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.
2, 3, 1, and 5 is (2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75, which truncates to 2.\n
Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2021/11/07/eg1.png\")
\n\nInput: nums = [7,4,3,9,1,8,5,2,6], k = 3\nOutput: [-1,-1,-1,5,4,4,-1,-1,-1]\nExplanation:\n- avg[0], avg[1], and avg[2] are -1 because there are less than k elements before each index.\n- The sum of the subarray centered at index 3 with radius 3 is: 7 + 4 + 3 + 9 + 1 + 8 + 5 = 37.\n Using integer division, avg[3] = 37 / 7 = 5.\n- For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4.\n- For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4.\n- avg[6], avg[7], and avg[8] are -1 because there are less than k elements after each index.\n\n\n
Example 2:
\n\n\nInput: nums = [100000], k = 0\nOutput: [100000]\nExplanation:\n- The sum of the subarray centered at index 0 with radius 0 is: 100000.\n avg[0] = 100000 / 1 = 100000.\n\n\n
Example 3:
\n\n\nInput: nums = [8], k = 100000\nOutput: [-1]\nExplanation: \n- avg[0] is -1 because there are less than k elements before and after index 0.\n\n\n
\n
Constraints:
\n\nn == nums.length1 <= n <= 1050 <= nums[i], k <= 105Compiled with clang 11 using the latest C++ 20 standard.
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Most standard library headers are already included automatically for your convenience.
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