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Formatted question description: https://leetcode.ca/all/2090.html

2090. K Radius Subarray Averages (Medium)

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 = 3.75, which truncates to 3.

 

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

Similar Questions:

Solution 1. Sliding Window

Intuition: Keep a fixed-length sliding window of length 2k + 1. Slide it from left to right, and update the sum of the window on the fly.

Algorithm:

Let len = 2*k + 1.

If N < len, return all -1s directly because this array is too short to cover any window

Otherwise, loop i from 0 to N-1:

  • Push A[i] into the window. sum += A[i]
  • Pop A[i-len] out of the window. sum -= A[i - len]
  • Assign the average value. ans[i - k] = sum / len.
  • // OJ: https://leetcode.com/problems/k-radius-subarray-averages/
// Time: O(N)
// Space: O(1) extra space
class Solution {
public:
    vector<int> getAverages(vector<int>& A, int k) {
        long N = A.size(), len = 2 * k + 1, sum = 0; // `len` is the length of the window
        vector<int> ans(N, -1);
        if (N < len) return ans; // If the array is too short to cover a window, return all -1s
        for (int i = 0; i < N; ++i) {
            sum += A[i]; // push A[i] into the window
            if (i - len >= 0) sum -= A[i - len]; // pop A[i-len], if any, out of window
            if (i >= len - 1) ans[i - k] = sum / len; // the center of this window is at `i-k`
        }
        return ans;
    }
};

  • class Solution:
    def getAverages(self, nums: List[int], k: int) -> List[int]:
        s = 0
        ans = [-1] * len(nums)
        for i, v in enumerate(nums):
            s += v
            if i >= k * 2:
                ans[i - k] = s // (k * 2 + 1)
                s -= nums[i - k * 2]
        return ans

############

# 2090. K Radius Subarray Averages
# https://leetcode.com/problems/k-radius-subarray-averages/

class Solution:
    def getAverages(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        res = [-1] * n
        prefix = [0] + list(accumulate(nums))
        
        for i in range(k, n - k):
            x = (prefix[i + k + 1] - prefix[i - k]) // (k + k + 1)
            res[i] = x
        
        return res


  • 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;
    }
}

  • func getAverages(nums []int, k int) []int {
	ans := make([]int, len(nums))
	s := 0
	for i, v := range nums {
		ans[i] = -1
		s += v
		if i >= k*2 {
			ans[i-k] = s / (k*2 + 1)
			s -= nums[i-k*2]
		}
	}
	return ans
}

  • function getAverages(nums: number[], k: number): number[] {
    const n = nums.length;
    const ans: number[] = new Array(n).fill(-1);
    let s = 0;
    for (let i = 0; i < n; ++i) {
        s += nums[i];
        if (i >= k * 2) {
            ans[i - k] = Math.floor(s / (k * 2 + 1));
            s -= nums[i - k * 2];
        }
    }
    return ans;
}

Solution 2. Prefix Sum

  • // OJ: https://leetcode.com/problems/k-radius-subarray-averages/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> getAverages(vector<int>& A, int k) {
        int N = A.size(), len = 2 * k + 1;
        vector<int> ans(N, -1);
        if (N < len) return ans; // If the array is too short to cover a window, return all -1s
        vector<long> sum(N + 1);
        for (int i = 0; i < N; ++i) sum[i + 1] = sum[i] + A[i];
        for (int i = k; i + k < N; ++i) ans[i] = (sum[i + k + 1] - sum[i - k]) / len;
        return ans;
    }
};

  • class Solution:
    def getAverages(self, nums: List[int], k: int) -> List[int]:
        s = 0
        ans = [-1] * len(nums)
        for i, v in enumerate(nums):
            s += v
            if i >= k * 2:
                ans[i - k] = s // (k * 2 + 1)
                s -= nums[i - k * 2]
        return ans

############

# 2090. K Radius Subarray Averages
# https://leetcode.com/problems/k-radius-subarray-averages/

class Solution:
    def getAverages(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        res = [-1] * n
        prefix = [0] + list(accumulate(nums))
        
        for i in range(k, n - k):
            x = (prefix[i + k + 1] - prefix[i - k]) // (k + k + 1)
            res[i] = x
        
        return res


  • 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;
    }
}

  • func getAverages(nums []int, k int) []int {
	ans := make([]int, len(nums))
	s := 0
	for i, v := range nums {
		ans[i] = -1
		s += v
		if i >= k*2 {
			ans[i-k] = s / (k*2 + 1)
			s -= nums[i-k*2]
		}
	}
	return ans
}

  • function getAverages(nums: number[], k: number): number[] {
    const n = nums.length;
    const ans: number[] = new Array(n).fill(-1);
    let s = 0;
    for (let i = 0; i < n; ++i) {
        s += nums[i];
        if (i >= k * 2) {
            ans[i - k] = Math.floor(s / (k * 2 + 1));
            s -= nums[i - k * 2];
        }
    }
    return ans;
}

Discuss

https://leetcode.com/problems/k-radius-subarray-averages/discuss/1599803/C%2B%2B-O(N)-time-Sliding-Window

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