You are given an integer array nums consisting of n elements, and an integer k.
Find a contiguous subarray whose length is greater than or 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:
- When the length is 4, averages are [0.5, 12.75, 10.5] and the maximum average is 12.75
- When the length is 5, averages are [10.4, 10.8] and the maximum average is 10.8
- When the length is 6, averages are [9.16667] and the maximum average is 9.16667
The maximum average is when we choose a subarray of length 4 (i.e., the sub array [12, -5, -6, 50]) which has the max average 12.75, so we return 12.75
Note that we do not consider the subarrays of length < 4.
Example 2:
Input: nums = [5], k = 1
Output: 5.00000
Constraints:
n == nums.length
1 <= k <= n <= 104
-104 <= nums[i] <= 104
Solution:
class Solution {
public double findMaxAverage(int[] nums, int k) {
double left = -10000;
double right = 10000;
double prevMid = 10000, error = 10000;
while (error >= 0.00001) {
double mid = (left + right) / 2;
if (canFindSubarray(nums, k, mid)) {
left = mid;
} else {
right = mid;
}
error = Math.abs(prevMid - mid);
prevMid = mid;
}
return prevMid;
}
private boolean canFindSubarray(int[] nums, int k, double avg) {
double sum = 0, prev = 0;
for (int i = 0; i < k; i ++) {
sum += nums[i] - avg;
}
if (sum >= 0) return true;
double minPrev = 0;
for (int i = k; i < nums.length; i ++) {
sum += nums[i] - avg;
prev += nums[i - k] - avg;
minPrev = Math.min(minPrev, prev);
if (sum >= minPrev) {
return true;
}
}
return false;
}
}