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Recognize that to maximize revenue, you should always sell the item with the current highest price. Use a max-heap to simulate this process, but since orders can be up to 1e13, you must batch sales by processing groups of equal prices. For each group, compute how many sales can be made before the price drops to the next distinct level, and sum the revenue using arithmetic series formulas.
Pro tip: Mention that you can avoid simulating each sale by using a priority queue of (price, count) pairs and processing in batches, which reduces time complexity to O(n log n). Also, emphasize the importance of using modulo 1e9+7 only at the end or carefully during summation to avoid overflow.
Clarify that each sale reduces the price of that item type by 1, and you must make exactly the given number of sales. Note the large constraints: orders up to 1e13, so O(orders) simulation is infeasible.
Use a max-heap (priority queue) to always access the current highest price. To handle large orders, store prices with their frequencies or process in batches.
While orders remain, pop the highest price group. Determine how many sales can be made before the price drops to the next highest price (or to zero). Compute the number of sales in this batch as min(orders, count * (price - next_price)).
For each batch, calculate the sum of an arithmetic series: if selling k units at decreasing prices from p down to p - (k-1), the sum is k*(2p - k + 1)/2. Apply modulo 1e9+7 carefully.
Ensure prices don't go below zero. If orders remain after all prices reach zero, additional sales yield zero revenue. Return the total revenue modulo 1e9+7.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Sliding window type problem but the exact-min-and-exact-max constraint makes it trickier than it looks.
Use a sliding window technique with two pointers to count subarrays where all elements are within [minVal, maxVal], then subtract counts for subarrays where all elements are within [minVal+1, maxVal] and [minVal, maxVal-1], and add back the count for [minVal+1, maxVal-1]. This inclusion-exclusion approach runs in O(n) time and O(1) space.
Pro tip: Clearly explain the inclusion-exclusion principle and how the sliding window counts subarrays with elements in a given range; this demonstrates strong problem-solving skills and attention to edge cases.
Create a function countSubarraysWithBounds(low, high) that counts subarrays where every element is between low and high inclusive, using a sliding window.
Iterate through the array, maintaining a window [left, right] where all elements are within [low, high]. For each right, move left past any element outside the range, then add (right - left + 1) to the count.
Compute total = countSubarraysWithBounds(minVal, maxVal) - countSubarraysWithBounds(minVal+1, maxVal) - countSubarraysWithBounds(minVal, maxVal-1) + countSubarraysWithBounds(minVal+1, maxVal-1).
If minVal > maxVal, return 0. Also ensure the helper function handles low > high by returning 0.
Return the computed total as the final answer.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
For each row, compute a canonical form that is invariant under column flips, such as XORing the row with its first element (or first 1) to normalize it. Then count frequencies of these canonical forms and sum C(freq, 2) for each form to get the number of unordered pairs. This works because two rows are equivalent iff their canonical forms are identical.
Pro tip: Emphasize that the canonical form must be a complete invariant: flipping columns is equivalent to XORing the row with a fixed mask, so normalizing by the first bit (or first 1) ensures uniqueness. Also mention that using a hash map with integer keys (since columns ≤ 30, the mask fits in a 32-bit integer) gives O(N) time and avoids sorting.
Two rows are equivalent if there exists a subset of columns to flip such that one row becomes the other. This is equivalent to saying the XOR of the two rows is either all 0s or all 1s (i.e., they are bitwise complements or identical).
For each row, compute a canonical form by XORing the row with a mask derived from a fixed bit (e.g., the first bit). If the first bit is 1, XOR with all 1s; otherwise keep as is. This ensures equivalent rows map to the same integer.
Use a hash map to count how many rows produce each canonical form. Since columns ≤ 30, the canonical form fits in a 32-bit integer, making hashing efficient.
For each frequency f, add f*(f-1)/2 to the total count. This counts all unordered pairs of rows that share the same canonical form, i.e., are equivalent.
Time complexity is O(N * C) for computing canonical forms (or O(N) if using bit operations on integers) and O(N) for hashing, so overall O(N). Space is O(N). Handle edge cases like empty matrix, single row, or all rows identical.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.