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: Hash Map

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const specialTriplets = function (nums) {
  const MODULO = 10 ** 9 + 7;
  const n = nums.length;
  const numMap = new Map();
  const partialMap = new Map();
  let result = 0;

  for (let index = 0; index < n; index++) {
    const num = nums[index];
    const count = numMap.get(num) ?? 0;

    numMap.set(num, count + 1);
  }

  for (let index = 0; index < n; index++) {
    const num = nums[index];
    const count = partialMap.get(num) ?? 0;
    const target = num * 2;
    const prefixCount = partialMap.get(target);

    partialMap.set(num, count + 1);

    if (prefixCount) {
      const suffixCount = numMap.get(target) - partialMap.get(target);

      result = (result + prefixCount * suffixCount) % MODULO;
    }
  }

  return result;
};