3885. Count Special Triplets¶

3885. Count Special Triplets

Medium

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

Solution¶

import java.util.HashMap;
class Solution {
    private int mod = (int)(1e9 + 7);
    public int specialTriplets(int[] nums) {
        int n = nums.length;
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++)
            map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);
        HashMap<Integer, Integer> pref_map = new HashMap<>();
        long count = 0;
        for (int i = 0; i < n; i++) {
            pref_map.put(nums[i], pref_map.getOrDefault(nums[i], 0) + 1);
            if (nums[i] == 0) {
                int left_count = pref_map.getOrDefault(0, 0) - 1;
                int right_count = map.getOrDefault(0, 0) - pref_map.getOrDefault(0, 0);
                count = (count + (left_count * 1L *  right_count) % mod) % mod;
                continue;
            }
            int left_count = pref_map.getOrDefault(nums[i] * 2, 0);
            int right_count = map.getOrDefault(nums[i] * 2, 0) - left_count;
            count = (count + (left_count * 1L * right_count) % mod) % mod;
        }
        return (int)(count);
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Explanation¶

[Add detailed explanation here]