Question

1995. Count Special Quadruplets

中文文档

Description

Given a 0-indexed integer array nums, return _the number of distinct quadruplets_ (a, b, c, d) _such that:_

• nums[a] + nums[b] + nums[c] == nums[d], and
• a < b < c < d

 

Example 1:

Input: nums = [1,2,3,6]
Output: 1
Explanation: The only quadruplet that satisfies the requirement is (0, 1, 2, 3) because 1 + 2 + 3 == 6.

Example 2:

Input: nums = [3,3,6,4,5]
Output: 0
Explanation: There are no such quadruplets in [3,3,6,4,5].

Example 3:

Input: nums = [1,1,1,3,5]
Output: 4
Explanation: The 4 quadruplets that satisfy the requirement are:
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5

 

Constraints:

• 4 <= nums.length <= 50
• 1 <= nums[i] <= 100

Solutions

Solution 1

class Solution:
  def countQuadruplets(self, nums: List[int]) -> int:
    ans, n = 0, len(nums)
    for a in range(n - 3):
      for b in range(a + 1, n - 2):
        for c in range(b + 1, n - 1):
          for d in range(c + 1, n):
            if nums[a] + nums[b] + nums[c] == nums[d]:
              ans += 1
    return ans

class Solution {
  public int countQuadruplets(int[] nums) {
    int ans = 0, n = nums.length;
    for (int a = 0; a < n - 3; ++a) {
      for (int b = a + 1; b < n - 2; ++b) {
        for (int c = b + 1; c < n - 1; ++c) {
          for (int d = c + 1; d < n; ++d) {
            if (nums[a] + nums[b] + nums[c] == nums[d]) {
              ++ans;
            }
          }
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  int countQuadruplets(vector<int>& nums) {
    int ans = 0, n = nums.size();
    for (int a = 0; a < n - 3; ++a)
      for (int b = a + 1; b < n - 2; ++b)
        for (int c = b + 1; c < n - 1; ++c)
          for (int d = c + 1; d < n; ++d)
            if (nums[a] + nums[b] + nums[c] == nums[d]) ++ans;
    return ans;
  }
};

func countQuadruplets(nums []int) int {
  ans, n := 0, len(nums)
  for a := 0; a < n-3; a++ {
    for b := a + 1; b < n-2; b++ {
      for c := b + 1; c < n-1; c++ {
        for d := c + 1; d < n; d++ {
          if nums[a]+nums[b]+nums[c] == nums[d] {
            ans++
          }
        }
      }
    }
  }
  return ans
}

Solution 2

class Solution:
  def countQuadruplets(self, nums: List[int]) -> int:
    ans, n = 0, len(nums)
    counter = Counter()
    for c in range(n - 2, 1, -1):
      counter[nums[c + 1]] += 1
      for a in range(c - 1):
        for b in range(a + 1, c):
          ans += counter[nums[a] + nums[b] + nums[c]]
    return ans

class Solution {
  public int countQuadruplets(int[] nums) {
    int ans = 0, n = nums.length;
    int[] counter = new int[310];
    for (int c = n - 2; c > 1; --c) {
      ++counter[nums[c + 1]];
      for (int a = 0; a < c - 1; ++a) {
        for (int b = a + 1; b < c; ++b) {
          ans += counter[nums[a] + nums[b] + nums[c]];
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  int countQuadruplets(vector<int>& nums) {
    int ans = 0, n = nums.size();
    vector<int> counter(310);
    for (int c = n - 2; c > 1; --c) {
      ++counter[nums[c + 1]];
      for (int a = 0; a < c - 1; ++a) {
        for (int b = a + 1; b < c; ++b) {
          ans += counter[nums[a] + nums[b] + nums[c]];
        }
      }
    }
    return ans;
  }
};

func countQuadruplets(nums []int) int {
  ans, n := 0, len(nums)
  counter := make([]int, 310)
  for c := n - 2; c > 1; c-- {
    counter[nums[c+1]]++
    for a := 0; a < c-1; a++ {
      for b := a + 1; b < c; b++ {
        ans += counter[nums[a]+nums[b]+nums[c]]
      }
    }
  }
  return ans
}

Solution 3

class Solution:
  def countQuadruplets(self, nums: List[int]) -> int:
    ans, n = 0, len(nums)
    counter = Counter()
    for b in range(n - 3, 0, -1):
      c = b + 1
      for d in range(c + 1, n):
        counter[nums[d] - nums[c]] += 1
      for a in range(b):
        ans += counter[nums[a] + nums[b]]
    return ans

class Solution {
  public int countQuadruplets(int[] nums) {
    int ans = 0, n = nums.length;
    int[] counter = new int[310];
    for (int b = n - 3; b > 0; --b) {
      int c = b + 1;
      for (int d = c + 1; d < n; ++d) {
        if (nums[d] - nums[c] >= 0) {
          ++counter[nums[d] - nums[c]];
        }
      }
      for (int a = 0; a < b; ++a) {
        ans += counter[nums[a] + nums[b]];
      }
    }
    return ans;
  }
}

class Solution {
public:
  int countQuadruplets(vector<int>& nums) {
    int ans = 0, n = nums.size();
    vector<int> counter(310);
    for (int b = n - 3; b > 0; --b) {
      int c = b + 1;
      for (int d = c + 1; d < n; ++d) {
        if (nums[d] - nums[c] >= 0) {
          ++counter[nums[d] - nums[c]];
        }
      }
      for (int a = 0; a < b; ++a) {
        ans += counter[nums[a] + nums[b]];
      }
    }
    return ans;
  }
};

func countQuadruplets(nums []int) int {
  ans, n := 0, len(nums)
  counter := make([]int, 310)
  for b := n - 3; b > 0; b-- {
    c := b + 1
    for d := c + 1; d < n; d++ {
      if nums[d] >= nums[c] {
        counter[nums[d]-nums[c]]++
      }
    }
    for a := 0; a < b; a++ {
      ans += counter[nums[a]+nums[b]]
    }
  }
  return ans
}