2025-12-09 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's December LeetCoding Challenge with cpp.

December LeetCoding Challenge 9

Description

Count Special Triplets

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) 
    <ul>
	<li><code>nums[0] = 8</code>, <code>nums[1] = 4</code>, <code>nums[3] = 8</code></li>
	<li><code>nums[0] = nums[1] * 2 = 4 * 2 = 8</code></li>
	<li><code>nums[3] = nums[1] * 2 = 4 * 2 = 8</code></li>
</ul>
</li>
<li><code>(i, j, k) = (1, 2, 4)</code>
<ul>
	<li><code>nums[1] = 4</code>, <code>nums[2] = 2</code>, <code>nums[4] = 4</code></li>
	<li><code>nums[1] = nums[2] * 2 = 2 * 2 = 4</code></li>
	<li><code>nums[4] = nums[2] * 2 = 2 * 2 = 4</code></li>
</ul>
</li>

 

Constraints:

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

Solution

class Solution {
  const int MOD = 1e9 + 7;
public:
  int specialTriplets(vector<int>& nums) {
    map<int, vector<int>> positions;
    int len = nums.size();
    for(int i = 0; i < len; ++i) {
      positions[nums[i]].push_back(i);
    }
    int answer = 0;
    for(auto &[jNum, jPos] : positions) {
      if(!jNum) {
        int sz = jPos.size();
        answer += 1LL * sz * (sz - 1) * (sz - 2) / 6 % MOD;
        continue;
      }
      if(!positions.count(jNum * 2) || positions[jNum * 2].size() < 2) continue;
      vector<int> &ik = positions[jNum * 2];
      int sz = ik.size();
      int k = upper_bound(ik.begin(), ik.end(), jPos[0]) - ik.begin();
      for(auto j : jPos) {
        k = upper_bound(ik.begin() + k, ik.end(), j) - ik.begin();
        if(k == sz) break;
        answer += 1LL * (k) * (sz - k) % MOD;
        answer %= MOD;
      }
      // cout << jNum << ' ' << answer << endl;
    }
    return answer;
  }
};

// Accepted
// 1121/1121 cases passed (725 ms)
// Your runtime beats 64.23 % of cpp submissions
// Your memory usage beats 61.68 % of cpp submissions (368.9 MB)
Algorithm