Problem Description

You're given an integer array nums which contains n elements, and an integer k. The task is to find a contiguous subarray whose length is at least k which has the maximum possible average value. It's important to note that the answer needs to be accurate but is allowed a small margin for error. Specifically, any answer within 10^-5 (0.00001) of the actual answer will be considered correct.

A subarray, in this context, is a sequence of elements from the array nums that lies next to each other, without any gaps. For example, if nums is [1, 3, 5, 7, 9] and k is 2, we have to find a contiguous sequence of length 2 or more that gives the highest average. The returned value should be the average of that subarray.

The essence of the problem is to find the optimal section of the list - it can be from k to n elements long - and calculate the average of the numbers within this range. The difficulty lies in doing this efficiently, as brute force methods that check all possible subarrays will be too slow when the array nums is large.

Intuition

The intuition behind the solution involves binary search and prefix sums. Instead of directly searching for the maximum average, we flip the problem on its head: for a given average v, determine if there's a subarray with an average greater than or equal to v. Then we use binary search on v, narrowing down the range until we find the maximum average to satisfy the condition.

Here's how we arrive at the solution approach:

• Initially, we know that the maximum average lies between the minimum and maximum values in our array nums. These are our initial lower (l) and upper (r) bounds for binary search.
• We define a check function to verify if a subarray exists that has an average at least as large as v. This involves some smart tricks:
  • Calculate the prefix sum of the array elements minus v times k. This transforms the problem into finding a subarray sum that is non-negative.
  • Use a running sum (s) to keep track of the current sum of the last k elements, and a minimum sum (mi) to keep the smallest sum of any k elements we've seen. We update these as we move the running sum forward through nums.
  • If, at any point, our running sum minus the minimum sum we've seen so far (mi) is non-negative, it means there is a subarray with an average of at least v.
• The binary search operates by repeatedly checking the midpoint of our l and r range against the check function. If the function returns true, we know an average of at least mid is possible, so we move the lower bound l to mid. Otherwise, we move the upper bound r to mid.
• We continue the binary search until our range is less than 10^-5, at which point we can return either l or r as our result, since they will be sufficiently close to the maximum average.

This solution approach is much more efficient than checking every possible subarray, as it takes advantage of the properties of averages and the structure of contiguous subarrays to reduce the problem to a binary search over possible average values.

Learn more about Binary Search and Prefix Sum patterns.

Solution Approach

The solution uses a combination of binary search and prefix sums to efficiently calculate the maximum average subarray. Here is a step-by-step explanation of how the implementation works:

1. Binary Search Initialization: We start by establishing our search range for the possible maximum average, with l being the minimum value in nums and r being the maximum value. These represent the absolute possible bounds for the average.

2. Helper Function - check(): This function is crucial; it checks if there exists a contiguous subarray with an average value of at least v. Here's how it operates:

   • Sum up the first k elements of nums, then subtract k*v from this sum to get s. This operation shifts the problem from finding a maximum average subarray to finding a non-negative sum subarray.
   • Initialize two variables, t and mi, that represent the total sum and minimum sum encountered so far, respectively. Initially, t is 0 and mi is the smallest possible value.
   • Iterate over the array starting from the k-th element. For each element at index i, add nums[i] - v to s and nums[i - k] - v to t, effectively moving the k-length window one step forward.
   • Then, update mi to be the minimum of itself and t, which is the sum of the first i-k elements of the array minus k*v. This tracks the smallest sum we have seen up to that point, before considering the current k-length window.
   • If s - mi >= 0, the function returns True, indicating that it's possible to have an average of at least v. This is because we found a contiguous subarray where the sum of the numbers minus k*v is at least zero, which means the average of that subarray is at least v.

3. Binary Search Loop: The binary search loop keeps iterating while r - l is greater than a very small epsilon value (1e-5). Within this loop:

   • Calculate the midpoint mid as (l + r) / 2.
   • Use the check() function with this midpoint value. If check(mid) returns True, we know that it's possible to have a subarray with an average greater than or equal to mid, so we move the lower bound up to mid. If the result is False, we make the upper bound r equal to mid. This continuously narrows the range of possible averages.

4. Getting the Result: Once the binary search loop exits, the value of l (or r, as they are very close to each other) is a sufficiently accurate representation of the maximum average, within an error margin of 10^-5.

The solution elegantly combines binary search, which is a classic divide and conquer algorithm, with a manipulation of sums to reframe the problem into one that is far more computationally efficient. The running time complexity is O(n * log(max(nums) - min(nums))), as we have to run our check once for each iteration of binary search, and the range of the binary search is dictated by the values within nums. The linear pass in the check function, where we iterate over the array updates running sums, takes O(n) time. The constant factor has been reduced significantly as compared to a brute force method that might consider every possible subarray individually, moving the solution from possibly quadratic time complexity to linearithmic.

Example Walkthrough

Let's illustrate the solution approach with a small example. Assume we have an array nums = [1, 12, 2, 6, 7] and k = 2.

Step 1: Binary Search Initialization

• The minimum value in nums is 1, and the maximum value is 12. So we initialize our binary search range with l = 1 and r = 12.

Step 2: Helper Function - check()

• For illustration purposes, let's assume the binary search is at a point where it's testing v = 6.5. We would initialize s as the sum of the first k elements minus k*v, which is 1 + 12 - 2*6.5 = 0. Initialize t and mi as well. Here, t will start at 0 and mi will be 0, as we have not seen any sum yet.

Step 3: Iterate and Check

• Start iterating from the k-th element of nums (index 1): For each element nums[i], where i >= k, do the following (example for i = 2):
  • Add nums[i] - v to s (running sum), so s += 2 - 6.5, and we get s = -4.5.
  • Add nums[i - k] - v to t. For i = 2, it's 1 - 6.5 (since k=2), so t = -5.5.
  • Update mi to be the minimum of mi and t. As this is the first update and t is less than our initial mi, we now set mi = t = -5.5.
• Continuing this process, when i = 4 (nums[i] = 7):
  • s += 7 - 6.5, now s = 0.5 - 4.5 = -4.
  • t has been updated as we moved along, let's say it is now -1.5.
  • mi was updated along the way and it's min(mi, t); we'll consider it -5.5.
• When we check s - mi for each i, we are looking for it to be >= 0. In this case, at i = 4, s - mi is -4 - (-5.5) = 1.5, which is >= 0. So, a subarray with an average at least as high as 6.5 exists.

Step 4: Binary Search Loop

• We'd repeat the above process for different values of v, adjusting l and r with each iteration. If check(6.5) returned True, we'd set l = 6.5. If it returned False, we'd set r = 6.5.
• Continue until r - l < 10^-5.

Step 5: Result

• Once l and r are within 10^-5 of each other, the binary search terminates, and either value gives us the maximum possible average subarray to within the accepted error margin.

In this example, let's say the binary search concluded with l = 6.49995 and r = 6.50000, with an error margin below 10^-5, either l or r would serve as the result for the maximum average, which is approximately 6.5.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the findMaxAverage function consists of two main parts: the binary search over the range of values and the sliding window check within the check function.

• The binary search runs in O(log((max(nums) - min(nums)) / eps)) time because it narrows down the range [min(nums), max(nums)] by half in each iteration until the range is smaller than eps (= 1e-5).

• Within each iteration of the binary search, the check function is called once. The check function uses a sliding window approach that takes O(n) time, where n is the length of the nums list.

Combining these two parts, the overall time complexity of the function is O(n * log((max(nums) - min(nums)) / eps)).

Space Complexity

The space complexity of the function is O(1).

• There are only a constant number of extra variables used (s, t, mi, eps, l, r, mid), and no additional space that scales with the input size is required.

Together, the function achieves linearithmic time complexity and constant space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.