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Formatted question description: https://leetcode.ca/all/643.html

643. Maximum Average Subarray I (Easy)

Given an array consisting of n integers, find the contiguous subarray of given length k that has the maximum average value. And you need to output the maximum average value.

Example 1:

Input: [1,12,-5,-6,50,3], k = 4
Output: 12.75
Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75

Note:

  1. 1 <= k <= n <= 30,000.
  2. Elements of the given array will be in the range [-10,000, 10,000].

Companies:
Amazon

Related Topics:
Array

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/maximum-average-subarray-i/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        int sum = 0, maxSum = INT_MIN;
        for (int i = 0; i < nums.size(); ++i) {
            if (i < k) {
                sum += nums[i];
                if (i == k - 1) maxSum = sum;
            } else {
                sum += nums[i] - nums[i - k];
                maxSum = max(maxSum, sum);
            }
        }
        return (double)maxSum / k;
    }
};

  • class Solution {
    public double findMaxAverage(int[] nums, int k) {
        int length = nums.length;
        int sum = 0;
        for (int i = 0; i < k; i++)
            sum += nums[i];
        int maxSum = sum;
        for (int i = k; i < length; i++) {
            sum -= nums[i - k];
            sum += nums[i];
            maxSum = Math.max(maxSum, sum);
        }
        return 1.0 * maxSum / k;
    }
}

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

class Solution {
    public double findMaxAverage(int[] nums, int k) {
        int s = 0;
        for (int i = 0; i < k; ++i) {
            s += nums[i];
        }
        int ans = s;
        for (int i = k; i < nums.length; ++i) {
            s += (nums[i] - nums[i - k]);
            ans = Math.max(ans, s);
        }
        return ans * 1.0 / k;
    }
}

  • // OJ: https://leetcode.com/problems/maximum-average-subarray-i/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        int sum = 0, maxSum = INT_MIN;
        for (int i = 0; i < nums.size(); ++i) {
            if (i < k) {
                sum += nums[i];
                if (i == k - 1) maxSum = sum;
            } else {
                sum += nums[i] - nums[i - k];
                maxSum = max(maxSum, sum);
            }
        }
        return (double)maxSum / k;
    }
};

  • class Solution:
    def findMaxAverage(self, nums: List[int], k: int) -> float:
        s = sum(nums[:k])
        ans = s
        for i in range(k, len(nums)):
            s += nums[i] - nums[i - k]
            ans = max(ans, s)
        return ans / k

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

from collections import deque


class Solution(object):
  def findMaxAverage(self, nums, k):
    """
    :type nums: List[int]
    :type k: int
    :rtype: float
    """
    s = 0
    ans = float("-inf")
    queue = deque([])
    for num in nums:
      queue.append(num)
      s += num
      if len(queue) > k:
        s -= queue.popleft()
      if len(queue) == k:
        ans = max(ans, float(s) / k)
    return ans


  • function findMaxAverage(nums: number[], k: number): number {
    let n = nums.length;
    let ans = 0;
    let sum = 0;
    // 前k
    for (let i = 0; i < k; i++) {
        sum += nums[i];
    }
    ans = sum;
    for (let i = k; i < n; i++) {
        sum += nums[i] - nums[i - k];
        ans = Math.max(ans, sum);
    }
    return ans / k;
}


  • impl Solution {
    pub fn find_max_average(nums: Vec<i32>, k: i32) -> f64 {
        let k = k as usize;
        let n = nums.len();
        let mut sum = nums.iter().take(k).sum::<i32>();
        let mut max = sum;
        for i in k..n {
            sum += nums[i] - nums[i - k];
            max = max.max(sum);
        }
        f64::from(max) / f64::from(k as i32)
    }
}

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