Algorithm
LeetCode 75
Maximum Average Subarray I

643. Maximum Average Subarray I

Tags

  • Array
  • Sliding Window

Link

https://leetcode.com/problems/maximum-average-subarray-i/description/?envType=study-plan-v2&envId=leetcode-75 (opens in a new tab)

Question

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

Answer

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var findMaxAverage = function (nums, k) {
  let averageValue = -Infinity;
  let left = 0;
  let sum = 0;
  for (let right = 0; right < nums.length; right++) {
    sum += nums[right];
 
    if (right - left + 1 === k) {
      averageValue = Math.max(averageValue, sum / k);
      sum -= nums[left];
      left++;
    }
  }
  return averageValue;
};
/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var findMaxAverage = function (nums, k) {
  let totalSum = 0;
  for (let i = 0; i < k; i++) {
    totalSum += nums[i];
  }
 
  let maxAvg = totalSum / k;
 
  for (let i = 1; i < nums.length - k + 1; i++) {
    totalSum = totalSum - nums[i - 1] + nums[k + i - 1];
    if (totalSum / k > maxAvg) maxAvg = totalSum / k;
  }
 
  return maxAvg;
};
/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
function findMaxAverage(nums, k) {
  let maxSum = 0;
  let currentSum = 0;
 
  for (let i = 0; i < k; i++) {
    currentSum += nums[i];
  }
 
  maxSum = currentSum;
 
  for (let i = k; i < nums.length; i++) {
    currentSum += nums[i] - nums[i - k];
    maxSum = Math.max(maxSum, currentSum);
  }
 
  return maxSum / k;
}
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