This one took me a while to see the right angle.
Reformulate the problem as finding the minimum total increments needed to eliminate all descents, where each operation can increment a contiguous subarray. Use a greedy strategy that processes the array from left to right, maintaining the current maximum value and adding the difference whenever an element is less than the maximum. This yields the sum of positive differences between consecutive elements after enforcing nondecreasing order.
Pro tip: Explain that the answer is simply the sum of max(0, a[i] - a[i-1]) for i from 1 to n-1, and justify why this greedy approach is optimal by arguing that each descent must be fixed by at least that many increments, and these increments can be applied independently without affecting other descents.
Clarify that each operation increments a contiguous subarray by 1, and the goal is to make the entire array nondecreasing with minimum operations.
Observe that for any index i where a[i] < a[i-1], the element a[i] must be increased by at least a[i-1] - a[i] to eliminate that descent.
Process the array from left to right, keeping track of the current maximum value seen so far. Whenever the current element is less than the maximum, add the difference to the total operations and update the element to the maximum.
Argue that each descent requires at least the computed number of increments, and these increments can be applied independently (e.g., by incrementing the suffix starting at the descent) without creating new descents, so the sum is both necessary and sufficient.
Sum all positive differences between consecutive elements after enforcing nondecreasing order, and return that sum as the minimum number of operations.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Sort and pair adjacent elements, basically.
Sort the array in ascending order. Then, use a greedy two-pointer approach: pair the largest possible backup with the largest possible primary by scanning from the end, ensuring the backup is at least as large as the primary. This maximizes the sum of primaries because each primary is as large as possible without wasting a larger server as backup.
Pro tip: Clarify that the greedy choice is optimal by exchanging arguments: if a smaller server could be primary instead of a larger one, swapping them would not decrease the total sum and maintains the backup constraint. This shows you understand why the greedy works, not just how to implement it.
Restate the problem: pair servers into (primary, backup) with backup >= primary, maximize sum of primaries. Note that each server can be used at most once, and pairs are disjoint.
Sort the capacities in ascending order. This allows efficient pairing by considering the largest servers first.
Use two pointers: one at the end (candidate backup) and one just before it (candidate primary). If the backup is >= primary, form a pair, add primary to sum, and move both pointers left by 2. Otherwise, move the backup pointer left by 1 (this server cannot be a backup for any smaller primary either).
Write code for the two-pointer approach, handling edge cases like odd length (one server left unpaired) and empty array. Test with examples to ensure correctness.
State time complexity O(n log n) due to sorting, space O(1) extra. Provide a brief exchange argument: any optimal solution can be transformed into the greedy one without reducing the sum.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.