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The sliding window technique allows us to keep track of the sum of a subarray of fixed length k by adding one new element and removing the first element of the previous window, thus achieving O(n) time complexity.
Time Complexity: O(n)
Space Complexity: O(1)
1#include <stdio.h>
2
3float findMaxAverage(int* nums, int numsSize, int k) {
4 int sum = 0;
5 for (int i = 0; i < k; i++) {
6 sum += nums[i];
7 }
8 int maxSum = sum;
9 for (int i = k; i < numsSize; i++) {
10 sum = sum - nums[i - k] + nums[i];
11 if (sum > maxSum) {
12 maxSum = sum;
13 }
14 }
15 return (float)maxSum / k;
16}
17
18int main() {
19 int nums[] = {1,12,-5,-6,50,3};
20 int k = 4;
21 printf("%f\n", findMaxAverage(nums, 6, k));
22 return 0;
23}The C code initializes a sum of the first k elements and iterates over the elements from k to the end of the array. It updates the sum to add the current element and subtract the element that is sliding out of the window, then updates maxSum if the new sum is greater. Finally, it returns the maximum average.
The brute force approach involves checking every possible subarray of length k and calculating its average, then keeping track of the maximum average found. However, this approach is not efficient for large inputs.
Time Complexity: O(n*k)
Space Complexity: O(1)
1#include <iostream>
2#include <vector>
using namespace std;
class Solution {
public:
double findMaxAverage(vector<int>& nums, int k) {
double maxAvg = -10001;
for (int i = 0; i <= nums.size() - k; ++i) {
int sum = 0;
for (int j = i; j < i + k; ++j) {
sum += nums[j];
}
double avg = static_cast<double>(sum) / k;
if (avg > maxAvg) {
maxAvg = avg;
}
}
return maxAvg;
}
};
int main() {
Solution sol;
vector<int> nums = {1, 12, -5, -6, 50, 3};
int k = 4;
cout << sol.findMaxAverage(nums, k) << endl;
return 0;
}The C++ solution applies a similar brute force strategy, iteratively calculating the sum and average for each possible subarray of length k, and compares it to the maximum average seen.