This is basically Huffman coding in disguise, and I knew that immediately which felt good.
Recognize this as the optimal merge pattern problem, which is solved using a min-heap (priority queue). Repeatedly extract the two smallest items, merge them, add the cost to the total, and insert the merged item back until one item remains. Explain that this greedy strategy is optimal and analyze the time and space complexity.
Pro tip: Mention that this problem is equivalent to building a Huffman tree and that the greedy choice is provably optimal by exchange argument. Also, discuss edge cases like empty list or single element, and note that using a heap is more efficient than sorting repeatedly.
Restate the problem: given a list of positive integers, repeatedly merge two items at a cost equal to their sum, and find the minimum total cost to reduce to one item. Clarify that merging order affects total cost.
Explain that to minimize total cost, we should always merge the two smallest items first. This is a classic greedy algorithm known as the optimal merge pattern, similar to Huffman coding.
Use a min-heap (priority queue) to efficiently extract the two smallest items and insert the merged item. This ensures O(log n) operations per merge.
Initialize a min-heap with all task sizes. While the heap has more than one element, pop two smallest, sum them, add to total cost, and push the sum back. Return the total cost.
Time complexity: O(n log n) due to heap operations. Space complexity: O(n) for the heap. Handle edge cases: empty list returns 0, single element returns 0.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, compute the total cost of the original array and the cost contributions of each adjacent pair. Then, for each possible removal of k consecutive elements, calculate the new cost by subtracting the costs of the removed internal pairs and the two boundary pairs, and adding the cost of the new pair formed by the elements adjacent to the removed block. Use prefix sums to achieve O(n) time.
Pro tip: Clarify edge cases upfront: if k equals the array length, the remaining array is empty, so cost is 0 and any starting index is valid; if k = n-1, the remaining array has one element, cost is 0. Also, mention that the solution handles negative strengths correctly since absolute differences are used.
Restate the problem: remove exactly k consecutive elements to minimize the sum of absolute differences between adjacent elements in the remaining array. Define cost as sum_{i=1}^{m-1} |arr[i] - arr[i+1]| for an array of length m.
Compute the total cost of the original array and store the absolute difference for each adjacent pair in an array diff of length n-1, where diff[i] = |arr[i] - arr[i+1]|.
Build a prefix sum array over diff to quickly compute the sum of any contiguous subarray of diff. This allows O(1) retrieval of the sum of differences within a removed block and the boundary differences.
For each start index i from 0 to n-k, compute the new cost: new_cost = total_cost - sum(diff[i..i+k-2]) - (if i>0 then diff[i-1] else 0) - (if i+k-1 < n-1 then diff[i+k-1] else 0) + (if i>0 and i+k<n then |arr[i-1] - arr[i+k]| else 0). Track the minimum new_cost and the corresponding i.
If k == n, return (0, 0) or any valid index. If k == n-1, return (0, 0). Otherwise, return the minimum cost and the starting index i that achieved it.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.