2021-08-02 Daily-Challenge

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Today I have done Mean of Array After Removing Some Elements and leetcode's August LeetCoding Challenge with cpp.

Mean of Array After Removing Some Elements

Description

Given an integer array arr, return the mean of the remaining integers after removing the smallest 5% and the largest 5% of the elements.

Answers within 10-5 of the actual answer will be considered accepted.

Example 1:

Input: arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]
Output: 2.00000
Explanation: After erasing the minimum and the maximum values of this array, all elements are equal to 2, so the mean is 2.

Example 2:

Input: arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0]
Output: 4.00000

Example 3:

Input: arr = [6,0,7,0,7,5,7,8,3,4,0,7,8,1,6,8,1,1,2,4,8,1,9,5,4,3,8,5,10,8,6,6,1,0,6,10,8,2,3,4]
Output: 4.77778

Example 4:

Input: arr = [9,7,8,7,7,8,4,4,6,8,8,7,6,8,8,9,2,6,0,0,1,10,8,6,3,3,5,1,10,9,0,7,10,0,10,4,1,10,6,9,3,6,0,0,2,7,0,6,7,2,9,7,7,3,0,1,6,1,10,3]
Output: 5.27778

Example 5:

Input: arr = [4,8,4,10,0,7,1,3,7,8,8,3,4,1,6,2,1,1,8,0,9,8,0,3,9,10,3,10,1,10,7,3,2,1,4,9,10,7,6,4,0,8,5,1,2,1,6,2,5,0,7,10,9,10,3,7,10,5,8,5,7,6,7,6,10,9,5,10,5,5,7,2,10,7,7,8,2,0,1,1]
Output: 5.29167

Constraints:

  • 20 <= arr.length <= 1000
  • arr.length is a multiple of 20.
  • 0 <= arr[i] <= 10^5

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  double trimMean(vector<int>& arr) {
    sort(arr.begin(), arr.end());
    int sz = arr.size();
    double sum = 0;
    for(int i = sz / 20; i < sz - sz / 20; ++i) {
      sum += arr[i];
    }
    return sum / (sz - sz / 10);
  }
};

// Accepted
// 50/50 cases passed (8 ms)
// Your runtime beats 71.57 % of cpp submissions
// Your memory usage beats 13.07 % of cpp submissions (9.6 MB)

August LeetCoding Challenge 2

Description

Two Sum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have *exactly* one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

  • 2 <= nums.length <= 104
  • -109 <= nums[i] <= 109
  • -109 <= target <= 109
  • Only one valid answer exists.

 

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  vector<int> twoSum(vector<int>& nums, int target) {
    unordered_map<int, int> pos;
    int p = 0;
    for(auto i : nums) {
      if(pos.count(target - i)) {
        return {pos[target - i], p};
      }
      pos[i] = p++;
    }
    return {};
  }
};

// Accepted
// 54/54 cases passed (4 ms)
// Your runtime beats 99.68 % of cpp submissions
// Your memory usage beats 57.34 % of cpp submissions (10.6 MB)
Algorithm