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643. Maximum Average Subarray I

Description

You are given an integer array nums consisting of n elements, and an integer k.

Find a contiguous subarray whose length is equal to k that has the maximum average value and return this value. Any answer with a calculation error less than 10-5 will be accepted.

 

Example 1:

Input: nums = [1,12,-5,-6,50,3], k = 4
Output: 12.75000
Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75

Example 2:

Input: nums = [5], k = 1
Output: 5.00000

 

Constraints:

  • n == nums.length
  • 1 <= k <= n <= 105
  • -104 <= nums[i] <= 104

Solutions

Slide window.

  • 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;
        }
    }
    
  • 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
    
    
  • 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;
    }
    
    
  • class Solution {
        /**
         * @param Integer[] $nums
         * @param Integer $k
         * @return Float
         */
        function findMaxAverage($nums, $k) {
            $sum = 0;
            for ($i = 0; $i < $k; $i++) {
                $sum += $nums[$i];
            }
            $max = $sum;
            for ($j = $k; $j < count($nums); $j++) {
                $sum = $sum - $nums[$j - $k] + $nums[$j];
                $max = max($max, $sum);
            }
            return $max / $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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