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1995. Count Special Quadruplets

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: Brute Force

Thinking

\(n\le 50\), so four nested loops over \(a<b<c<d\) checking the sum are acceptable.

Enumerate \(a < b < c < d\) with four nested loops.

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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

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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;
    }
}

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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;
    }
};

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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: Hash Table

Thinking

The quartic constant is large. Fix \(c\) from the right, count values of \(d\) in a map, and query \(nums[a]+nums[b]+nums[c]\) for \(a,b<c\), which is \(O(n^3)\).

Scan \(c\) from the right and count values of \(d\), then enumerate \(a,b\).

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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

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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;
    }
}

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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;
    }
};

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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: Hash Table (Optimized)

Thinking

One more reduction: walk \(b\) from the right, insert every \(d-c\) with \(c=b+1\) into a map, then query \(nums[a]+nums[b]\) for \(a<b\), in \(O(n^2)\).

Enumerate \(b\) while counting \(nums[d]-nums[c]\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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