为何我的LeetCode 3962最多K次交换最大子数组和解法失效？

Greedy Logic Analysis #

Your core greedy idea—replacing the smallest elements inside a fixed subarray with the largest elements outside (when the outside element is larger than the inside one)—is theoretically correct for maximizing the sum gain of that subarray. The flaw likely lies in implementation details rather than the logic itself:

• Unnecessary swaps: If the largest outside element is smaller than the smallest inside element, swapping will reduce the sum. Failing to skip such non-beneficial swaps will lead to incorrect results.
• Edge case handling: When the subarray covers the entire array, there are no outside elements to swap with—your code must avoid attempting swaps here.

Counterexample to Invalid Implementation #

Consider nums = [5,4,3,2,1], k=2. For the subarray [0,1] (sum = 9), the outside elements are [3,2,1]. The largest outside element (3) is smaller than the smallest inside element (4). If your code proceeds with swaps anyway, it will compute a sum of 5+3=8 (worse than the original 9), leading to an incorrect result.

Optimized Solution Approach #

Enumerating all subarrays and sorting inside/outside elements each time results in an O(n³ logn) complexity, which is too slow for n=1500. Instead, use a sliding window with sorted data structures to reduce complexity to O(n² logn):

Step-by-Step Plan #

1. Sliding Window Iteration: For each starting index l, expand the window to include elements from l to r (ending index).
2. Maintain Sorted Sets:
   • Use a sorted multiset (or Fenwick Tree with coordinate compression) to track elements inside the current window, allowing quick access to the smallest elements and their sums.
   • Maintain another sorted multiset for elements outside the window, enabling quick access to the largest elements and their sums.
3. Calculate Maximum Gain: For each window, compute the cumulative gain by swapping the smallest inside elements with largest outside elements (only when the swap provides a positive gain), up to k times.
4. Track Maximum Sum: For each window, calculate the total sum (original window sum + maximum gain) and keep track of the overall maximum.

Key Optimizations #

• Prefix Sums: Precompute prefix sums to quickly calculate the original sum of any window.
• Sorted Multisets: Insert/delete elements in O(logn) time, and maintain prefix sums of sorted elements to get the sum of smallest t inside elements or largest t outside elements in O(1) time.

This approach avoids redundant sorting and reduces the number of operations drastically, making it feasible for n=1500.

内容的提问来源于stack exchange，提问作者Narendra Kumawat