2021-08-02 Daily-Challenge
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 of20
.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)