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The Question
Coding

Create Sorted Array through Minimum Cost Insertion

You are given an integer array instructions. You need to build a sorted collection by inserting elements from instructions one by one in the given order. The cost of inserting an element instructions[i] into the current collection is defined as the minimum of: 1. The number of elements currently in the collection that are strictly less than instructions[i]. 2. The number of elements currently in the collection that are strictly greater than instructions[i]. Calculate the total cost of inserting all elements from instructions. Since the total cost can be very large, return the result modulo $10^9 + 7$. Constraints: - 1 <= instructions.length <= 10^5 - 1 <= instructions[i] <= 10^5
Java
Fenwick Tree
Prefix Sum
Questions & Insights

Clarifying Questions

What is the range of values in the `instructions` array? Knowing the maximum value helps determine if a Binary Indexed Tree (BIT) or Segment Tree can be used directly on the value range or if coordinate compression is required.
What is the maximum length of the `instructions` array? This dictates the required time complexity (e.g., O(n \log n) is likely necessary for n=10^5).
How should duplicate values be handled? The problem states "strictly less than" and "strictly greater than," implying that duplicates of the current element already in the container do not contribute to either cost metric.
Is the modulo $10^9+7$ applied at each step or only at the final sum? Standard practice is to sum and apply modulo, or apply during summation to prevent overflow, although long in Java can handle the sum of 10^5 elements each up to 10^5 (10^{10} total), which fits in a 64-bit integer.

Assumptions

The maximum value in instructions is 10^5.
The maximum length of instructions is 10^5.
A Binary Indexed Tree (Fenwick Tree) will be used over the range of values [1, 100000] to efficiently calculate prefix sums (counts of elements).

Thinking Process

Dynamic Frequency Tracking: To calculate the number of elements strictly less than or greater than a value x, we need to maintain the frequencies of all previously processed elements.
Prefix Sum Strategy: The number of elements "strictly less than x" is the prefix sum of frequencies from 1 to x-1. The number of elements "strictly greater than x" is the total number of elements processed so far minus the prefix sum up to x.
Efficient Updates and Queries: Since we process elements one by one, we need a data structure that supports both increment(value) and query_prefix_sum(value) in O(\log M) time, where M is the maximum value.
Optimization: A Fenwick Tree (Binary Indexed Tree) is ideal here due to its low constant factor and ease of implementation compared to a Segment Tree or a Balanced BST.
Implementation Breakdown

Problem Set

Functional Requirements:
Iterate through instructions once.
For each element, compute count(val < instructions[i]) and .
Add the minimum of these two values to the total cost.
Update the frequency of the current element.
Return total cost modulo 10^9 + 7.
Constraints:
instructions.length up to 10^5.
instructions[i] up to 10^5.
Time Limit: Usually 1-2 seconds (requires O(N \log M)).

Approach

Algorithm: Fenwick Tree (Binary Indexed Tree)
Data Structure: Array-based BIT
Complexity:
Time: O(N \log M), where N is the number of instructions and M is the maximum value in the input.
Space: O(M) to store the frequency tree.

Implementation

Wrap Up

Advanced Topics

Coordinate Compression: If the values in instructions could be up to 10^9 but the number of elements is only 10^5, we would sort the unique values and map them to the range [1, 10^5] to keep the BIT size manageable.
Segment Tree Alternative: A Segment Tree could solve this as well. While it has a slightly higher constant factor, it is more flexible if we needed range maximums or more complex updates.
Parallelization: While the problem is inherently sequential (each insertion depends on the state created by previous ones), one could theoretically use a Merge Sort based approach (similar to counting inversions) to solve this in a divide-and-conquer manner, which is more amenable to parallelization.
Space Optimization: If M is much larger than N, and we don't want to use coordinate compression, a Dynamic Segment Tree or a balanced BST (like a Treap or AVL tree) could be used to only store nodes for values that actually appear.