Welcome to Subscribe On Youtube

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)
        }
    }
    
    

All Problems

All Solutions