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.lengthnums[i] == nums[j] * 2nums[k] == nums[j] * 2Return 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] = 6nums[0] = nums[1] * 2 = 3 * 2 = 6nums[2] = nums[1] * 2 = 3 * 2 = 6Example 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] = 0nums[0] = nums[2] * 2 = 0 * 2 = 0nums[3] = nums[2] * 2 = 0 * 2 = 0Example 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] = 8nums[0] = nums[1] * 2 = 4 * 2 = 8nums[3] = nums[1] * 2 = 4 * 2 = 8(i, j, k) = (1, 2, 4)
nums[1] = 4, nums[2] = 2, nums[4] = 4nums[1] = nums[2] * 2 = 2 * 2 = 4nums[4] = nums[2] * 2 = 2 * 2 = 4
Constraints:
3 <= n == nums.length <= 1050 <= nums[i] <= 105class Solution {
public int specialTriplets(int[] nums) {
Map<Integer, Integer> freqTotal = new HashMap<>();
int MOD = 1_000_000_007, ans = 0;
for (int i : nums)
freqTotal.put(i, freqTotal.getOrDefault(i, 0) + 1);
Map<Integer, Integer> freqRunning = new HashMap<>();
for (int i : nums) {
if (i == 0)
continue;
freqRunning.put(i, freqRunning.getOrDefault(i, 0) + 1);
int target = 2 * i;
ans = (ans + freqRunning.getOrDefault(target, 0)
* (freqTotal.getOrDefault(target, 0) - freqRunning.getOrDefault(target, 0))) % MOD;
}
return (ans + Math.max(0, freqTotal.getOrDefault(0, 0) - 2)) % MOD;
}
}