You are given a 0-indexed integer array nums. In one operation you can replace any element of the array with any two elements that sum to it.
nums = [5,6,7]. In one operation, we can replace nums[1] with 2 and 4 and convert nums to [5,2,4,7].Return the minimum number of operations to make an array that is sorted in non-decreasing order.
Example 1:
Input: nums = [3,9,3] Output: 2 Explanation: Here are the steps to sort the array in non-decreasing order: - From [3,9,3], replace the 9 with 3 and 6 so the array becomes [3,3,6,3] - From [3,3,6,3], replace the 6 with 3 and 3 so the array becomes [3,3,3,3,3] There are 2 steps to sort the array in non-decreasing order. Therefore, we return 2.
Example 2:
Input: nums = [1,2,3,4,5] Output: 0 Explanation: The array is already in non-decreasing order. Therefore, we return 0.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 109This approach works by iterating from the end of the array to the beginning. The idea is to ensure each element is not bigger than the rest of the previously sorted part of the array. This can be optimized using a priority queue to store the required values efficiently, making it easier to decide where to split large numbers.
Whenever we find a number that is greater than the required maximum, we split it and keep track of splits using the priority queue (min-heap in this case).
The solution uses a priority queue to manage elements efficiently. By iterating the array from end to start, we can ensure that each number is processed in a priority-wise manner, allowing us to minimize operations by only splitting when necessary.
JavaScript
Time Complexity: O(N log N), primarily due to the heap operations.
Space Complexity: O(N) due to the additional space used by the priority queue.
This approach relies on a greedy approach combined with a two-pointer technique. The algorithm will scan the array while managing smaller replacements for larger numbers. The idea is to minimize the change needed by checking previous and current indices simultaneously to determine necessary operations.
The C++ solution uses a vector to maintain the smaller elements found during the traversal from end to start of the array, making controlled adjustments to ensure non-decreasing order.
We maintain order using sorting and check conditions dynamically while iterating.
C#
Time Complexity: O(N^2) with sorting for maintaining order.
Space Complexity: O(N) due to the auxiliary vector space used.
| Approach | Complexity |
|---|---|
| Reverse Approach with Priority Queue | Time Complexity: O(N log N), primarily due to the heap operations. |
| Greedy Strategy with Two-Pointer Technique | Time Complexity: O(N^2) with sorting for maintaining order. |
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