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

April LeetCoding Challenge 3

Description

Maximum Value of an Ordered Triplet II

You are given a 0-indexed integer array nums.

Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0.

The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].

Example 1:

Input: nums = [12,6,1,2,7]
Output: 77
Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
It can be shown that there are no ordered triplets of indices with a value greater than 77. 

Example 2:

Input: nums = [1,10,3,4,19]
Output: 133
Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
It can be shown that there are no ordered triplets of indices with a value greater than 133.

Example 3:

Input: nums = [1,2,3]
Output: 0
Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.

Constraints:

• 3 <= nums.length <= 105
• 1 <= nums[i] <= 106

Solution

class Solution {
  using rmap = map<int, int, greater<>>;
public:
  long long maximumTripletValue(vector<int>& nums) {
    long long answer = 0;
    rmap i;
    rmap k;
    i[nums[0]] += 1;
    for(int i = 1; i < nums.size(); ++i) {
      k[nums[i]] += 1;
    }
    for(int j = 1; j < nums.size() - 1; ++j) {
      k[nums[j]] -= 1;
      if(!k[nums[j]]) k.erase(nums[j]);
      answer = max(answer, (i.begin()->first - nums[j]) * 1LL * k.begin()->first);
      i[nums[j]] += 1;
    }
    return answer;
  }
};

// Accepted
// 599/599 cases passed (691 ms)
// Your runtime beats 5.86 % of cpp submissions
// Your memory usage beats 5.28 % of cpp submissions (186.2 MB)