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

June LeetCoding Challenge 20

Description

K Radius Subarray Averages

You are given a 0-indexed array nums of n integers, and an integer k.

The k-radius average for a subarray of nums centered at some index i with the radius k is the average of all elements in nums between the indices i - k and i + k (inclusive). If there are less than k elements before or after the index i, then the k-radius average is -1.

Build and return an array avgs of length n where avgs[i] is the k-radius average for the subarray centered at index i.

The average of x elements is the sum of the x elements divided by x, using integer division. The integer division truncates toward zero, which means losing its fractional part.

• For example, the average of four elements 2, 3, 1, and 5 is (2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75, which truncates to 2.

Example 1:

Input: nums = [7,4,3,9,1,8,5,2,6], k = 3
Output: [-1,-1,-1,5,4,4,-1,-1,-1]
Explanation:
- avg[0], avg[1], and avg[2] are -1 because there are less than k elements before each index.
- The sum of the subarray centered at index 3 with radius 3 is: 7 + 4 + 3 + 9 + 1 + 8 + 5 = 37.
  Using integer division, avg[3] = 37 / 7 = 5.
- For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4.
- For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4.
- avg[6], avg[7], and avg[8] are -1 because there are less than k elements after each index.

Example 2:

Input: nums = [100000], k = 0
Output: [100000]
Explanation:
- The sum of the subarray centered at index 0 with radius 0 is: 100000.
  avg[0] = 100000 / 1 = 100000.

Example 3:

Input: nums = [8], k = 100000
Output: [-1]
Explanation: 
- avg[0] is -1 because there are less than k elements before and after index 0.

Constraints:

• n == nums.length
• 1 <= n <= 105
• 0 <= nums[i], k <= 105

Solution

class Solution {
public:
  vector<int> getAverages(vector<int>& nums, int k) {
    if(2 * k + 1 > nums.size()) return vector<int>(nums.size(), -1);
    vector<int> answer(nums.size());
    long long sum = 0;
    for(int i = 0; i < k; ++i) {
      answer[i] = -1;
      sum += nums[2 * i];
      sum += nums[2 * i + 1];
    }
    for(int i = k; i < nums.size(); ++i) {
      if(i + k >= nums.size()) {
        answer[i] = -1;
        continue;
      }
      sum += nums[i + k];
      answer[i] = sum / (2 * k + 1);
      sum -= nums[i - k];
    }

    return answer;
  }
};

// Accepted
// 39/39 cases passed (251 ms)
// Your runtime beats 70.05 % of cpp submissions
// Your memory usage beats 66.09 % of cpp submissions (130 MB)