Asked in: DataTroops
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All Testcases Passed ✔
Solution
def solutioon(n,x,arr):
arr.sort()
i = 0
j = 0
// ... rest of solution available after purchaseExplanation
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To approach this problem, you need to think about how to distribute packets of toffees to x students such that the range (difference between the maximum and minimum toffees in the selected packets) is as small as possible. The constraints indicate you have more packets than students, so not all packets will be distributed. The challenge is to find the subset of size x from the given packets where the toffee count difference is minimized.
Step 1: Understanding the problem
You are given n packets with varying toffee counts, and you want to select exactly x of these packets for the students. The selection should minimize the gap between the packet with the largest number of toffees and the packet with the smallest number of toffees in that selection.
This boils down to finding a contiguous segment of packets (after sorting) whose maximum and minimum values have the least difference.
Step 2: Key insight - Sorting
Sorting the array of toffee counts is crucial. Once the packets are sorted, all packets with similar toffee counts will be adjacent to each other. Because the difference between max and min is what matters, the optimal selection will be found among contiguous segments in the sorted array.
Why? Because any non-contiguous selection would inherently have a larger gap between min and max than some contiguous segment.
Step 3: Sliding window over sorted array
After sorting, you want to consider every contiguous subarray of length x. Each subarray represents a potential distribution of packets to the students.
For each subarray of length x, calculate the difference between the last element (max in the subarray) and the first element (min in the subarray). Keep track of the minimum such difference encountered.
Step 4: Efficient comparison
You only need to check n - x + 1 such windows because that’s how many contiguous subarrays of length x exist in an array of length n.
Iterate from i = 0 to i = n - x, and at each step:
- Find difference = arr[i + x - 1] - arr[i]
- Update minimum difference if this difference is smaller than the previously recorded minimum
Step 5: Return the result
After checking all possible windows, the minimum difference found will be the smallest possible gap between maximum and minimum toffees assigned to any student.
Step 6: Handle edge cases and constraints
- Make sure to check that n > x, as specified.
- If the input size is small, this method still works efficiently since sorting is O(n log n), and sliding over the array is O(n).
- Because all values are positive and the number of toffees can vary widely, sorting ensures you do not miss any optimal grouping.
Step 7: Explanation using the example
Given arr = [7, 3, 2, 4, 9, 12, 56] and x = 3:
- Sort the array: [2, 3, 4, 7, 9, 12, 56]
- Possible contiguous groups of length 3:
- [2, 3, 4], difference = 4 - 2 = 2
- [3, 4, 7], difference = 7 - 3 = 4
- [4, 7, 9], difference = 9 - 4 = 5
- [7, 9, 12], difference = 12 - 7 = 5
- [9, 12, 56], difference = 56 - 9 = 47
- Minimum difference is 2, which occurs in the first window. That’s the minimal gap achievable.
Summary:
- Sort the array of toffee counts.
- Use a sliding window of size x to scan through the sorted packets.
- For each window, compute difference between max and min (endpoints of the window).
- Track and return the minimum such difference.
This approach effectively leverages sorting to reduce the problem complexity and find the optimal selection efficiently.
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