Asked in: AMAZON
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All Testcases Passed ✔
Solution
import bisect
def findMaxUnloaded(boxes, space):
boxes.sort()
n = len(boxes)
// ... rest of solution available after purchaseExplanation
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To approach this problem, the core challenge is to identify a subset of boxes whose maximum size is at most `space` times the minimum size, and to minimize how many boxes we remove to achieve that condition.
Let's break down the problem and think about the key insights and strategies.
1. **Understanding the condition:**
The condition `max(boxes) ≤ space * min(boxes)` means that the ratio between the largest and smallest box sizes in the remaining set should be at most `space`. If the boxes in the subset are tightly clustered in size (within a multiplicative factor of `space`), the condition is met.
2. **Goal interpretation:**
We want to find the *largest subset* of boxes that satisfy this ratio condition, because minimizing removals is equivalent to maximizing the size of the valid subset.
3. **Sort the boxes:**
Since the condition involves max and min, sorting the array of box sizes is a natural first step. After sorting, the minimum box in any contiguous subset is the first element of that subset, and the maximum is the last element.
This reduces the problem to finding a contiguous subsequence (or subarray) in the sorted array where `max ≤ space * min`. Since the array is sorted, the minimum is the left endpoint, and the maximum is the right endpoint.
4. **Using two pointers or sliding window:**
Because we want to find the longest contiguous subsequence satisfying the condition, a sliding window approach is appropriate:
- Maintain two pointers, `left` and `right`, initially both at the start of the sorted array.
- Expand `right` pointer step-by-step to include boxes.
- For each new position of `right`, check if the current window `[left, right]` satisfies the condition:
- Check if `boxes[right] ≤ space * boxes[left]`.
- If it does, update the maximum window size found.
- If it does not, move `left` forward to reduce the window size until the condition holds again.
5. **Why contiguous windows after sorting?**
Sorting arranges boxes by size, so the minimum of any subset is at the left end, and the maximum at the right. Because the condition is about the ratio of max to min, checking contiguous subarrays in sorted order covers all possibilities without missing any subsets.
6. **Compute result:**
After processing the entire array, the maximum size of the valid window is known. The minimum boxes to unload is simply `n - max_window_size`, since the rest must be removed.
7. **Edge cases:**
- If `space` is very large, possibly the entire array satisfies the condition, so zero removals.
- If all boxes are the same size, condition trivially holds.
- If the array is very diverse, some boxes must be removed, but the sliding window ensures the largest possible valid subset.
8. **Time complexity:**
Sorting takes O(n log n). The sliding window is O(n) since each pointer moves at most n times. So overall O(n log n) time.
9. **Intuition summary:**
By sorting, you create a natural order to check for contiguous valid ranges. The two-pointer window technique efficiently finds the longest window where the largest box is at most `space` times the smallest. Then the difference between total boxes and this window size is the minimal removal count.
In summary, think about transforming the problem into finding the longest subarray in sorted order that satisfies a multiplicative ratio condition on the first and last elements, then use two pointers to identify this maximal subarray efficiently.
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