Problem Description

In this problem, we are given a sorted 2D integer array fruits where each entry fruits[i] consists of two integers, position_i and amount_i. position_i represents a unique position on an infinite x-axis where fruits are located, and amount_i is the number of fruits available at that position. Additionally, we have a starting position on the axis, startPos, and an integer k which represents the maximum number of steps we can take in total. The goal is to find the maximum total number of fruits one can harvest.

One can walk either left or right on the x-axis, and upon visiting a position, all fruits there are harvested and thus removed from that position. Walking one unit on the x-axis counts as one step, and the total number of steps cannot exceed k. The task is to devise a strategy for harvesting the most fruits under these constraints.

Intuition

To arrive at the solution, we first need to understand that walking too far in one direction might prevent collecting fruits in the other direction because of the step limit k. To maximize the harvest, we should consider walking in one direction to collect fruits and then potentially changing direction to collect more if the steps allow.

The intuition behind the solution is to use a sliding window technique to keep track of the total amount of fruits that can be collected within k steps from startPos. We slide the window across the sorted positions while ensuring the total number of steps including the return to startPos does not exceed k. If at any point the total distance of our sliding window exceeds k, we move the starting point of our sliding window to the right to shrink the harvesting range back within the allowed steps.

We calculate the maximum number of fruits that can be collected within this range and update our answer accordingly. At each step, we consider reaching the farthest point in our window from startPos and then returning to the nearest point. The heart of this approach relies on the optimal substructure of the problem—maximizing fruits within a range naturally contributes to maximizing the overall harvest.

Learn more about Binary Search, Prefix Sum and Sliding Window patterns.

Solution Approach

The solution implements a sliding window approach to maintain a range of positions we can visit to collect fruits within k steps. The fruits array gives us the positions and the number of fruits at each position, sorted by the positions. The implementation iterates over this array, expanding and shrinking the window to find the optimal total amount of fruits that can be harvested.

The maxTotalFruits function starts by initializing important variables: ans to store the maximum number of fruits we can harvest, i to denote the start of the sliding window, s to keep the sum of fruits within the window, and j to iterate over the fruit positions.

Here is a step-by-step explanation of the algorithm:

1. Iterate through the fruit array using the index j and for each position pj with fruit count fj, add fj to the sum s.
2. Check if the current window [i, j] exceeds k steps. This is done by calculating the distance pj - fruits[i][0] (the width of the window) and adding the smaller distance from startPos to either end of the window.
3. If the window is too wide (exceeds k steps), we shrink the window from the left by incrementing i, effectively removing fruits at fruits[i][1] from the sum s.
4. At each iteration, after adjusting the window, update ans with the maximum of itself or the current sum s.
5. Continue this process until all positions have been checked.

The core concept here leverages the fact that, since the fruit positions are sorted and unique, we can efficiently determine the range of positions to harvest by only considering the endpoints of the current window.

The algorithm uses constant space for variables and O(n) time where n is the number of positions since it scans through the positions once.

Here is a portion of the code explaining the critical part of the algorithm:

for j, (pj, fj) in enumerate(fruits):
    s += fj
    while i <= j and pj - fruits[i][0] + min(abs(startPos - fruits[i][0]), abs(startPos - pj)) > k:
        s -= fruits[i][1]
        i += 1
    ans = max(ans, s)

The loop adds fruits to s and potentially contracts the window by advancing i if the condition (pj - fruits[i][0] + min(abs(startPos - fruits[i][0]), abs(startPos - pj))) > k is met, ensuring the total steps do not exceed k. The answer ans is updated to the maximum sum s found within the constraints.

This elegant approach effectively balances expanding and contracting the harvesting range on the fly to maximize fruit collection within the step limit.

Example Walkthrough

Let's consider a small example to illustrate the solution approach. Suppose we have the following input:

• fruits array: [[0,3], [2,1], [5,2]]
• startPos: 1
• k: 4

Here, we can move a maximum of 4 steps from the starting position, and positions [0, 2, 5] have [3, 1, 2] fruits respectively. We want to maximize the number of fruits we can pick within those 4 steps.

Here's how the solution approach would work on this example:

1. Initialize ans as 0 (maximum number of fruits harvested), i as 0 (start of the sliding window), and s as 0 (sum of fruits within the window).
2. Start iterating through fruits array with index j. As we add fj to sum s, the sum of fruits becomes 3 when j = 0.
3. There is no need to shrink the window yet since we are within k steps range (the max steps we can move is 4).
4. Move to the next item in the array, j = 1. The sum of fruits s becomes 3 + 1 = 4. The window [i, j] is now [0, 2], and the total steps required are abs(startPos - fruits[i][0]) + (fruits[j][0] - fruits[i][0]) = abs(1 - 0) + (2 - 0) = 3 steps to go from startPos to pj, which still does not exceed k.
5. Move to the next item in the array, j = 2. Add fruit count at that position to s, so now s = 4 + 2 = 6. The window [i, j] is now [0, 5], and we check the steps: abs(startPos - fruits[i][0]) + (fruits[j][0] - fruits[i][0]) = abs(1 - 0) + (5 - 0) = 6. This exceeds k, so we need to shrink the window from the left by incrementing i.
6. The new window [i, j] is now [2, 5]. We update s by subtracting the fruits at fruits[i][1] and calculate steps again: s = 6 - fruits[i][1] = 6 - 3 = 3. The steps are abs(startPos - fruits[i][0]) + (fruits[j][0] - fruits[i][0]) = abs(1 - 2) + (5 - 2) = 4, which is equal to k so we keep the window.
7. As i moves up, s decreases to 2, having removed the fruits at position 0. Since this is still less than k, we continue.
8. Now, ans is updated to a maximum of itself or current sum s, so ans = max(0, 3) = 3.
9. No more positions left, we end with ans = 3, which is the amount of fruit we harvested.

Therefore, following this approach with our example, the maximum number of fruits that can be harvested is 3.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(N), where N represents the number of pairs in the fruits list. Although the code features a nested loop, each fruit in the fruits list is processed only once due to the use of two-pointer technique. The inner while loop only advances the i pointer and does not perform excessive iterations for each j index in the outer loop. Therefore, every element is visited at most twice, keeping the overall time complexity linear with respect to the number of fruit positions.

Space Complexity

The space complexity of the given code is O(1). The only extra space used is for variables to keep track of the current maximum fruit total (ans), the current sum of fruits (s), and the pointers (i, j) used for traversing the fruits list. This use of space does not scale with the size of the input, and as such, remains constant.

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