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3583. Count Special Triplets

Description

You are given an integer array nums.

A special triplet is defined as a triplet of indices (i, j, k) such that:

  • 0 <= i < j < k < n, where n = nums.length
  • nums[i] == nums[j] * 2
  • nums[k] == nums[j] * 2

Return the total number of special triplets in the array.

Since the answer may be large, return it modulo 109 + 7.

 

Example 1:

Input: nums = [6,3,6]

Output: 1

Explanation:

The only special triplet is (i, j, k) = (0, 1, 2), where:

  • nums[0] = 6, nums[1] = 3, nums[2] = 6
  • nums[0] = nums[1] * 2 = 3 * 2 = 6
  • nums[2] = nums[1] * 2 = 3 * 2 = 6

Example 2:

Input: nums = [0,1,0,0]

Output: 1

Explanation:

The only special triplet is (i, j, k) = (0, 2, 3), where:

  • nums[0] = 0, nums[2] = 0, nums[3] = 0
  • nums[0] = nums[2] * 2 = 0 * 2 = 0
  • nums[3] = nums[2] * 2 = 0 * 2 = 0

Example 3:

Input: nums = [8,4,2,8,4]

Output: 2

Explanation:

There are exactly two special triplets:

  • (i, j, k) = (0, 1, 3)
    • nums[0] = 8, nums[1] = 4, nums[3] = 8
    • nums[0] = nums[1] * 2 = 4 * 2 = 8
    • nums[3] = nums[1] * 2 = 4 * 2 = 8
  • (i, j, k) = (1, 2, 4)
    • nums[1] = 4, nums[2] = 2, nums[4] = 4
    • nums[1] = nums[2] * 2 = 2 * 2 = 4
    • nums[4] = nums[2] * 2 = 2 * 2 = 4

 

Constraints:

  • 3 <= n == nums.length <= 105
  • 0 <= nums[i] <= 105

Solutions

Solution 1: Enumerate Middle Number + Hash Table

We can enumerate the middle number $\textit{nums}[j]$, and use two hash tables, $\textit{left}$ and $\textit{right}$, to record the occurrence counts of numbers to the left and right of $\textit{nums}[j]$, respectively.

First, we add all numbers to $\textit{right}$. Then, we traverse each number $\textit{nums}[j]$ from left to right. During the traversal:

  1. Remove $\textit{nums}[j]$ from $\textit{right}$.
  2. Count the occurrences of the number $\textit{nums}[i] = \textit{nums}[j] * 2$ to the left of $\textit{nums}[j]$, denoted as $\textit{left}[\textit{nums}[j] * 2]$.
  3. Count the occurrences of the number $\textit{nums}[k] = \textit{nums}[j] * 2$ to the right of $\textit{nums}[j]$, denoted as $\textit{right}[\textit{nums}[j] * 2]$.
  4. Multiply $\textit{left}[\textit{nums}[j] * 2]$ and $\textit{right}[\textit{nums}[j] * 2]$ to get the number of special triplets with $\textit{nums}[j]$ as the middle number, and add the result to the answer.
  5. Add $\textit{nums}[j]$ to $\textit{left}$.

Finally, return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$.

  • class Solution {
    public int specialTriplets(int[] nums) {
        Map<Integer, Integer> left = new HashMap<>();
        Map<Integer, Integer> right = new HashMap<>();
        for (int x : nums) {
            right.merge(x, 1, Integer::sum);
        }
        long ans = 0;
        final int mod = (int) 1e9 + 7;
        for (int x : nums) {
            right.merge(x, -1, Integer::sum);
            ans = (ans + 1L * left.getOrDefault(x * 2, 0) * right.getOrDefault(x * 2, 0) % mod)
                % mod;
            left.merge(x, 1, Integer::sum);
        }
        return (int) ans;
    }
}

  • class Solution {
public:
    int specialTriplets(vector<int>& nums) {
        unordered_map<int, int> left, right;
        for (int x : nums) {
            right[x]++;
        }
        long long ans = 0;
        const int mod = 1e9 + 7;
        for (int x : nums) {
            right[x]--;
            ans = (ans + 1LL * left[x * 2] * right[x * 2] % mod) % mod;
            left[x]++;
        }
        return (int) ans;
    }
};


  • class Solution:
    def specialTriplets(self, nums: List[int]) -> int:
        left = Counter()
        right = Counter(nums)
        ans = 0
        mod = 10**9 + 7
        for x in nums:
            right[x] -= 1
            ans = (ans + left[x * 2] * right[x * 2] % mod) % mod
            left[x] += 1
        return ans


  • func specialTriplets(nums []int) int {
	left := make(map[int]int)
	right := make(map[int]int)
	for _, x := range nums {
		right[x]++
	}
	ans := int64(0)
	mod := int64(1e9 + 7)
	for _, x := range nums {
		right[x]--
		ans = (ans + int64(left[x*2])*int64(right[x*2])%mod) % mod
		left[x]++
	}
	return int(ans)
}


  • function specialTriplets(nums: number[]): number {
    const left = new Map<number, number>();
    const right = new Map<number, number>();
    for (const x of nums) {
        right.set(x, (right.get(x) || 0) + 1);
    }
    let ans = 0;
    const mod = 1e9 + 7;
    for (const x of nums) {
        right.set(x, (right.get(x) || 0) - 1);
        const lx = left.get(x * 2) || 0;
        const rx = right.get(x * 2) || 0;
        ans = (ans + ((lx * rx) % mod)) % mod;
        left.set(x, (left.get(x) || 0) + 1);
    }
    return ans;
}

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