Welcome to Subscribe On Youtube

2090. K Radius Subarray Averages

Description

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

Solutions

  • class Solution {
    public int[] getAverages(int[] nums, int k) {
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        int[] ans = new int[n];
        Arrays.fill(ans, -1);
        for (int i = 0; i < n; ++i) {
            if (i - k >= 0 && i + k < n) {
                ans[i] = (int) ((s[i + k + 1] - s[i - k]) / (k << 1 | 1));
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<int> getAverages(vector<int>& nums, int k) {
        int n = nums.size();
        long s[n + 1];
        s[0] = 0;
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        vector<int> ans(n, -1);
        for (int i = 0; i < n; ++i) {
            if (i - k >= 0 && i + k < n) {
                ans[i] = (s[i + k + 1] - s[i - k]) / (k << 1 | 1);
            }
        }
        return ans;
    }
};

  • class Solution:
    def getAverages(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        ans = [-1] * n
        s = list(accumulate(nums, initial=0))
        for i in range(n):
            if i - k >= 0 and i + k < n:
                ans[i] = (s[i + k + 1] - s[i - k]) // (k << 1 | 1)
        return ans


  • func getAverages(nums []int, k int) []int {
	n := len(nums)
	s := make([]int, n+1)
	for i, v := range nums {
		s[i+1] = s[i] + v
	}
	ans := make([]int, n)
	for i := 0; i < n; i++ {
		ans[i] = -1
		if i-k >= 0 && i+k < n {
			ans[i] = (s[i+k+1] - s[i-k]) / (k<<1 | 1)
		}
	}
	return ans
}

  • function getAverages(nums: number[], k: number): number[] {
    const n = nums.length;
    const s = new Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + nums[i];
    }
    const ans: number[] = new Array(n).fill(-1);
    for (let i = 0; i < n; ++i) {
        if (i - k >= 0 && i + k < n) {
            ans[i] = Math.floor((s[i + k + 1] - s[i - k]) / ((k << 1) | 1));
        }
    }
    return ans;
}

All Problems

All Solutions