Problem Description

Given an array of integers called nums, and two other integers known as left and right, the problem asks us to calculate how many non-empty subarrays (continuous parts of the array) have a maximum element that falls within the range [left, right]. To clarify, for a subarray to be counted, its largest number must be greater than or equal to left and less than or equal to right. Essentially, we're tasked with identifying portions of the array where if you pick out the biggest number, it wouldn't be smaller than left or larger than right.

Intuition

To solve this problem, we can use the concept of counting subarrays with elements less than or equal to a given maximum. The provided solution uses a helper function f(x) which counts subarrays where all elements are less than or equal to x. The idea is that we can count all subarrays that have elements under or equal to right, which captures all subarrays that could potentially meet our condition. Then, we subtract the count of subarrays with elements under or equal to left - 1, which represents subarrays that don't meet our condition (since their maximum would be too small).

The reasoning here is that when we have the count of subarrays with maximums not surpassing right, we include both valid subarrays (those we are looking for) and subarrays that have maximums that are too small. We don't want the latter, so we find out how many of them there are (using the same count technique but with a maximum of left - 1) and remove them from our total. The difference gives us exactly the number of valid subarrays we're seeking.

The f(x) function works by scanning through the array and using a temporary count t to keep track of valid subarrays ending at the current position. If a value is encountered that is not greater than x, it can be added to existing valid subarrays to form new ones; hence, t is incremented by 1. If a value is too large, t is reset to 0, as no new valid subarrays can end at this position. The variable cnt accumulates the count of all valid subarrays as we iterate through nums.

Learn more about Two Pointers patterns.

Solution Approach

The approach taken in the given solution involves a clever use of prefix sums, a common technique in which we construct an array of sums that represent the sum of elements up to a certain index. However, in this specific implementation, rather than directly computing a sum, we are counting the possible subarrays up to a certain index that respect the maximum value condition.

Let's break down the f(x) function, which is at the heart of the approach:

1. Initialize cnt and t to 0. cnt will hold the total count of subarrays that do not exceed the maximum allowed element x, and t will count such subarrays ending at the current index in nums.
2. Iterate over each value v in the nums array.
3. If v is greater than x, then any subarray including v cannot be part of our solution, so we reset t to 0. This essentially says "you can't start or continue a valid subarray from this point."
4. If v is less than or equal to x, then we can extend every subarray counted by t with this new value v plus one extra: the subarray that only includes v itself. Thus, we increment t by 1.
5. For every iteration, we add t to cnt which accumulates our count.

The outer function numSubarrayBoundedMax makes use of f(x) to apply our subarray counting method twice:

• First, we count all valid subarrays with maximums not more than right, which gives us f(right).
• Then, we count the subarrays that are strictly smaller than left, which is f(left - 1).

The subarrays we want to exclude from our count are the ones where the maximum element value is strictly less than 'left'. By counting these with f(left - 1), we get the count of subarrays with maximum values in the range [0, left-1]. Thus, when we subtract f(left - 1) from f(right), we remove these invalid subarrays from our total, leaving us with the number of valid subarrays that have their maximum element between left and right, inclusive.

This elegant subtraction of prefix counts helps to efficiently compute the final desired count without having to evaluate each subarray independently, thereby reducing the time complexity.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach:

Suppose we have the array nums = [2, 1, 4, 3] and the integers left = 2 and right = 3. We want to count the number of subarrays where the maximum is at least left and at most right.

First, let's consider the f(x) function that counts the subarrays with maximum elements less than or equal to x:

• Using f(right), which is f(3):

  1. Initialize cnt = 0 and t = 0.
  2. For v = 2 (first value), t is incremented to 1 (t += 1), so cnt = 1.
  3. For v = 1 (second value), t is incremented to 2 (t += 1), so cnt = 3.
  4. For v = 4 (third value), t is reset to 0, as 4 is greater than 3.
  5. For v = 3 (fourth value), t is incremented to 1 (t += 1), so cnt = 4.

  f(3) gives us a count of 4 subarrays.

• Using f(left - 1), which is f(1):

  1. Initialize cnt = 0 and t = 0.
  2. For v = 2 (first value), t is reset to 0, since 2 is greater than 1.
  3. For v = 1 (second value), t is incremented to 1 (t += 1), so cnt = 1.
  4. For v = 4 (third value), t is reset to 0, as it's greater than 1.
  5. For v = 3 (fourth value), t is reset again to 0, since 3 is also greater than 1.

  f(1) gives us a count of 1 subarray.

Now, we calculate the number of subarrays we are looking for by subtracting f(left - 1) from f(right):

• Count of subarrays we are seeking = f(right) - f(left - 1) = 4 - 1 = 3.

So, there are 3 subarrays where the maximum element is between 2 and 3, inclusive. These subarrays are [2], [2, 1], and [3]. This method allows us to efficiently calculate the desired count without having to examine each subarray individually.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code is determined by the function f(x) which is called twice within the numSubarrayBoundedMax method. This function iterates through each element of the nums array exactly once. The number of elements in the nums array is represented as n.

Given that there are no nested loops and each operation inside the loop takes constant time, the time taken by f(x) is proportional to the number of elements in the array, which is O(n). Since f(x) is called twice, the total time complexity is O(2n), which simplifies to O(n).

Space Complexity

The space complexity of the code can be analyzed by looking at the space taken by variables that could scale with the input. The function f(x) uses two variables cnt and t that do not scale with the size of the input array and are thus constant extra space (O(1)).

Since there are no additional data structures utilized that grow proportionally with the input size, the space complexity remains constant, irrespective of the size of the input array nums. Therefore, the space complexity of the code is O(1).

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