
  }
};

func countQuadruplets(nums []int) int64 {
  n := len(nums)
  f := make([][]int, n)
  g := make([][]int, n)
  for i := range f {
    f[i] = make([]int, n)
    g[i] = make([]int, n)
  }
  for j := 1; j < n-2; j++ {
    cnt := 0
    for l := j + 1; l < n; l++ {
      if nums[l] > nums[j] {
        cnt++
      }
    }
    for k := j + 1; k < n-1; k++ {
      if nums[j] > nums[k] {
        f[j][k] = cnt
      } else {
        cnt--
      }
    }
  }
  ans := 0
  for k := 2; k < n-1; k++ {
    cnt := 0
    for i := 0; i < k; i++ {
      if nums[i] < nums[k] {
        cnt++
      }
    }
    for j := k - 1; j > 0; j-- {
      if nums[j] > nums[k] {
        g[j][k] = cnt
        ans += f[j][k] * g[j][k]
      } else {
        cnt--
      }
    }
  }
  return int64(ans)
}