K Radius Subarray Averages

LeetCode 2211 | Difficulty: Medium

Medium

Problem 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 <= 10^5`

- `0 <= nums[i], k <= 10^5`

Topics: Array, Sliding Window

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Approach

Sliding Window

Maintain a window [left, right] over the array/string. Expand right to include new elements, and shrink left when the window violates constraints. Track the optimal answer as the window slides.

When to use

Finding subarrays/substrings with a property (max length, min length, exact count).

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Solutions

Solution 1: C# (Best: 424 ms)

Metric	Value
Runtime	424 ms
Memory	51.5 MB
Date	2021-12-02

Solution

public class Solution {
    public int[] GetAverages(int[] nums, int k) {
        int n = nums.Length;
    int windowSize = 2*k+1;
    long sum = 0;
    int left=0, right=0;
    int[] result = new int[n];
    while(right<n)
    {
        sum += nums[right];
        result[right] = -1;
        if(right-left+1 >= windowSize)
        {
            result[left+k] = (int) (sum/windowSize);
            sum -= nums[left];
            left++;
        }
        right++;
    }
    
    return result;
    }
}

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Complexity Analysis

Approach	Time	Space
Sliding Window	$O(n)$	$O(k)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Clarify what makes a window "valid" and what triggers expansion vs shrinking.
• LeetCode provides 3 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: To calculate the average of a subarray, you need the sum and the K. K is already given. How could you quickly calculate the sum of a subarray?

Hint 2: Use the Prefix Sums method to calculate the subarray sums.

Hint 3: It is possible that the sum of all the elements does not fit in a 32-bit integer type. Be sure to use a 64-bit integer type for the prefix sum array.